1) Find all real solutions of: 4 2 2 x + x =3 2 x 4 + x23=0 4 2 2 2 x 2 x +3 x 3=0 x ( 21)=0 2 x 2 ( x 21)+ 3 ( 2 x2 +3 ) ( x 21 ) =0 ( 2 x2 +3 ) =0 2 2 x =3 3 x 2= 2 3 x= 2 3 x= i 2 2) Given the complex numbers. Find and simplify z w , , zw : w z z=5+iw=4 +2i z 5+i = w 4+2 i Multiply by conjugate of the Denominator; 5+i 42i z 4 +2 i = w 42i z 5(42 i)+i(42 i) = w 4(42i)+ 2i(42 i) z 20+10 i4 i2i 2 = w 162i 2 z 20+6 i+2 = w 16 +2 z 22+ 6 i = w 18 z 11+3 i = w 9 z 11 1 = + i w 9 3 w 4+2 i = z 5+i Multiply by conjugate of the Denominator; 4 +2 i 5i w 5+i = z 5i w 4 (5i ) +2 i (5i ) = z 5 (5i ) +i (5i ) w 20+4 i10i2i = 2 z 25i 2 w 20+6 i+2 = z 25+1 w 22+ 6 i = z 26 w 11+3 i = z 13 w 11+3 i = z 13 w 11 3 = + i z 13 13 zw= ( 5+ i )*( 4+2 i ) 5 ( 4+2 i )+ i ( 2010 i4 i+2i 2 2014 i2 zw=1814 i 3) Solve the inequality: 9 x 2+16 24 x 2 9 x +1624 x 0 2 9 x 24 x +16 0 2 9 x 12 x12 x +16 0 3 x ( 3 x4 )4(3 x4) 0 (3 x4 ) ( 3 x4 ) 0 ( 3 x4 ) 0 3 x4 x 4 3 4+2 i ) 4) Skippy wishes to plant a vegetable garden along one side of his house. In his garage, he found 32 linear feet of fencing. Since one side of the garden will border the house, Skippy doesn't need fencing along that side. What are the dimensions of the garden, which will maximize the area of the garden? What is the maximum area of the garden? Maximize : Area= A=xy Constraint :fence=x+ 2 y =32 x=322 y Substitute A=xy A= (322 y ) y 2 A=32 y2 y Find the Derivative ... A ' ( y ) =324 y 4 y=32 y=8 ft x=322 ( 8 ) =3216 x=16 ft Dimensions=8 ft by 16 ft Maximum Area=816 2 Maximum Area=128 ft 5) The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior's Lemonade Stand is C ( x )=18 x+ 240, x 0 and the price-demand function, in cents per cup, is p ( x )=903 x , 0 x 30 Find the maximum profit. Let Upper Case P ( x ) =Profit Function, P ( x )=x . p ( x )C (x) P ( x )=x .(903 x)(18 x+ 240) P ( x )= ( 90 x3 x2 ) 18 x240 P ( x )=3 x 2+ 72 x24 0 To maximize the profit function, take its first derivative and set equal to zero. d =6 x +72=0 dx 6 x=72 x=9 P ( 9 )=3 ( 9 )2 +72(9)24 0 P ( 9 )=165 Section 0.10: Complex Numbers from Precalculus Prerequisites a.k.a. 'Chapter 0' by Carl Stitz, PhD, and Jeff Zeager, PhD, is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. 2013, Carl Stitz. 126 Prerequisites 0.10 Complex Numbers We conclude our Prerequisites chapter with a review the set of Complex Numbers. As you may recall, the complex numbers fill an algebraic gap left by the real numbers. There is no real number x with x 2 = 1, sincepfor any real number x 2 0. However, we could formally extract p square roots and write x = 1. We build the complex numbers by relabeling the quantity 1 as i, 1 the unfortunately misnamed imaginary unit. The number i, while not a real number, is defined so that it plays along well with real numbers and acts very much like any other radical expression. For instance, 3(2i) = 6i, 7i 3i = 4i, (2 7i) + (3 + 4i) = 5 3i, and so forth. The key properties which distinguish i from the real numbers are listed below. Definition 0.18. The imaginary unit i satisfies the two following properties: 1. i 2 = 1 2. If c is a real number with c 0 then p p c=i c Property 1 in Definition 0.18 establishes that i does act as a square root2 of 1, and property 2 establishes what we mean by the 'principal square root' of a negative real number. In property 2, to remember theprestriction p on c. For example, it is perfectly acceptable to say p p it is important 4 = i 4 = i(2) = 2i. However, ( 4) 6= i 4, otherwise, we'd get 2= p 4= p ( 4) = i p 4 = i(2i) = 2i 2 = 2( 1) = 2, which is unacceptable. The moral of this story is that the general properties of radicals do not apply for even roots of negative quantities. With Definition 0.18 in place, we are now in position to define the complex numbers. Definition 0.19. A complex number is a number of the form a + bi, where a and b are real numbers and i is the imaginary unit. The set of complex numbers is denoted C. p Complex numbers include things you'd normally expect, like 3 + 2i and 25 i 3. However, don't forget that a or b could be zero, which means numbers like 3i and 6 are also complex numbers. p In other words, don't forget that the complex numbers include the real numbers,3 so 0 and 21 are both considered complex numbers. The arithmetic of complex numbers is as you would expect. The only things you need to remember are the two properties in Definition 0.18. The next example should help recall how these animals behave. 1 Some Technical Mathematics textbooks label it 'j'. While it carries the adjective 'imaginary', these numbers have essential real-world implications. For example, every electronic device owes its existence to the study of 'imaginary' numbers. 2 Note the use of the indefinite article 'a'. Whatever beast is chosen to be i, i is the other square root of 1. 3 To use the language of Section 0.1.2, R C. 0.10 Complex Numbers 127 Example 0.10.1. Perform the indicated operations. 1. (1 4. p 2i) (3 + 4i) 2. (1 p 3 12 2i)(3 + 4i) 3. p ( 3)( 12) 5. Solution. 1 3 6. (x 2i 4i [1 + 2i])(x [1 2i]) 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We distribute and combine like terms: (1 2i) (3 + 4i) = 1 2i 3 = 2 6i 4i Distribute Gather like terms Technically, we'd have to rewrite our answer 2 6i as ( 2) + ( 6)i to be (in the strictest sense) 'in the form a + bi'. That being said, even pedants have their limits, and we'll consider 2 6i good enough. 2. Using the Distributive Property (a.k.a. F.O.I.L.), we get (1 2i)(3 + 4i) = = = = = (1)(3) + (1)(4i) (2i)(3) 3 + 4i 6i 8i 2 3 2i 8( 1) 3 2i + 8 11 2i (2i)(4i) F.O.I.L. i2 = 1 3. How in the world are we supposed to simplify 13 2i 4i ? Well, we deal with the denominator 3 4i as we would any other denominator containing two terms, one of which is a square root: we and multiply both numerator and denominator by 3 + 4i, the (complex) conjugate of 3 4i. Doing so produces 1 3 2i 4i = = = = = (1 2i)(3 + 4i) (3 4i)(3 + 4i) 3 + 4i 6i 8i 2 9 16i 2 3 2i 8( 1) 9 16( 1) 11 2i 25 11 2 i 25 25 Equivalent Fractions F.O.I.L. i2 = 1 4. We use property 2 of Definition then apply the rules of radicals applicable to real p0.18 first, p 2p p p p numbers to get 3 12 = i 3 i 12 = i 3 12 = 36 = 6. 128 Prerequisites 5. We adhere p to the order p of operations here and perform the multiplication before the radical to get ( 3)( 12) = 36 = 6. 6. We can brute force multiply using the distributive property and see that (x [1 + 2i])(x [1 2i]) = = = = x2 x2 x2 x2 x[1 2i] x[1 + 2i] + [1 2i][1 + 2i] F.O.I.L. 2 x + 2ix x 2ix + 1 2i + 2i 4i Distribute 2x + 1 4( 1) Gather like terms 2x + 5 i2 = 1 This type of factoring will be revisited in Section 3.4. In the previous example, we used the 'conjugate' idea from Section 0.9 to divide two complex numbers. More generally, the complex conjugate of a complex number a + bi is the number a bi. The notation commonly used for complex conjugation is a 'bar': a + bi = a bi. For example, 3 + 2i = 3 2i and 3 2i = 3 + 2i. To find 6, we note that 6 = 6 + 0i = 6 0i = 6, so 6 = 6. p p p Similarly, 4i = 4i, since 4i = 0 + 4i = 0 4i = 4i. Note that 3 + 5 = 3 + 5, not 3 5, since p p p p 3 + 5 = 3 + 5 + 0i = 3 + 5 0i = 3p + 5. Here, the conjugation specified by the 'bar' notation p involves reversing the sign before i = 1, not before 5. The properties of the conjugate are summarized in the following theorem. Theorem 0.12. Properties of the Complex Conjugate: Let z and w be complex numbers. z=z z +w =z +w zw = z w z n = (z)n , for any natural number n z is a real number if and only if z = z. Essentially, Theorem 0.12 says that complex conjugation works well with addition, multiplication and powers. The proofs of these properties can best be achieved by writing out z = a + bi and w = c + di for real numbers a, b, c and d. Next, we compute the left and right sides of each equation and verify that they are the same. The proof of the first property is a very quick exercise.4 To prove the second property, we compare z + w with z + w. We have z + w = a + bi + c + di = a bi + c di. To find z + w, we first compute z + w = (a + bi) + (c + di) = (a + c) + (b + d)i so z + w = (a + c) + (b + d)i = (a + c) 4 Trust us on this. (b + d)i = a + c bi di = a bi + c di = z + w 0.10 Complex Numbers 129 As such, we have established z + w = z + w. The proof for multiplication works similarly. The proof that the conjugate works well with powers can be viewed as a repeated application of the product rule, and is best proved using a technique called Mathematical Induction.5 The last property is a characterization of real numbers. If z is real, then z = a + 0i, so z = a 0i = a = z. On the other hand, if z = z, then a + bi = a bi which means b = b so b = 0. Hence, z = a + 0i = a and is real. We now return to the business of solving quadratic equations. Consider x 2 2x + 5 = 0. The discriminant b2 4ac = 16 is negative, so we knowp by Theorem 0.10 there are no real solutions, since the Quadratic Formula would involve the term 16. Complex numbers, however, are built just for such situations, so we can go ahead and apply the Quadratic Formula to get: p p ( 2) ( 2)2 4(1)(5) 2 16 2 4i x= = = = 1 2i. 2(1) 2 2 Example 0.10.2. Find the complex solutions to the following equations.6 1. 2x =x +3 x +1 2. 2t 4 = 9t 2 + 5 3. z 3 + 1 = 0 Solution. 1. Clearing fractions yields a quadratic equation so we then proceed as in Section 0.7. 2x x +1 2x 2x 2x 0 = x +3 = = = = (x + 3)(x + 1) Multiply by (x + 1) to clear denominators x 2 + x + 3x + 3 F.O.I.L. 2 x + 4x + 3 Gather like terms x 2 + 2x + 3 Subtract 2x From here, we apply the Quadratic Formula p 2 22 4(1)(3) x = p 2(1) 2 8 = 2p 2i 8 = 2 p 2 i2 2 = 2 p 2( 1 i 2) = 2p = 1i 2 5 6 Quadratic Formula Simplify Definition of i Product Rule for Radicals Factor and reduce See Section 9.3. Remember, all real numbers are complex numbers, so 'complex solutions' means both real and non-real answers. 130 Prerequisites p p We get two answers: x = 1 + i 2 and its conjugate x = 1 i 2. Checking both of these answers reviews all of the salient points about complex number arithmetic and is therefore strongly encouraged. 2. Since we have three terms, and the exponent on one term ('4' on t 4 ) is exactly twice the exponent on the other ('2' on t 2 ), we have a Quadratic in Disguise. We proceed accordingly. 2t 4 = 5 = (2t 2 + 1)(t 2 5) = 2t 2 + 1 = 0 or 2t 4 9t 2 From 2t 2 + 1 = 0 we get 2t 2 = t = r 9t 2 + 5 0 Subtract 9t 2 and 5 0 Factor 2 t = 5 Zero Product Property 1 2. 1, or t 2 = 1 = i 2 r We extract square roots as follows: p p 1 1 1 i 2 = i p = i p = , 2 2 2 2 p where we have rationalized the denominator per convention. From t 2 = 5, we get t p = 5. p In total, we have four complex solutions - two real: t = 5 and two non-real: t = i 22 . 3 + 1 = 0, we can subtract the 1 from both sides and extract 3. To find the real solutions to zp 3 cube roots: z = 1, so z = 3 1 = 1. It turns out there are two more non-real complex number solutions to this equation. To get at these, we factor: (z + z3 + 1 = 0 z + 1) = 0 z + 1 = 0 or z 2 1)(z 2 Factor (Sum of Two Cubes) z +1=0 From z + 1 = 0, we get our real solution z = Formula to get: z= ( 1) p ( 1)2 2(1) 1. From z 2 4(1)(1) = 1 z + 1 = 0, we apply the Quadratic p 2 3 p 1i 3 = 2 p Thus we get three solutions to z 3 + 1 = 0 - one real: z = 1 and two non-real: z = 1i2 3 . As always, the reader is encouraged to test their algebraic mettle and check these solutions. It is no coincidence that the non-real solutions to the equations in Example 0.10.2 appear in complex conjugate pairs. Any time we use the Quadratic Formula to solve an equation with real coefficients, the answers will form a complex conjugate pair owing to the in the Quadratic Formula. This leads us to a generalization of Theorem 0.10 which we state on the next page. 0.10 Complex Numbers 131 Theorem 0.13. Discriminant Theorem: Given a Quadratic Equation AX 2 + BX + C = 0, where A, B and C are real numbers, let D = B 2 4AC be the discriminant. If D > 0, there are two distinct real number solutions to the equation. If D = 0, there is one (repeated) real number solution. Note: 'Repeated' here comes from the fact that 'both' solutions B0 2A reduce to B 2A . If D < 0, there are two non-real solutions which form a complex conjugate pair. We will have much more to say about complex solutions to equations in Section 3.4 and we will revisit Theorem 0.13 then. 132 Prerequisites 0.10.1 Exercises In Exercises 1 - 10, use the given complex numbers z and w to find and simplify the following. z +w z2 zw 1 z z z w w z (z)2 zz 1. z = 2 + 3i, w = 4i 2. z = 1 + i, w = i 3. z = i, w = 4. z = 4i, w = 2 2i 1 + 2i 5. z = 3 5i, w = 2 + 7i p p p p 7. z = 2 i 2, w = 2 + i 2 p p 1 3 1 3 9. z = + i, w = + i 2 2 2 2 6. z = 5 + i, w = 4 + 2i p p 8. z = 1 i 3, w = 1 i 3 p p p 2 2 2 10. z = + i, w = 2 2 2 In Exercises 11 - 18, simplify the quantity. p p 11. 49 12. 9 p p p 15. 9 16 16. ( 9)( 16) 13. 17. p p 25 p p 2 i 2 p ( 25)( 4) p 18. ( 9) 4 14. ( 9) We know that i 2 = 1 which means i 3 = i 2 i = ( 1) i = i and i 4 = i 2 i 2 = ( 1)( 1) = 1. In Exercises 19 - 26, use this information to simplify the given power of i. 19. i 5 20. i 6 21. i 7 22. i 8 23. i 15 24. i 26 25. i 117 26. i 304 In Exercises 27 - 35, find all complex solutions. 27. 3x 2 + 6 = 4x 30. 2 1 w 33. x = p 28. 15t 2 + 2t + 5 = 3t(t 2 + 1) =w 31. 2 5 x 34. 36. Multiply and simplify: x [3 y 3 3 =y y 5y 4 + 1 = 3y 2 y2 1 p p i 23] x [3 + i 23] 29. 3y 2 + 4 = y 4 32. x3 x = 2x 1 3 35. z 4 = 16 0.10 Complex Numbers 0.10.2 133 Answers 1. For z = 2 + 3i and w = 4i z + w = 2 + 7i 1 z = 2 13 3 13 z=2 zw = i 3i = 1 2 1 2 z=1 i i i w z = 12 13 z w = 1+i w z = 1 + 3i zw = z w 2 5 z2 = i 1 5 zz = 1 (z)2 = zw = 8 + 8i z2 = 4i 5. For z = 3 i 1 z = i 1 2 i 2i 1 =2+i 1 z w = 1+i w z = zz = 16 (z)2 = zw = 41 + 11i z2 = 16 1 2 16 5i and w = 2 + 7i z + w = 5 + 2i 1 2 2i z + w = 2 + 2i z= i w z i 4. For z = 4i and w = 2 = 2 z= 1 4 12i 1 + 2i = 1 2 (z)2 = i 1 z 5 i z 2 = 2i i = 8 13 + (z)2 = zz = 2 z +w = 1 z 1 2 zw = 1 3. For z = i and w = 3 4 5 + 12i i z +w =1 1 z = zz = 13 2. For z = 1 + i and w = z w z2 = 12 + 8i 3 34 + 5 34 z = 3 + 5i i z w = 29 53 zz = 34 31 53 i w z = (z)2 = 16 29 34 + 30i 31 34 i 16 + 30i 134 Prerequisites 6. For z = 5 + i and w = 4 + 2i z +w = 1 z 1 + 3i 5 26 = z= 1 26 5 i i z w 22 9 10 = + z 2 = 24 6i 7 10 i 1 z z= 2 4 p p 2 4 + i p 2+i 2 p i 3 and w = p z + w = 2i 3 1 z = 1 4 + p 3 4 1 2 1 z + = 1 2 z= 1 2 10. For z = 3 2 1 z = z= 11. 7i + p 3 2 + p 2 2 p 2 2 i p 2 2 p z w = 1 2 z2 = + p 3 2 i 2 2 i p 2 2 i p z w p 2 2 2 2 = p 1 2 3 2 i p 1 2 3 2 i p 2 + 2i 3 z w 1 2 w z = 1 2 + + p i z2 = = i w z i =i (z)2 = i 13. 10 p 3 2 1 2 (z)2 = zz = 1 12. 3i = z2 = 1 zw = 1 p w z p 2i 3 2 (z)2 = zz = 1 2 =i i i and w = w z 4i (z)2 = 4i 4 zw = 3 2 i p 3 2 i z +w = 1 2 p 2 2 i zz = 4 p p = zw = i i and w = p z +w =i 3 7 13 p i 3 1 p z =1+i 3 9. For z = z w z2 = zz = 4 8. For z = 1 9 13 = 2 p = w z 10i (z)2 = 24 + 10i zz = 26 p p p i 2 and w = 2 + i 2 p z +w =2 2 zw = 4 7. For z = p zw = 14. 10 3 2 i i p 3 2 i 0.10 Complex Numbers 15. 12 135 16. 12 19. i 5 = i 4 i = 1 i = i 21. i 7 = i 4 i 3 = 1 ( i) = 23. i 15 = i 4 25. i 117 = i 4 3 36. x 2 6x + 32 22. i 8 = i 4 i 4 = i 4 i i =1i =i p 2 i 14 27. x = 3 p 1i 7 30. w = 2 p p 5i 3 33. x = 2 18. 20. i 6 = i 4 i 2 = 1 ( 1) = i 3 = 1 ( i) = 29 17. 3 i 24. i 26 = i 4 26. i 304 = i 4 p i 3 28. t = 5, 3 p 3i 2 31. y = 2 p i 2 34. y = i, 2 6 2 3i 1 = (1)2 = 1 i 2 = 1 ( 1) = 76 1 = 176 = 1 29. y = 2, i p 1i 2 32. x = 0, 3 35. z = 2, 2i Section 0.7: Quadratic Equations from Precalculus Prerequisites a.k.a. 'Chapter 0' by Carl Stitz, PhD, and Jeff Zeager, PhD, is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. 2013, Carl Stitz. 0.7 Quadratic Equations 0.7 83 Quadratic Equations In Section 0.6.1, we reviewed how to solve basic non-linear equations by factoring. The astute reader should have noticed that all of the equations in that section were carefully constructed so that the polynomials could be factored using the integers. To demonstrate just how contrived the equations had to be, we can solve 2x 2 + 5x 3 = 0 by factoring, (2x 1)(x + 3) = 0, from which we obtain x = 12 and x = 3. If we change the 5 to a 6 and try to solve 2x 2 + 6x 3 = 0, however, we find that this polynomial doesn't factor over the integers and we are stuck. It turns out that there are two real number solutions to this equation, but they are irrational numbers, and our aim in this section is to review the techniques which allow us to find these solutions.1 In this section, we focus our attention on quadratic equations. Definition 0.15. An equation is said to be quadratic in a variable X if it can be written in the form AX 2 + BX + C = 0 where A, B and C are expressions which do not involve X and A 6= 0. Think of quadratic equations as equations that are one degree up from linear equations - instead of the highest power of X being just X = X 1 , it's X 2 . The simplest class of quadratic equations to solve are the ones in which B = 0. In that case, we have the following. Solving Quadratic Equations by Extracting Square Roots p If c is a real number with c 0, the solutions to X 2 = c are X = c. Note: If c < 0, X 2 = c has no real number solutions. There are a couple different ways to see why Extracting Square Roots works, both of which are demonstrated by solving the equation x 2 = 3. If we follow the procedure outlined in the previous 2 section, we subtract 3 from both sides to get x 2 3 = 0 and we now try to factor 3. As p x2 2 2 mentioned in the remarks following Definition 0.14, we could think of x 3 = x ( 3) and p p p p apply 2 the Difference of Squares formula to factor x 3 = (x 3)(x + 3). We solve (x 3)(x + p3) = 0 by usingpthe Zero Product Property as before by setting each factor equal to zero: x 3 = 0 p p and x + 3 0.pWe get the p answerspx = 3. In general, if c 0, then c is a real number, so x 2 c = x 2 ( c)2 = (x c)(x + c). Replacing the '3' with 'c' in the above discussion gives the general result. 2 Another way to view this result is p pto visualize 'taking the square root' of both sides: since xp= c, p 2 2 x2 p = c. How do we simplify p a bit of caution here. Note that (5) p x ? We have to exercise and ( 5)2 both simplify to 25 = 5. In both cases, x 2 returned a positive number, p since the negative in 5 was 'squared away' before we took the squareproot. In other words, x 2 is x if x is positive, or, if x is negative, we make x positive - that is, x 2 = |x|, the absolute value p p of x. 2 = So from x 2 = 3, we 'take the square root' of both sides of the equation to get x 3. This p p p simplifies to |x| = 3, which by Theorem 0.3 is equivalent to x = 3 or x = 3. Replacing the '3' in the previous argument with 'c,' gives the general result. 1 While our discussion in this section departs from factoring, we'll see in Chapter 3 that the same correspondence between factoring and solving equations holds whether or not the polynomial factors over the integers. 84 Prerequisites As you might expect, Extracting Square Roots can be applied to more complicated equations. Consider the equation below. We can solve it by Extracting Square Roots provided we first isolate the perfect square quantity: 3 2 2 x+ 2 2 x+ x+ 15 = 0 2 3 2 15 = 2 2 2 3 15 = 2 4r 3 15 x+ = 2 p 4 3 15 x+ = 2 2 p 3 15 x = 2 p2 3 15 x = 2 Add 15 2 Divide by 2 Extract Square Roots Property of Radicals Subtract 3 2 Add fractions Let's return to the equation 2x 2 + 6x 3 = 0 from the beginning of the section. We leave it to the reader to show that 3 2 15 2 x+ = 2x 2 + 6x 3. 2 2 (Hint: Expand the left side.) In other words, we can solve 2x 2 + 6x 3 = 0 by transforming into an equivalent equation. This process, you may recall, is called 'Completing the Square.' We'll revisit Completing the Square in Section 2.3 in more generality and for a different purpose but for now we revisit the steps needed to complete the square to solve a quadratic equation. Solving Quadratic Equations: Completing the Square To solve a quadratic equation AX 2 + BX + C = 0 by Completing the Square: 1. Subtract the constant C from both sides. 2. Divide both sides by A, the coefficient of X 2 . (Remember: A 6= 0.) 2 B 3. Add 2A to both sides of the equation. (That's half the coefficient of X , squared.) 2 B . 4. Factor the left hand side of the equation as X + 2A 5. Extract Square Roots. 6. Subtract B 2A from both sides. 0.7 Quadratic Equations 85 To refresh our memories, we apply this method to solve 3x 2 3x 2 x2 24x + 5 = 0 24x = 5 5 2 x 8x = 3 5 8x + 16 = + 16 3 43 (x 4)2 = 3r 43 x 4 = 3 r 43 x = 4 3 3x 2 24x + 5 = 0: Subtract C = 5 Divide by A = 3 2 B Add 2A = ( 4)2 = 16 Factor: Perfect Square Trinomial Extract Square Roots Add 4 At this point, we use properties of fractions and radicals to 'rationalize' the denominator:2 r r p p 43 43 3 129 129 = = p = 3 33 3 9 We can now get a common (integer) denominator which yields: r p p 43 129 12 129 x =4 =4 = 3 3 3 The key to Completing the Square is that the procedure always produces a perfect square trinomial. To see why this works every single time, we start with AX 2 + BX + C = 0 and follow the procedure: AX 2 + BX + C = 0 AX 2 + BX = C Subtract C BX C X2 + = Divide by A 6= 0 A A 2 2 2 BX B C B B 2 X + + = + Add A 2A A 2A 2A (Hold onto the line above for a moment.) Here's the heart of the method - we need to show that 2 BX B B 2 2 X + + = X+ A 2A 2A To show this, we start with the right side of the equation and apply the Perfect Square Formula from Theorem 0.7 2 2 B 2 B B BX B 2 2 X+ =X +2 X+ =X + + X 2A 2A 2A A 2A 2 Recall that this means we want to get a denominator with rational (more specifically, integer) numbers. 86 Prerequisites With just a few more steps we can solve the general equation AX 2 + BX + C = 0 so let's pick up the story where we left off. (The line on the previous page we told you to hold on to.) X2 2 BX B + + A 2A B 2 X+ 2A B 2 X+ 2A B 2 X+ 2A B X+ 2A B X+ 2A = = = = = = X = X = 2 C B + A 2A C B2 + A 4A2 4AC B2 + 4A2 4A2 2 B 4AC 2 r4A 2 B 4AC 4A2 p B 2 4AC 2A p B B 2 4AC 2A p 2A 2 B B 4AC 2A Factor: Perfect Square Trinomial Get a common denominator Add fractions Extract Square Roots Properties of Radicals Subtract B 2A Add fractions. Lo and behold, we have derived the legendary Quadratic Formula! Theorem 0.9. Quadratic Formula: The solution to AX 2 + BX + C = 0 with A 6= 0 is: p B B 2 4AC X = 2A We can check our earlier solutions to 2x 2 + 6x 3 = 0 and 3x 2 24x + 5 = 0 using the Quadratic Formula. For 2x 2 + 6x 3 = 0, we identify A = 2, B = 6 and C = 3. The quadratic formula gives: p p p 6 62 4(2)( 3) 6 36 + 24 6 60 x= = 2(2) 4 4 p p 2( 3 15) 3 15 = . 4 2 p 3 15 3 , as required. 2 p p Using properties of radicals ( 60 = 2 15), this reduces to reader to show these two answers are the same as For 3x 2 24x + 5 = 0, we identify A = 3, B = 24 and C = 5. Here, we get: p p ( 24) ( 24)2 4(3)(5) 24 516 x= = 2(3) 6 p p p Since 516 = 2 129, this reduces to x = 123 129 . 3 Think about what (3 p 15) is really telling you. We leave it to the 0.7 Quadratic Equations 87 It is worth noting that the Quadratic Formula applies to all quadratic equations - even ones we could solve using other techniques. For example, to solve 2x 2 + 5x 3 = 0 we identify A = 2, B = 5 and C = 3. This yields: p p 5 52 4(2)( 3) 5 49 57 x= = = 2(2) 4 4 At this point, we have x = 5+7 = 12 and x = 54 7 = 412 = 3 - the same two answers we obtained 4 factoring. We can also use it to solve x 2 = 3, if we wanted to. From x 2 3 = 0, we have A = 1, B = 0 and C = 3. The Quadratic Formula produces p p p p 0 02 4(1)(3) 12 2 3 x= = = = 3 2(1) 2 2 As this last example illustrates, while the Quadratic Formula can be used to solve every quadratic equation, that doesn't mean it should be used. Many times other methods are more efficient. We now provide a more comprehensive approach to solving Quadratic Equations. Strategies for Solving Quadratic Equations If the variable appears in the squared term only, isolate it and Extract Square Roots. Otherwise, put the nonzero terms on one side of the equation so that the other side is 0. - Try factoring. - If the expression doesn't factor easily, use the Quadratic Formula. The reader is encouraged to pause for a moment to think about why 'Completing the Square' doesn't appear in our list of strategies despite the fact that we've spent the majority of the section so far talking about it.4 Let's get some practice solving quadratic equations, shall we? Example 0.7.1. Find all real number solutions to the following equations. 1. 3 (2w 4. 5(25 1)2 = 0 21x) = 59 4 2. 5x 25x 2 5. x(x 3) = 7 p 4.9t 2 + 10t 3 + 2 = 0 3. (y y +2 3 1)2 = 2 6. 2x 2 = 3x 4 6 Solution. 1. Since 3 (2w 1)2 = 0 contains a perfect square, we isolate it first then extract square roots: 3 4 1)2 p3 p3 1 p3 1 3 2 (2w = = = = 0 (2w 1)2 Add (2w 1)2 2w 1 Extract Square Roots 2w Add 1 = w Divide by 2 Unacceptable answers include \"Jeff and Carl are mean\" and \"It was one of Carl's Pedantic Rants\". 88 Prerequisites p We find our two answers w = 12 3 . The reader is encouraged to check both answers by substituting each into the original equation.5 2. To solve 5x x(x side equal to 0. 3) = 7, we begin performing the indicated operations and getting one 5x x(x 3) 5x x 2 + 3x x 2 + 8x 2 x + 8x 7 = = = = 7 7 Distribute 7 Gather like terms 0 Subtract 7 At this point, we attempt to factor and find x 2 + 8x 7 = (x 1)( x + 7). Using the Zero Product Property, we get x 1 = 0 or x + 7 = 0. Our answers are x = 1 or x = 7, both of which are easy to check. 3. Even though we have a perfect square in (y 1)2 = 2 y+2 3 , Extracting Square Roots won't help matters since we have a y on the other side of the equation. Our strategy here is to perform the indicated operations (and clear the fraction for good measure) and get 0 on one side of the equation. (y y2 3(y 2 3y 2 3y 2 y +2 3 y +2 2 Perfect Square Trinomial 3 y +2 3 2 Multiply by 3 3 y +2 6 3 Distribute 3 6 (y + 2) 0 Subtract 6, Add (y + 2) 0 1)2 = 2 2y + 1 = 2y + 1) = 6y + 3 = 3y 2 6y + 3 = 6y + 3 6 + (y + 2) = 3y 2 5y 1 = A cursory attempt at factoring bears no fruit, so we run this through the Quadratic Formula with A = 3, B = 5 and C = 1. p ( 5) ( 5)2 4(3)( 1) y = 2(3) p 5 25 + 12 y = p6 5 37 y = 6 p p Since 37 is prime, we have no way to reduce 37. Thus, our final answers are y = 56 37 . The reader is encouraged to supply the details of the challenging verification of the answers. 5 It's excellent practice working with radicals fractions so we really, really want you to take the time to do it. 0.7 Quadratic Equations 89 4. We proceed as before; our aim is to gather the nonzero terms on one side of the equation. 59 25x 2 4 59 125 105x = 25x 2 4 59 2 4(125 105x) = 4 25x 4 500 420x = 59 100x 2 5(25 500 21x) = Distribute Multiply by 4 Distribute 420x 59 + 100x 2 = 0 Subtract 59, Add 100x 2 100x 2 420x + 441 = 0 Gather like terms With highly composite numbers like 100 and 441, factoring seems inefficient at best,6 so we apply the Quadratic Formula with A = 100, B = 420 and C = 441: x = = = = = = ( 420) 420 p p ( 420)2 2(100) 176000 200 4(100)(441) 176400 p 420 0 200 420 0 200 420 200 21 10 To our surprise and delight we obtain just one answer, x = 21 10 . p 5. Our next equation 4.9t 2 + 10t p 3 + 2 = 0, already has 0 on one side of the equation, but with coefficients like 4.9 and 10 3, factoring with integers is not an option. We could make things a bit easier on the eyes by clearing the decimal (by multiplying through by 10) p p to get 49t 2 + 100t 3 + 20 = 0 but we simply cannot rid ourselves of the irrational number 3. The p Quadratic Formula is our only recourse. With A = 49, B = 100 3 and C = 20 we get: 6 This is actually the Perfect Square Trinomial (10x 21)2 . 90 Prerequisites t = = = = = = = = q p p 100 3 (100 3)2 4( 49)(20) 2( 49) p p 100 3 30000 + 3920 98 p p 100 3 33920 98 p p 100 3 8 530 98 p p 2( 50 3 4 530) 2( 49) p p 50 3 4 530 49 p p ( 50 3 4 530) 49 p p 50 3 4 530 49 Reduce Properties of Negatives Distribute You'll note that when we 'distributed' the negative in the last step, we changed the '' to a '.' While this is technically correct, at thepend pof the day both symbols mean 'plus or minus',7 530 so we can write our answers as t = 50 34 . Checking these answers are a true test of 49 arithmetic mettle. 6. At first glance, the equation 2x 2 = 3x 4 6 seems misplaced. The highest power of the variable x here is 4, not 2, so this equation isn't a quadratic equation - at least not in terms of the variable x. It is, however, an example of an equation that is quadratic 'in disguise.'8 We introduce a new variable u to help us see the pattern - specifically we let u = x 2 . Thus u 2 = (x 2 )2 = x 4 . So in terms of the variable u, the equation 2x 2 = 3x 4 6 is 2u = 3u 2 6. The latter is a quadratic equation, which we can solve using the usual techniques: 2u = 3u 2 0 = 3u 2 6 2u 6 Subtract 2u After a few attempts at factoring, we resort to the Quadratic Formula with A = 3, B = 7 2, There are instances where we need both symbols, however. For example, the Sum and Difference of Cubes Formulas (page 71) can be written as a single formula: a3 b3 = (a b)(a2 ab + b2 ). In this case, all of the 'top' symbols are read to give the sum formula; the 'bottom' symbols give the difference formula. 8 More formally, quadratic in form. Carl likes 'Quadratics in Disguise' since it reminds him of the tagline of one of his beloved childhood cartoons and toy lines. 0.7 Quadratic Equations C= 91 6 and get: u = = = = = = = ( 2) 2 p p ( 2)2 2(3) 4(3)( 6) 4 + 72 6 p 2 76 6 p 2 4 19 6 p 2 2 19 6 p 2(1 19) 2(3) p 1 19 3 Properties of Radicals Factor Reduce We've solved the equation for u, but what we still need to solve the original equation9 - which means we need to find the corresponding values of x. Since u = x 2 , we have two equations: x2 = 1+ p 19 3 or x2 = 1 p 19 3 We can solve the first equation by extracting square roots to get x = p q p 1+ 19 3 . The second equation, however, has no real number solutions because 1 3 19 is a negative number. For our final answers we can rationalize the denominator10 to get: x = s 1+ p 3 19 = s 1+ p 19 3 = 3 3 p p 3 + 3 19 3 As with the previous exercise, the very challenging check is left to the reader. Our last example above, the 'Quadratic in Disguise', hints that the Quadratic Formula is applicable to a wider class of equations than those which are strictly quadratic. We give some general guidelines to recognizing these beasts in the wild on the next page. 9 10 Or, you've solved the equation for 'you' (u), now you have to solve it for your instructor (x). We'll say more about this technique in Section 0.9. 92 Prerequisites Identifying Quadratics in Disguise An equation is a 'Quadratic in Disguise' if it can be written in the form: AX 2m + BX m + C = 0. In other words: There are exactly three terms, two with variables and one constant term. The exponent on the variable in one term is exactly twice the variable on the other term. To transform a Quadratic in Disguise to a quadratic equation, let u = X m so u 2 = (X m )2 = X 2m . This transforms the equation into Au 2 + Bu + C = 0. For example, 3x 6 2x 3 + 1 = 0 is a Quadratic in Disguise, since 6 = 2 3. If we let u = x 3 , we get u 2 = (x 3 )2 = x 6 , so the equation becomes 3u 2 2u + 1 = 0. However, 3x 6 2x 2 + 1 = 0 is not a Quadratic in Disguise, since 6 6= 2 2. The substitution u = x 2 yields u 2 = (x 2 )2 = x 4 , not x 6 as required. We'll see more instances of 'Quadratics in Disguise' in later sections. We close this section with a review of the discriminant of a quadratic equation as defined below. Definition 0.16. The Discriminant: Given a quadratic equation AX 2 + BX + C = 0, the quantity B 2 4AC is called the discriminant of the equation. The discriminant is the radicand of the square root in the quadratic formula: p B B 2 4AC X = 2A It discriminates between the nature and number of solutions we get from a quadratic equation. The results are summarized below. Theorem 0.10. Discriminant Theorem: Given a Quadratic Equation AX 2 + BX + C = 0, let D = B 2 4AC be the discriminant. If D > 0, there are two distinct real number solutions to the equation. If D = 0, there is one repeated real number solution. Note: 'Repeated' here comes from the fact that 'both' solutions B0 2A reduce to B 2A . If D < 0, there are no real solutions. For example, x 2 + x 1 = 0 has two real number solutions since the discriminant works out to be p (1)2 4(1)( 1) = 5 > 0. This results in a 5 in the Quadratic Formula, generating two different answers. On the other hand, x 2 + x + 1 = 0 haspno real solutions since here, the discriminant is (1)2 4(1)(1) = 3 < 0 which generates a 3 in the Quadratic Formula. The equation p x 2 + 2x + 1 = 0 has discriminant (2)2 4(1)(1) = 0 so in the Quadratic Formula we get a 0 = 0 thereby generating just one solution. More can be said p as well. For example, the discriminant of 6x 2 x 40 = 0 is 961. This is a perfect square, 961 = 31, which means our solutions are 0.7 Quadratic Equations 93 rational numbers. When our solutions are rational numbers, the quadratic actually factors nicely. In our example 6x 2 x 40 = (2x + 5)(3x 8). Admittedly, if you've already computed the discriminant, you're most of the way done with the problem and probably wouldn't take the time to experiment with factoring the quadratic at this point - but we'll see another use for this analysis of the discriminant in the next section.11 11 Specifically in Example 0.8.1. 94 Prerequisites 0.7.1 Exercises In Exercises 1 - 21, find all real solutions. Check your answers, as directed by your instructor. 1. 3 x 1 2 4. x 2 + x 7. 2 = 5 12 2. 4 5. 3w 2 = 2 1=0 z = 4z 2 2 13. w 4 + 3w 2 11. (x 1=0 16. 3x 4 + 6x 2 = 15x 3 19. y 2 p 8y = 3. 3(y 2 w p 18y 1 3)2 2 = 10 6. y(y + 4) = 1 8. 0.1v 2 + 0.2v = 0.3 1 t = 2(t + 1)2 10. 3 (5t + 3)2 = 3 9. x 2 = x 3)2 = x 2 + 9 12. (3y 1 1)(2y + 1) = 5y 14. 2x 4 + x 2 = 3 15. (2 17. 6p + 2 = p2 + 3p3 18. 10v = 7v 3 v 5 p v2 v 3 21. = +1 3 2 p p p 20. x 2 3 = x 6 + 12 y)4 = 3(2 y )2 + 1 In Exercises 22 - 27, find all real solutions and use a calculator to approximate your answers, rounded to two decimal places. p 22. 5.542 + b2 = 36 23. r 2 = 37 24. 54 = 8r 2 + r 2 25. 4.9t 2 + 100t = 410 26. x 2 = 1.65(3 x)2 27. (0.5+2A)2 = 0.7(0.1 A)2 In Exercises 28 - 30, use Theorem 0.3 along with the techniques in this section to find all real solutions to the following. 28. |x 2 3x| = 2 29. |2x x 2 | = |2x 1| 30. |x 2 x + 3| = |4 31. Prove that for every nonzero number p, x 2 + xp + p2 = 0 has no real solutions. 32. Solve for t: 1 2 gt + vt + h = 0. Assume g > 0, v 2 0 and h 0. x 2| 0.7 Quadratic Equations 0.7.2 1. 4. 7. 10. 13. 16. 19. 22. 24. 26. 28. 95 Answers p p 3 5 4 2 x= 2. t = , 3. y = 1, 5 6 5 5 p p 1 5 2 x= 5. w = 1, 6. y = 2 5 2 3 p 1 65 z= 8. v = 3, 1 9. No real solution. 16 p p 5 33 2 10 t= 11. x = 0 12. y = 4 6 p rp p 13 3 4 6 + 2 13 w = 14. x = 1 15. y = 2 2 p p p 5 17 1 p x = 0, 17. p = , 2 18. v = 0, 2, 5 2 3 p p p p p 5 2 46 2 10 3 p y= 20. x = 21. v = ,2 3 2 2 2 p r 13271 37 b= 2.30 23. r = 3.43 50 p p p 500 10 491 4 2 54 + 32 25. t = , t 5.68, 14.73 r= , r 6.32, 2.72 49 p p 99 6 165 107 7 70 x= , x 1.69, 13.54 27. A = , A 0.50, 0.15 13 330 p p 3 17 1 x = 1, 2, 29. x = 1, 2 3 30. x = , 1, 7 2 2 31. The discriminant is: D = p2 p v v 2 + 2gh 32. t = g 4p2 = 3p2 < 0. Since D < 0, there are no real solutions. Section 2.3: Quadratic Functions, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 license. 2013, Carl Stitz. 188 2.3 Linear and Quadratic Functions Quadratic Functions You may recall studying quadratic equations in Intermediate Algebra. In this section, we review those equations in the context of our next family of functions: the quadratic functions. Definition 2.5. A quadratic function is a function of the form f (x) = ax2 + bx + c, where a, b and c are real numbers with a 6= 0. The domain of a quadratic function is ( 1, 1). The most basic quadratic function is f (x) = x2 , whose graph appears below. Its shape should look familiar from Intermediate Algebra - it is called a parabola. The point (0, 0) is called the vertex of the parabola. In this case, the vertex is a relative minimum and is also the where the absolute minimum value of f can be found. y ( 2, 4) (2, 4) 4 3 2 ( 1, 1) 2 1 (1, 1) 1 (0, 0) f (x) = 1 2 x x2 Much like many of the absolute value functions in Section 2.2, knowing the graph of f (x) = x2 enables us to graph an entire family of quadratic functions using transformations. Example 2.3.1. Graph the following functions starting with the graph of f (x) = x2 and using transformations. Find the vertex, state the range and find the x- and y-intercepts, if any exist. 1. g(x) = (x + 2)2 3 2. h(x) = 2(x 3)2 + 1 Solution. 1. Since g(x) = (x + 2)2 3 = f (x + 2) 3, Theorem 1.7 instructs us to first subtract 2 from each of the x-values of the points on y = f (x). This shifts the graph of y = f (x) to the left 2 units and moves ( 2, 4) to ( 4, 4), ( 1, 1) to ( 3, 1), (0, 0) to ( 2, 0), (1, 1) to ( 1, 1) and (2, 4) to (0, 4). Next, we subtract 3 from each of the y-values of these new points. This moves the graph down 3 units and moves ( 4, 4) to ( 4, 1), ( 3, 1) to ( 3, 2), ( 2, 0) to ( 2, 3), ( 1, 1) to ( 1, 2) and (0, 4) to (0, 1). We connect the dots in parabolic fashion to get 2.3 Quadratic Functions 189 y ( 2, 4) y (2, 4) 4 ( 4, 1) 3 1 4 3 2 x 1 1 2 ( 1, 1) (0, 1) (1, 1) 1 ( 3, 2) ( 1, 2) 3 2 1 (0, 0) 1 x 2 ( 2, ! f (x) = x2 3) 3 = (x + 2)2 g(x) = f (x + 2) 3 From the graph, we see that the vertex has moved from (0, 0) on the graph of y = f (x) to ( 2, 3) on the graph of y = g(x). This sets [ 3, 1) as the range of g. We see that the graph of y = g(x) crosses the x-axis twice, so we expect two x-intercepts. To find these, we set y = g(x) = 0 and solve. Doing so yields the equation p (x + 2)2 3 = 0, or p (x + 2)2 = p 3. Extracting square roots gives p x + 2 = 3, or x = 2 3. Our x-intercepts are ( 2 3, 0) ( 3.73, 0) and ( 2 + 3, 0) ( 0.27, 0). The y-intercept of the graph, (0, 1) was one of the points we originally plotted, so we are done. 2. Following Theorem 1.7 once more, to graph h(x) = 2(x 3)2 + 1 = 2f (x 3) + 1, we first start by adding 3 to each of the x-values of the points on the graph of y = f (x). This eects a horizontal shift right 3 units and moves ( 2, 4) to (1, 4), ( 1, 1) to (2, 1), (0, 0) to (3, 0), (1, 1) to (4, 1) and (2, 4) to (5, 4). Next, we multiply each of our y-values first by 2 and then add 1 to that result. Geometrically, this is a vertical stretch by a factor of 2, followed by a reflection about the x-axis, followed by a vertical shift up 1 unit. This moves (1, 4) to (1, 7), (2, 1) to (2, 1), (3, 0) to (3, 1), (4, 1) to (4, 1) and (5, 4) to (5, 7). y (3, 1) 1 1 1 (2, 2 1) 3 4 5 (4, x 1) 2 y 3 ( 2, 4) ( 1, 1) 2 1 4 (2, 4) 4 3 5 2 6 (0, 0) f (x) = (1, (1, 1) 1 1 x2 2 x ! 7) h(x) = = (5, 7) 2f (x 3) + 1 2(x 3)2 + 1 The vertex is (3, 1) which makes the range of h ( 1, 1]. From our graph, we know that there are two x-intercepts, so we set y = h(x) = 0 and solve. We get 2(x 3)2 + 1 = 0 190 Linear and Quadratic Functions p 2 3 = p2 , so that when we p 6 2 6 2 2 add 3 to each side, we get x = 2 . Hence, our x-intercepts are 2 , 0 (2.29, 0) and p 6+ 2 2 , 0 (3.71, 0). Although our graph doesn't show it, there is a y-intercept which can which gives (x 3)2 = 1 2. Extracting square roots1 gives x be found by setting x = 0. With h(0) = is (0, 17). 2(0 3)2 + 1 = 17, we have that our y-intercept A few remarks about Example 2.3.1 are in order. First note that neither the formula given for g(x) nor the one given for h(x) match the form given in Definition 2.5. We could, of course, convert both g(x) and h(x) into that form by expanding and collecting like terms. Doing so, we find g(x) = (x + 2)2 3 = x2 + 4x + 1 and h(x) = 2(x 3)2 + 1 = 2x2 + 12x 17. While these 'simplified' formulas for g(x) and h(x) satisfy Definition 2.5, they do not lend themselves to graphing easily. For that reason, the form of g and h presented in Example 2.3.2 is given a special name, which we list below, along with the form presented in Definition 2.5. Definition 2.6. Standard and General Form of Quadratic Functions: Suppose f is a quadratic function. The general form of the quadratic function f is f (x) = ax2 + bx + c, where a, b and c are real numbers with a 6= 0. The standard form of the quadratic function f is f (x) = a(x k are real numbers with a 6= 0. h)2 + k, where a, h and It is important to note at this stage that we have no guarantees that every quadratic function can be written in standard form. This is actually true, and we prove this later in the exposition, but for now we celebrate the advantages of the standard form, starting with the following theorem. Theorem 2.2. Vertex Formula for Quadratics in Standard Form: For the quadratic function f (x) = a(x h)2 + k, where a, h and k are real numbers with a 6= 0, the vertex of the graph of y = f (x) is (h, k). We can readily verify the formula given Theorem 2.2 with the two functions given in Example 2.3.1. After a (slight) rewrite, g(x) = (x + 2)2 3 = (x ( 2))2 + ( 3), and we identify h = 2 and k = 3. Sure enough, we found the vertex of the graph of y = g(x) to be ( 2, 3). For h(x) = 2(x 3)2 + 1, no rewrite is needed. We can directly identify h = 3 and k = 1 and, sure enough, we found the vertex of the graph of y = h(x) to be (3, 1). To see why the formula in Theorem 2.2 produces the vertex, consider the graph of the equation y = a(x h)2 + k. When we substitute x = h, we get y = k, so (h, k) is on the graph. If x 6= h, then x h 6= 0 so (x h)2 is a positive number. If a > 0, then a(x h)2 is positive, thus y = a(x h)2 + k is always a number larger than k. This means that when a > 0, (h, k) is the lowest point on the graph and thus the parabola must open upwards, making (h, k) the vertex. A similar argument 1 2 and rationalizing denominators! and get common denominators! 2.3 Quadratic Functions 191 shows that if a < 0, (h, k) is the highest point on the graph, so the parabola opens downwards, and (h, k) is also the vertex in this case. Alternatively, we can apply the machinery in Section 1.7. Since the vertex of y = x2 is (0, 0), we can determine the vertex of y = a(x h)2 +k by determining the final destination of (0, 0) as it is moved through each transformation. To obtain the formula f (x) = a(x h)2 + k, we start with g(x) = x2 and first define g1 (x) = ag(x) = ax2 . This is results in a vertical scaling and/or reflection.3 Since we multiply the output by a, we multiply the y-coordinates on the graph of g by a, so the point (0, 0) remains (0, 0) and remains the vertex. Next, we define g2 (x) = g1 (x h) = a(x h)2 . This induces a horizontal shift right or left h units4 moves the vertex, in either case, to (h, 0). Finally, f (x) = g2 (x) + k = a(x h)2 + k which eects a vertical shift up or down k units5 resulting in the vertex moving from (h, 0) to (h, k). In addition to verifying Theorem 2.2, the arguments in the two preceding paragraphs have also shown us the role of the number a in the graphs of quadratic functions. The graph of y = a(x h)2 +k is a parabola 'opening upwards' if a > 0, and 'opening downwards' if a < 0. Moreover, the symmetry enjoyed by the graph of y = x2 about the y-axis is translated to a symmetry about the vertical line x = h which is the vertical line through the vertex.6 This line is called the axis of symmetry of the parabola and is dashed in the figures below. vertex vertex a>0 a<0 h)2 + k. graphs of y = a(x without a doubt, the standard form quadratic function, coupled with machinery in section 1.7, allows us to list attributes such functions quickly and elegantly. what remains be shown, however, is fact that every function can written form. convert given general into form, we employ ancient rite 'completing square'. remind reader how this done our next example. example 2.3.2. below from find vertex, axis symmetry any x- or y-intercepts. graph each determine its range. 1. f (x) =x2 3 4x 3. 2. g(x) =6 just scaling if> 0. If a < 0, there is a reflection involved. Right if h > 0, left if h < 0. 5 Up if k > 0, down if k < 0 6 You should use transformations to verify this! 4 x x2 192 Linear and Quadratic Functions Solution. 1. To convert from general form to standard form, we complete the square.7 First, we verify that the coefficient of x2 is 1. Next, we find the coefficient of x, in this case 4, and take half of it to get 12 ( 4) = 2. This tells us that our target perfect square quantity is (x 2)2 . To get an expression equivalent to (x 2)2 , we need to add ( 2)2 = 4 to the x2 4x to create a perfect square trinomial, but to keep the balance, we must also subtract it. We collect the terms which create the perfect square and gather the remaining constant terms. Putting it all together, we get f (x) = x2 (Compute 12 ( 4) = 4x + 3 = x2 4x + 4 = x2 4x + 4 = (x 2)2 2.) 4 + 3 (Add and subtract ( 2)2 = 4 to (x2 + 4x).) 4+3 (Group the perfect square trinomial.) 1 (Factor the perfect square trinomial.) Of course, we can always check our answer by multiplying out f (x) = (x 2)2 1 to see that it simplifies to f (x) = x2 4x 1. In the form f (x) = (x 2)2 1, we readily find the vertex to be (2, 1) which makes the axis of symmetry x = 2. To find the x-intercepts, we set y = f (x) = 0. We are spoiled for choice, since we have two formulas for f (x). Since we recognize f (x) = x2 4x + 3 to be easily factorable,8 we proceed to solve x2 4x + 3 = 0. Factoring gives (x 3)(x 1) = 0 so that x = 3 or x = 1. The x-intercepts are then (1, 0) and (3, 0). To find the y-intercept, we set x = 0. Once again, the general form f (x) = x2 4x + 3 is easiest to work with here, and we find y = f (0) = 3. Hence, the y-intercept is (0, 3). With the vertex, axis of symmetry and the intercepts, we get a pretty good graph without the need to plot additional points. We see that the range of f is [ 1, 1) and we are done. 2. To get started, we rewrite g(x) = 6 x x2 = x2 x + 6 and note that the coefficient of x2 is 1, not 1. This means our first step is to factor out the ( 1) from both the x2 and x terms. We then follow the completing the square recipe as above. g(x) = x2 x + 6 = ( 1) x2 + x + 6 = ( 1) x2 + x + 14 = ( 1) x2 + x + = x+ 1 2 2 + 1 4 1 4 (Factor the coefficient of x2 from x2 and x.) +6 + ( 1) 1 4 +6 (Group the perfect square trinomial.) 25 4 7 If you forget why we do what we do to complete the square, start with a(x step, and then reverse the process. 8 Experience pays o, here! h)2 + k, multiply it out, step by 2.3 Quadratic Functions 193 2 1 25 From g(x) = x + 12 + 25 and the axis of symmetry to 4 , we get the vertex to be 2, 4 1 be x = 2 . To get the x-intercepts, we opt to set the given formula g(x) = 6 x x2 = 0. Solving, we get x = 3 and x = 2 , so the x-intercepts are ( 3, 0) and (2, 0). Setting x = 0, we find g(0) = 6, so the y-intercept is (0, 6). Plotting these points gives us the graph below. We see that the range of g is 1, 25 4 . 1 25 , 2 4 y 8 6 7 5 6 4 x=2 5 (0, 3) (0, 6) 3 4 2 3 x= ( 3, 0) 2 3 (1, 0) 1 y 2 1 2 (2, 0) 1 1 2 x (3, 0) 1 2 3 4 5 x 1 (2, f (x) = x2 1) 4x + 3 g(x) = 6 x x2 With Example 2.3.2 fresh in our minds, we are now in a position to show that every quadratic function can be written in standard form. We begin with f (x) = ax2 + bx + c, assume a 6= 0, and complete the square in complete generality. f (x) = ax2 + bx + c b 2 = a x + x +c (Factor out coefficient of x2 from x2 and x.) a b b2 b2 2 = a x + x+ 2 +c a 4a 4a2 2 b b2 b 2 = a x + x+ 2 a +c (Group the perfect square trinomial.) a 4a 4a2 b 2 4ac b2 = a x+ + (Factor and get a common denominator.) 2a 4a b Comparing this last expression with the standard form, we identify (x h) with x + 2a so that 2 2 b 4ac b b 4ac b h = 2a . Instead of memorizing the value k = 4a , we see that f 2a = 4a . As such, we have derived a vertex formula for the general form. We summarize both vertex formulas in the box at the top of the next page. 194 Linear and Quadratic Functions Equation 2.4. Vertex Formulas for Quadratic Functions: Suppose a, b, c, h and k are real numbers with a 6= 0. h)2 + k, the vertex of the graph of y = f (x) is the point (h, k). b b 2 If f (x) = ax + bx + c, the vertex of the graph of y = f (x) is the point ,f . 2a 2a If f (x) = a(x There are two more results which can be gleaned from the completed-square form of the general form of a quadratic function, b f (x) = ax + bx + c = a x + 2a 2 2 + 4ac b2 4a We have seen that the number a in the standard form of a quadratic function determines whether the parabola opens upwards (if a > 0) or downwards (if a < 0). We see here that this number a is none other than the coefficient of x2 in the general form of the quadratic function. In other words, it is the coefficient of x2 alone which determines this behavior - a result that is generalized in Section 3.1. The second treasure is a re-discovery of the quadratic formula. Equation 2.5. The Quadratic Formula: If a, b and c are real numbers with a 6= 0, then the solutions to ax2 + bx + c = 0 are p b b2 4ac x= . 2a Assuming the conditions of Equation 2.5, the solutions to ax2 + bx + c = 0 are precisely the zeros of f (x) = ax2 + bx + c. Since b 2 4ac b2 f (x) = ax2 + bx + c = a x + + 2a 4a the equation ax2 + bx + c = 0 is equivalent to b a x+ 2a Solving gives 2 + 4ac b2 = 0. 4a 2.3 Quadratic Functions 2 195 4ac b2 4a b 2 a x+ 2a " # 1 b 2 a x+ a 2a b 2 x+ 2a b a x+ 2a + b x+ 2a b x+ 2a = 0 = = = 4ac b2 4a 1 b2 4ac a 4a b2 = = x = x = 4ac 4a2 r p b2 b2 4ac 4a2 extract square roots 4ac 2a p b b2 4ac 2a 2a p b b2 4ac 2a In our discussions of domain, we were warned against having negative numbers underneath the p square root. Given that b2 4ac is part of the Quadratic Formula, we will need to pay special attention to the radicand b2 4ac. It turns out that the quantity b2 4ac plays a critical role in determining the nature of the solutions to a quadratic equation. It is given a special name. Definition 2.7. If a, b and c are real numbers with a 6= 0, then the discriminant of the quadratic equation ax2 + bx + c = 0 is the quantity b2 4ac. The discriminant 'discriminates' between the kinds of solutions we get from a quadratic equation. These cases, and their relation to the discriminant, are summarized below. Theorem 2.3. Discriminant Trichotomy: Let a, b and c be real numbers with a 6= 0. If b2 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions. If b2 4ac = 0, the equation ax2 + bx + c = 0 has exactly one real solution. If b2 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions. The proof of Theorem 2.3 stems from the position of the discriminant in the quadratic equation, and is left as a good mental exercise for the reader. The next example exploits the fruits of all of our labor in this section thus far. 196 Linear and Quadratic Functions Example 2.3.3. Recall that the profit (defined on page 82) for a product is defined by the equation Profit = Revenue Cost, or P (x) = R(x) C(x). In Example 2.1.7 the weekly revenue, in dollars, made by selling x PortaBoy Game Systems was found to be R(x) = 1.5x2 + 250x with the restriction (carried over from the price-demand function) that 0 x 166. The cost, in dollars, to produce x PortaBoy Game Systems is given in Example 2.1.5 as C(x) = 80x + 150 for x 0. 1. Determine the weekly profit function P (x). 2. Graph y = P (x). Include the x- and y-intercepts as well as the vertex and axis of symmetry. 3. Interpret the zeros of P . 4. Interpret the vertex of the graph of y = P (x). 5. Recall that the weekly price-demand equation for PortaBoys is p(x) = 1.5x + 250, where p(x) is the price per PortaBoy, in dollars, and x is the weekly sales. What should the price per system be in order to maximize profit? Solution. 1. To find the profit function P (x), we subtract P (x) = R(x) C(x) = 1.5x2 + 250x (80x + 150) = 1.5x2 + 170x 150. Since the revenue function is valid when 0 x 166, P is also restricted to these values. 2. To find the x-intercepts, we set P (x) = 0 and solve 1.5x2 + 170x 150 = 0. The mere thought of trying to factor the left hand side of this equation could do serious psychological damage, so we resort to the quadratic formula, Equation 2.5. Identifying a = 1.5, b = 170, and c = 150, we obtain p b2 4ac 2ap 170 1702 4( 1.5)( 150) = 2( 1.5) p 170 28000 = 3p 170 20 70 = 3 p p 70 170+20 70 We get two x-intercepts: 170 20 , 0 and , 0 . To find the y-intercept, we set 3 3 x = 0 and find y = P (0) = 150 for a y-intercept of (0, 150). To find the vertex, we use the fact that P (x) = 1.5x2 + 170x 150 is in the general form of a quadratic function and 170 appeal to Equation 2.4. Substituting a = 1.5 and b = 170, we get x = 2(170 1.5) = 3 . x = b 2.3 Quadratic Functions 197 To find the y-coordinate of the vertex, we compute P 170 = 14000 and find that our vertex 3 3 170 14000 is 3 , 3 . The axis of symmetry is the vertical line passing through the vertex so it is the line x = 170 3 . To sketch a reasonable graph, we approximate the x-intercepts, (0.89, 0) and (112.44, 0), and the vertex, (56.67, 4666.67). (Note that in order to get the x-intercepts and the vertex to show up in the same picture, we had to scale the x-axis dierently than the y-axis. This results in the left-hand x-intercept and the y-intercept being uncomfortably close to each other and to the origin in the picture.) y 4000 3000 2000 1000 10 20 30 40 50 60 70 80 90 100 110 120 x 3. The zeros of P are the solutions to P (x) = 0, which we have found to be approximately 0.89 and 112.44. As we saw in Example 1.5.3, these are the 'break-even' points of the profit function, where enough product is sold to recover the cost spent to make the product. More importantly, we see from the graph that as long as x is between 0.89 and 112.44, the graph y = P (x) is above the x-axis, meaning y = P (x) > 0 there. This means that for these values of x, a profit is being made. Since x represents the weekly sales of PortaBoy Game Systems, we round the zeros to positive integers and have that as long as 1, but no more than 112 game systems are sold weekly, the retailer will make a profit. 4. From the graph, we see that the maximum value of P occurs at the vertex, which is approximately (56.67, 4666.67). As above, x represents the weekly sales of PortaBoy systems, so we can't sell 56.67 game systems. Comparing P (56) = 4666 and P (57) = 4666.5, we conclude that we will make a maximum profit of $4666.50 if we sell 57 game systems. 5. In the previous part, we found that we need to sell 57 PortaBoys per week to maximize profit. To find the price per PortaBoy, we substitute x = 57 into the price-demand function to get p(57) = 1.5(57) + 250 = 164.5. The price should be set at $164.50. Our next example is another classic application of quadratic functions. Example 2.3.4. Much to Donnie's surprise and delight, he inherits a large parcel of land in Ashtabula County from one of his (e)strange(d) relatives. The time is finally right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough money for 200 linear feet of fencing material. If he makes the pasture adjacent to a stream (so no fencing is required on that side), what are the dimensions of the pasture which maximize the area? What is the maximum area? If an average alpaca needs 25 square feet of grazing area, how many alpaca can Donnie keep in his pasture? 198 Linear and Quadratic Functions Solution. It is always helpful to sketch the problem situation, so we do so below. river pasture w w l We are tasked to find the dimensions of the pasture which would give a maximum area. We let w denote the width of the pasture and we let l denote the length of the pasture. Since the units given to us in the statement of the problem are feet, we assume w and l are measured in feet. The area of the pasture, which we'll call A, is related to w and l by the equation A = wl. Since w and l are both measured in feet, A has units of feet2 , or square feet. We are given the total amount of fencing available is 200 feet, which means w + l + w = 200, or, l + 2w = 200. We now have two equations, A = wl and l + 2w = 200. In order to use the tools given to us in this section to maximize A, we need to use the information given to write A as a function of just one variable, either w or l. This is where we use the equation l + 2w = 200. Solving for l, we find l = 200 2w, and we substitute this into our equation for A. We get A = wl = w(200 2w) = 200w 2w2 . We now have A as a function of w, A(w) = 200w 2w2 = 2w2 + 200w. Before we go any further, we need to find the applied domain of A so that we know what values of w make sense in this problem situation.9 Since w represents the width of the pasture, w > 0. Likewise, l represents the length of the pasture, so l = 200 2w > 0. Solving this latter inequality, we find w < 100. Hence, the function we wish to maximize is A(w) = 2w2 +200w for 0 < w < 100. Since A is a quadratic function (of w), we know that the graph of y = A(w) is a parabola. Since the coefficient of w2 is 2, we know that this parabola opens downwards. This means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we find w = 2(2002) = 50, and A(50) = 2(50)2 + 200(50) = 5000. Since w = 50 lies in the applied domain, 0 < w < 100, we have that the area of the pasture is maximized when the width is 50 feet. To find the length, we use l = 200 2w and find l = 200 2(50) = 100, so the length of the pasture is 100 feet. The maximum area is A(50) = 5000, or 5000 square feet. If an average alpaca requires 25 square feet of pasture, Donnie can raise 5000 25 = 200 average alpaca. We conclude this section with the graph of a more complicated absolute value function. Example 2.3.5. Graph f (x) = |x2 x 6|. Solution. Using the definition of absolute value, Definition 2.4, we have x2 x 6 , if x2 x 6 < 0 f (x) = x2 x 6, if x2 x 6 0 The trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities such as x2 x 6 < 0. You won't have to wait long; this is one of the main topics of Section 2.4. 9 Donnie would be very upset if, for example, we told him the width of the pasture needs to be 50 feet. 2.3 Quadratic Functions 199 Nevertheless, we can attack this problem graphically. To that end, we graph y = g(x) = x2 x 6 using the intercepts and the vertex. To find the x-intercepts, we solve x2 x 6 = 0. Factoring gives (x 3)(x + 2) = 0 so x = 2 or x = 3. Hence, ( 2, 0) and (3, 0) are x-intercepts. The b 1 y-intercept (0, 6) is found by setting x = 0. To plot the vertex, we find x = 2a = 2(1) = 12 , and 2 1 y = 12 6 = 25 2 4 = 6.25. Plotting, we get the parabola seen below on the left. To obtain points on the graph of y = f (x) = |x2 x 6|, we can take points on the graph of g(x) = x2 x 6 and apply the absolute value to each of the y values on the parabola. We see from the graph of g that for x 2 or x 3, the y values on the parabola are greater than or equal to zero (since the graph is on or above the x-axis), so the absolute value leaves these portions of the graph alone. For x between 2 and 3, however, the y values on the parabola are negative. For example, the point (0, 6) on y = x2 x 6 would result in the point (0, | 6|) = (0, ( 6)) = (0, 6) on the graph of f (x) = |x2 x 6|. Proceeding in this manner for all points with x-coordinates between 2 and 3 results in the graph seen below on the right. y y 7 7 6 6 5 5 4 4 3 3 2 2 1 3 2 1 1 1 1 2 3 x 3 2 1 1 2 2 3 3 4 4 5 5 6 6 y = g(x) = x2 x 6 1 y = f (x) = |x2 2 3 x x 6| If we take a step back and look at the graphs of g and f in the last example, we notice that to obtain the graph of f from the graph of g, we reflect a portion of the graph of g about the x-axis. We can see this analytically by substituting g(x) = x2 x 6 into the formula for f (x) and calling to mind Theorem 1.4 from Section 1.7. g(x), if g(x) < 0 f (x) = g(x), if g(x) 0 The function f is defined so that when g(x) is negative (i.e., when its graph is below the x-axis), the graph of f is its refection across the x-axis. This is a general template to graph functions of the form f (x) = |g(x)|. From this perspective, the graph of f (x) = |x| can be obtained by reflecting the portion of the line g(x) = x which is below the x-axis back above the x-axis creating the characteristic '_' shape. 200 Linear and Quadratic Functions 2.3.1 Exercises In Exercises 1 - 9, graph the quadratic function. Find the x- and y-intercepts of each graph, if any exist. If it is given in general form, convert it into standard form; if it is given in standard form, convert it into general form. Find the domain and range of the function and list the intervals on which the function is increasing or decreasing. Identify the vertex and the axis of symmetry and determine whether the vertex yields a relative and absolute maximum or minimum. 1. f (x) = x2 + 2 4. f (x) = 2(x + 1)2 + 4 7. f (x) = x2 + x + 1 2. f (x) = (x + 2)2 5. f (x) = 2x2 8. f (x) = 4x 3. f (x) = x2 1 3x2 + 5x + 4 6. f (x) = 2x 8 3x2 + 4x 9.10 f (x) = x2 1 x 100 7 1 In Exercises 10 - 14, the cost and price-demand functions are given for dierent scenarios. For each scenario, Find the profit function P (x). Find the number of items which need to be sold in order to maximize profit. Find the maximum profit. Find the price to charge per item in order to maximize profit. Find and interpret break-even points. 10. The cost, in dollars, to produce x \"I'd rather be a Sasquatch\" T-Shirts is C(x) = 2x + 26, x 0 and the price-demand function, in dollars per shirt, is p(x) = 30 2x, 0 x 15. 11. The cost, in dollars, to produce x bottles of 100% All-Natural Certified Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x 0 and the price-demand function, in dollars per bottle, is p(x) = 35 x, 0 x 35. 12. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior's Lemonade Stand is C(x) = 18x + 240, x 0 and the price-demand function, in cents per cup, is p(x) = 90 3x, 0 x 30. 13. The daily cost, in dollars, to produce x Sasquatch Berry Pies is C(x) = 3x + 36, x the price-demand function, in dollars per pie, is p(x) = 12 0.5x, 0 x 24. 0 and 14. The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is C(x) = 20x + 1000, x 0 and the price-demand function, in hundreds of dollars per scooter, is p(x) = 140 2x, 0 x 70. 10 We have already seen the graph of this function. It was used as an example in Section 1.6 to show how the graphing calculator can be misleading. 2.3 Quadratic Functions 201 15. The International Silver Strings Submarine Band holds a bake sale each year to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking x cookies is C(x) = 0.1x + 25 and that the demand function for their cookies is p = 10 .01x. How many cookies should they bake in order to maximize their profit? 16. Using data from Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = 0.0076t2 + 0.45t + 16, 0 t 28, where t is the number of years since 1980. Find and interpret the coordinates of the vertex of the graph of y = F (t). 17. The temperature T , in degrees Fahrenheit, t hours after 6 AM is given by: T (t) = 1 2 t + 8t + 32, 2 0 t 12 What is the warmest temperature of the day? When does this happen? 18. Suppose C(x) = x2 10x + 27 represents the costs, in hundreds, to produce x thousand pens. How many pens should be produced to minimize the cost? What is this minimum cost? 19. Skippy wishes to plant a vegetable garden along one side of his house. In his garage, he found 32 linear feet of fencing. Since one side of the garden will border the house, Skippy doesn't need fencing along that side. What are the dimensions of the garden which will maximize the area of the garden? What is the maximum area of the garden? 20. In the situation of Example 2.3.4, Donnie has a nightmare that one of his alpaca herd fell into the river and drowned. To avoid this, he wants to move his rectangular pasture away from the river. This means that all four sides of the pasture require fencing. If the total amount of fencing available is still 200 linear feet, what dimensions maximize the area of the pasture now? What is the maximum area? Assuming an average alpaca requires 25 square feet of pasture, how many alpaca can he raise now? 21. What is the largest rectangular area one can enclose with 14 inches of string? 22. The height of an object dropped from the roof of an eight story building is modeled by h(t) = 16t2 + 64, 0 t 2. Here, h is the hei