Faith Marsh Mr. Ishiu MTH 151 3 November 2016 Writing Assignment: Problem 1. Let f(x) = x2. (i) For every a, nd an equation of the tangent line of y = f(x) at x = a. First, if x=a, then f(x) = f(a), and, by substitution into the original problem, f (a)= a2 . Then, the derivative can be found using power rule: f ' (a)=2 a . Substituting f ' (a) for slope and (a, a2) for the (x,y) value into the point-slope formula, the equation for the tangent line can be found: ya2=2 a( xa) 2 2 y=2 ax2 a + a y=2 axa2 Thus the equation of the tangent line of y=f(x) at x=a is y=2 axa 2 . (ii) Show that there exists a tangent line of y = f(x) that passes through(1, 8). (Hint: Find a such that the tangent line of y = f(x) at a passes through (1, 8)) To find the tangent line of y = f(x) that passes through (1,-8), thus proving its existance, you must substitute (1,-8) into the equation for the tangent line (see part i). 2 y=2 axa (8)=2 a(1)a 2 8=2 aa2 2 0=a + 2a+ 8 0=(a4)( a+2) a=4a=2 In this problem there are two possible answers, both of which are correct (due to the fact that f(x)=x^2 is a parabola--see graph on left). In order to find the tangent lines, substitute the value of a into the original equation for the tangent line of y=f(x) through x=a: or 2 2 2 4 y=2(2) x y=2( 4) x y=8 x16 y=4 x4 Thus, there does exist a (or rather, two) tangent line(s) of y=f(x) that passes through (1,8). (iii) Show that there exists no tangent line of y = f(x) that passes through(1, 8). (Hint: Try to do the same as (ii) and observe why it fails). In order to prove the NON-existence of a tangent line of y=f(x) that passes through (1,8), you must attempt to prove it in order to show its deficiency. So similarly to part ii, you must first substitute (1,8) into the equation for the tangent line (see part i). 2 Because for a. y=2 axa (8)=2 a(1)a2 8=2aa 2 2 0=a + 2a8 2 a +2 a8 is not factorable, you must use the quadratic equation to solve a= ( 2) a= a= 2 2 2 a=i * *technically, this step would be negative sign. a=i , but the cancels out the effect of the Because a is an imaginary number, there is no value for which a exists for (1,8), and thus, there is no tangent line for y=f(x) that passes through (1,8) (iv) Show that for every point (x0, y0) in the plane, there exists a tangent line of y = f(x) that passes through (x0, y0) if and only if y0 < x02 y=2 axa2 using (x0, y0). However, this is impossible without more information because 0=y 0 +2 axa2 is not in the form of 0=a x2 +bx +c , and thus the quadratic equation cannot be used, and a cannot be solved for. In order to prove this, you must be able solve the equation If y 0 < x 0 , then the slope of the equation must be less that 0. We already know that the slope of f(x)=x^2 is equal to 2a (see i) when the equation of the tangent line is equal to 2 y=2 axa . I'm pretty sure both of these are wrong. I was just at a loss as to what I could do