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mathematics
precalculus

Precalculus 9th edition Michael Sullivan - Solutions

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 at its base. (a) Find the equation of a parabola with the same dimensions. Let x equal the
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 at its center.
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 the height of the arch at distances of 10, 30, and 50 feet from the center. 125 ft 120 ft
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 opening and is 6 feet deep, where will the heat source be concentrated? Sun's rays 20 6'
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 width of the opening be?
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 searchlight be?
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 towers, what is the height of the cable at a point 50 feet from the center of the bridge?
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 midway between the towers, what is the height of the cable from the road at a point 150 feet from the
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 headlight at its opening?
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 will be reflected parallel to the axis?
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 located. If the dish is 10 feet across at its opening and 4 feet deep at its center, at what position
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 - 1)(H) (x + 1)2 = -4(y + 1) У -3 х (-1, –1) 2 2.
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 - 1)(H) (x + 1)2 = -4(y + 1) 2 x -2 (-1, –2) -2F
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 - 1)(H) (x + 1)2 = -4(y + 1) (1, 2) -2 2 X -2-
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 - 1)(H) (x + 1)2 = -4(y + 1) У (-1, -1) -2 2 х -2
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 - 1)(H) (x + 1)2 = -4(y + 1) УА -2 2 х (-2, -1) т: 2.
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 - 1)(H) (x + 1)2 = -4(y + 1) УА 2 (1, 1) -2 -2
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 - 1)(H) (x + 1)2 = -4(y + 1) (1, 1) -2
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 - 1)(H) (x + 1)2 = -4(y + 1) У л (2, 1) -2 2 х -2F 2.
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 will be free hanging so that the ends of the cable will make equal angles with the ceiling. If the
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.
Decompose v into its vertical and horizontal components.P1 = (3√2, 7√2) and P2 = (8√2, 2√2)
P1 = (3√2, 7√2) and P2 = (8√2, 2√2)Find the angle between v and i.
P1 = (3√2, 7√2) and P2 = (8√2, 2√2)Find the unit vector in the direction of v.
P1 = (3√2, 7√2) and P2 = (8√2, 2√2)Find |v|.
P1 = (3√2, 7√2) and P2 = (8√2, 2√2)Find the position vector equal to P1P2(vector).

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