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mathematics
precalculus

Questions and Answers of Precalculus

The distance d from P1 = (2, -5) to P2 = (4, -2) is d = ______.
Show that the graph of an equation of the form Cy2 + Dx + Ey + F = 0, C ≠ 0,(a) Is parabola if D ≠ 0.(b) Is a vertical line if D = 0 and E2 - 4CF = 0.(c) Is two vertical lines if
Show that the graph of an equation of the form Ax2 + Dx + Ey + F = 0, A ≠ 0,(a) Is parabola if E ≠ 0.(b) Is a vertical line if E = 0 and D2 - 4AF = 0.(c) Is two vertical lines if
Show that an equation of the form Cy2 + Dx = 0, C ≠ 0, D ≠ 0is the equation of a parabola with vertex at and axis of symmetry the x-axis. Find its focus and directrix.
Show that an equation of the form Ax2 + Ey = 0 A ≠ 0, E ≠ 0is the equation of a parabola with vertex at and axis of symmetry the y-axis. Find its focus and directrix.
The Gateway Arch in St. Louis is often mistaken to be parabolic in shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 625 feet high and 598 feet wide
A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10 feet. Find the height of the arch
A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a maximum height of 25 feet. See the illustration. Choose a suitable rectangular coordinate system and find
A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 4 inches across at its opening and is 3 inches deep, where will the collected light be concentrated?
A mirror is shaped like a paraboloid of revolution and will be used to concentrate the rays of the sun at its focus, creating a heat source. See the figure. If the mirror is 20 feet across at its
A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the depth of the searchlight is 4 feet, what should the
A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, how deep should the
The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 400 feet apart and 100 feet high. If the cables are at a height of 10 feet midway between the
The cables of a suspension bridge are in the shape of a parabola, as shown in the figure. The towers supporting the cable are 600 feet apart and 80 feet high. If the cables touch the road surface
A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is 1 inch from the vertex. If the depth is to be 2 inches, what is the diameter of the
The reflector of a flashlight is in the shape of a paraboloid of revolution. Its diameter is 4 inches and its depth is 1 inch. How far from the vertex should the light bulb be placed so that the rays
A cable TV receiving dish is in the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 6 feet across at its opening and 2 feet deep.
A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is
Write an equation for each parabola. 2 |(0, 1) (1, 0) 2 х -2 -2F
Write an equation for each parabola. y. 2 (0, 1), (-2, 0) 2 х -2
Write an equation for each parabola. 2 (0, 1) 2 x -2 (1, –1) -1) -2F
Write an equation for each parabola. УА 2 (2, 2) (0, 1) -2 х -2
Write an equation for each parabola. Уд 2 (2, 0) х (0, –1) -2F
Write an equation for each parabola. 2 (2, 1) -2 (1, 0) х -2
Write an equation for each parabola. УА (1, 2) (2, 1) 2 -2 х -2- 2.
Write an equation for each parabola. У 2 (1, 2) (0, 1) -2 х -2Н 2.
Find the vertex, focus, and directrix of each parabola. Graph the equation. y2 + 12y = -x + 1
Find the vertex, focus, and directrix of each parabola. Graph the equation. x2 – 4x = y + 4
Find the vertex, focus, and directrix of each parabola. Graph the equation. x2 – 4x = 2y
Find the vertex, focus, and directrix of each parabola. Graph the equation. y2 + 2y – x = 0
Find the vertex, focus, and directrix of each parabola. Graph the equation. y2 – 2y = 8x - 1
Find the vertex, focus, and directrix of each parabola. Graph the equation. x2 + 8x = 4y - 8
Find the vertex, focus, and directrix of each parabola. Graph the equation. x2 + 6x – 4y + 1 = 0
Find the vertex, focus, and directrix of each parabola. Graph the equation. y2 – 4y + 4x + 4 = 0
Find the vertex, focus, and directrix of each parabola. Graph the equation. (x – 2)2 = 4(y – 3)
Find the vertex, focus, and directrix of each parabola. Graph the equation. ( y + 3)2 = 8(x – 2)
Find the vertex, focus, and directrix of each parabola. Graph the equation. (y + 1)2 = -4(x – 2)
Find the vertex, focus, and directrix of each parabola. Graph the equation. ( x- 3)2 = -(y + 1)
Find the vertex, focus, and directrix of each parabola. Graph the equation. (x + 4)2 = 16(y + 2)
Find the vertex, focus, and directrix of each parabola. Graph the equation. (y – 2)2 = 8(x + 1)
Find the vertex, focus, and directrix of each parabola. Graph the equation. x2 = -4y
Find the vertex, focus, and directrix of each parabola. Graph the equation. y2 = -16x
Find the vertex, focus, and directrix of each parabola. Graph the equation. y2 = 8x
Find the vertex, focus, and directrix of each parabola. Graph the equation. x2 = 4y
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (-4, 4); directrix the line y = -2.
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (-3, -2); directrix the line x = 1.
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (2, 4); directrix the line y = -4.
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (-3, 4); directrix the line y = 2.
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (3, 0); vertex at (3, -2)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (-1, -2); vertex at (0, -2)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (4, -2); vertex at (6, -2)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (2, -3); vertex at (2, -5)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Vertex at (0, 0) axis of symmetry the x-axis; containing the point (2, 3)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Vertex at (0, 0); axis of symmetry the y-axis; containing the point (2, 3)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Directrix the line x = -1/2; vertex at (0, 0)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Directrix the line y = -1/2; vertex at (0, 0)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (0, -1); directix the line y = 1
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (-2, 0); directix the line x =2
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (-4, 0); vertex at (0, 0)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (0, -3); vertex at (0, 0)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (0, 2); vertex at (0, 0)
Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.Focus at (4, 0); vertex at (0, 0)
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
The graph of a parabola is given. Match each graph to its equation.(A) y2 = 4x(B) x2 = 4y(C) y2 = -4x(D) x2 = -4y(E) (y - 1)2 = 4(x – 1)(F) (x + 1)2 = 4(y + 1)(G) (y - 1)2 = -4(x -
If a = 4 then the equation of the directrix is _______. У+ V= (3, 2) х
If a = 4 then the coordinates of the focus are ______. У+ V= (3, 2) х
The coordinates of the vertex are _______. У+ V= (3, 2) х
If a > 0 the equation of the parabola is of the form (a) (y – k)2 = 4a(x – h)(b) (y – k)2 = - 4a(x – h)(c) (x – h)2 = 4a(y – k)(d) (x – h)2 = - 4a(y – k) У+ V= (3, 2) х
A(n) ________ is the collection of all points in the plane such that the distance from each point to a fixed point equals its distance to a fixed line
To graph y = ( x – 3)2 + 1 shift the graph of y = x2 to the right ______ units and then 1unit.
The point that is symmetric with respect to the x-axis to the point (-2,5) is _____.
Use the Square Root Method to find the real solutions of (x + 4)2 = 9.
To complete the square of x2 – 4x add ________.
The formula for the distance d from P1 = (x1, y1) is P2 = (x2, y2) is d ______.
What is the amplitude and period of y = -4 cos(πx).
Graph the equations r = 2 and θ = π/3 on the same set of polar coordinates.
Graph the equations x = 3 and y = 4 on the same set of rectangular coordinates.
Graph the function y = sin-1(-1/2).
Graph the function y = sin|x|.
Graph the function y = |sinx|.
Graph the function y = |ln x|.
Test the equation x2 + y3 = 2x4 for symmetry with respect to the x-axis, the y-axis, and the origin.
What is the domain of the function f(x) = ln(1 -2x)?
Find an equation for the circle with center at the point (0,1) and radius 3. Graph this circle.
Find an equation for the line containing the origin that makes an angle of 30° with the positive x-axis.
Find the real solutions, if any, of the equation -9 = 1
A 1200-pound chandelier is to be suspended over a large ballroom; the chandelier will be hung on two cables of equal length whose ends will be attached to the ceiling, 16 feet apart. The chandelier
Use the vectors u = 2i - 3j + k and v = -i + 3j + 2k.Find the area of the parallelogram that has and as adjacent sides.
Use the vectors u = 2i - 3j + k and v = -i + 3j + 2k.Find the direction angles for u.
Use the vectors u = 2i - 3j + k and v = -i + 3j + 2k.Find u x v.
v1 = (4,6), v2 = (-3,-6), v3 = (-8,4), v4 = (10,15)Find the angle between vectors v1 and v2.
v1 = (4,6), v2 = (-3,-6), v3 = (-8,4), v4 = (10,15)Which two vectors are orthogonal?
v1 = (4,6), v2 = (-3,-6), v3 = (-8,4), v4 = (10,15)Which two vectors are parallel?
v1 = (4,6), v2 = (-3,-6), v3 = (-8,4), v4 = (10,15)Find the vector v1 + 2v2 – v3.

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