Question 1

Find an equation for the ellipse described. Graph the equation.
-Vertices at $( 5 , - 4 )$ and $( 5,8 )$; length of minor axis is 6

Accepted Answer

A) 
 
$\frac { ( x - 5 ) ^ { 2 } } { 36 } + \frac { ( y - 2 ) ^ { 2 } } { 9 } = 1$ 
B) 
 
$$\frac { ( x + 5 ) ^ { 2 } } { 9 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$$ 
C) 
 
$\frac { ( x + 5 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$ 
D) 
 
$$\frac { ( x - 5 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$$ 
A) 
 
$\frac { ( x - 5 ) ^ { 2 } } { 36 } + \frac { ( y - 2 ) ^ { 2 } } { 9 } = 1$ 
B) 
 
$$\frac { ( x + 5 ) ^ { 2 } } { 9 } + \frac { ( y + 2 ) ^ { 2 } } { 36 } = 1$$ 
C) 
 
$\frac { ( x + 5 ) ^ { 2 } } { 36 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$ 
D) 
 
$$\frac { ( x - 5 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 36 } = 1$$

Question 2

Rotate the axes so that the new equation contains no xy-term. Graph the new equation.
-$31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$

Accepted Answer

A) 
  
B) 
  
A) 
  
B)

Question 3

Find the vertex, focus, and directrix of the parabola. Graph the equation.
-$(x-3)^{2}=(y-3)$

Accepted Answer

A)  vertex: $ (3,3) $
focus: $ (3,3.25) $
directrix: $ \mathrm{y}=2.75 $
  
B)  vertex: $ (-3,-3) $
focus: $ (-2.75,-3) $
directrix: $ x=-3.25 $

C) 
vertex: $ (3,3) $
focus: $ (3.25,3) $
directrix: $ x=2.75 $

D) 
vertex: $ (-3,-3) $
focus: $ (-3,-2.75) $
directrix: $ \mathrm{y}=-3.25 $

A)  vertex: $ (3,3) $
focus: $ (3,3.25) $
directrix: $ \mathrm{y}=2.75 $
  
B)  vertex: $ (-3,-3) $
focus: $ (-2.75,-3) $
directrix: $ x=-3.25 $

C) 
vertex: $ (3,3) $
focus: $ (3.25,3) $
directrix: $ x=2.75 $

D) 
vertex: $ (-3,-3) $
focus: $ (-3,-2.75) $
directrix: $ \mathrm{y}=-3.25 $

Question 4

Find the asymptotes of the hyperbola.
-$$\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$$

Accepted Answer

A)  $y - 1 = \frac { 5 } { 2 } ( x - 2 )$ and $y - 1 = - \frac { 5 } { 2 } ( x - 2 )$ 
B)  $y - 1 = \frac { 2 } { 5 } ( x - 2 )$ and $y - 1 = - \frac { 2 } { 5 } ( x - 2 )$ 
C)  $y - 2 = \frac { 5 } { 2 } ( x - 1 )$ and $y - 2 = - \frac { 5 } { 2 } ( x - 1 )$ 
D)  $y = \frac { 5 } { 2 } ( x - 2 )$ and $y = - \frac { 5 } { 2 } ( x - 2 )$ 
A)  $y - 1 = \frac { 5 } { 2 } ( x - 2 )$ and $y - 1 = - \frac { 5 } { 2 } ( x - 2 )$ 
B)  $y - 1 = \frac { 2 } { 5 } ( x - 2 )$ and $y - 1 = - \frac { 2 } { 5 } ( x - 2 )$ 
C)  $y - 2 = \frac { 5 } { 2 } ( x - 1 )$ and $y - 2 = - \frac { 5 } { 2 } ( x - 1 )$ 
D)  $y = \frac { 5 } { 2 } ( x - 2 )$ and $y = - \frac { 5 } { 2 } ( x - 2 )$

Question 5

Solve the problem.
-A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 24 inches across at its opening and is 4 feet deep, where will the light be concentrated?

Accepted Answer

A)  0.1 in. from the vertex 
B)  10.1 in. from the vertex 
C)  0.2 in. from the vertex 
D)  0.8 in. from the vertex 
A)  0.1 in. from the vertex 
B)  10.1 in. from the vertex 
C)  0.2 in. from the vertex 
D)  0.8 in. from the vertex

Question 6

Discuss the equation and graph it.
-$$\mathrm { r } = \frac { 3 } { 2 + 4 \sin \theta }$$

Accepted Answer

A)  ellipse, directrix parallel to the polar axis $ \frac{3}{2} $ unit below the pole vertices $ \left(\frac{3}{2}, \frac{\pi}{2}\right),\left(\frac{1}{2}, \frac{3 \pi}{2}\right) $
  
B) ellipse, directrix perpendicular to the polar axis $ \frac{3}{2} $ unit left of the pole vertices $ \left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, \pi\right) $
  
C)  hyperbola; directrix parallel to the polar axis $ \frac{3}{4} $ unit above the pole vertices $ \left(\frac{1}{2}, \frac{\pi}{2}\right),\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) $
  
D) hyperbola, directrix perpendicular to the polar axis $ \frac{3}{4} $ unit right of the pole vertices $ \left(\frac{1}{2}, 0\right),\left(-\frac{3}{2}, \pi\right) $
  
A)  ellipse, directrix parallel to the polar axis $ \frac{3}{2} $ unit below the pole vertices $ \left(\frac{3}{2}, \frac{\pi}{2}\right),\left(\frac{1}{2}, \frac{3 \pi}{2}\right) $
  
B) ellipse, directrix perpendicular to the polar axis $ \frac{3}{2} $ unit left of the pole vertices $ \left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, \pi\right) $
  
C)  hyperbola; directrix parallel to the polar axis $ \frac{3}{4} $ unit above the pole vertices $ \left(\frac{1}{2}, \frac{\pi}{2}\right),\left(-\frac{3}{2}, \frac{3 \pi}{2}\right) $
  
D) hyperbola, directrix perpendicular to the polar axis $ \frac{3}{4} $ unit right of the pole vertices $ \left(\frac{1}{2}, 0\right),\left(-\frac{3}{2}, \pi\right) $

Question 7

Determine the appropriate rotation formulas to use so that the new equation contains no xy-term.
-$$3 x ^ { 2 } + 5 x y + 3 y ^ { 2 } - 8 x + 8 y = 0$$

Accepted Answer

A)  $x = \frac { 1 } { 2 } x ^ { \prime } - \frac { \sqrt { 3 } } { 2 } y ^ { \prime }$ and $y = \frac { \sqrt { 3 } } { 2 } x ^ { \prime } + \frac { 1 } { 2 } y ^ { \prime }$ 
B)  $x = \frac { \sqrt { 2 + \sqrt { 2 } } } { 2 } x ^ { \prime } - \frac { \sqrt { 2 - \sqrt { 2 } } } { 2 } y ^ { \prime }$ and $y = \frac { \sqrt { 2 - \sqrt { 2 } } } { 2 } x ^ { \prime } + \frac { \sqrt { 2 + \sqrt { 2 } } } { 2 } y ^ { \prime }$ 
C)  $x = \frac { \sqrt { 2 } } { 2 } \left( x ^ { \prime } - y ^ { \prime } \right)$ and $y = \frac { \sqrt { 2 } } { 2 } \left( x ^ { \prime } + y ^ { \prime } \right)$ 
D)  $x = - y ^ { \prime }$ and $y = x ^ { \prime }$ 
A)  $x = \frac { 1 } { 2 } x ^ { \prime } - \frac { \sqrt { 3 } } { 2 } y ^ { \prime }$ and $y = \frac { \sqrt { 3 } } { 2 } x ^ { \prime } + \frac { 1 } { 2 } y ^ { \prime }$ 
B)  $x = \frac { \sqrt { 2 + \sqrt { 2 } } } { 2 } x ^ { \prime } - \frac { \sqrt { 2 - \sqrt { 2 } } } { 2 } y ^ { \prime }$ and $y = \frac { \sqrt { 2 - \sqrt { 2 } } } { 2 } x ^ { \prime } + \frac { \sqrt { 2 + \sqrt { 2 } } } { 2 } y ^ { \prime }$ 
C)  $x = \frac { \sqrt { 2 } } { 2 } \left( x ^ { \prime } - y ^ { \prime } \right)$ and $y = \frac { \sqrt { 2 } } { 2 } \left( x ^ { \prime } + y ^ { \prime } \right)$ 
D)  $x = - y ^ { \prime }$ and $y = x ^ { \prime }$

Question 8

Convert the polar equation to a rectangular equation.
-$$\mathrm { r } = \frac { 2 } { 2 - 2 \cos \theta }$$

Accepted Answer

A)  $y ^ { 2 } = - 4 x + 4$ 
B)  $x ^ { 2 } = - 4 y + 4$ 
C)  $y ^ { 2 } = 4 x + 4$ 
D)  $x ^ { 2 } = 4 y + 4$ 
A)  $y ^ { 2 } = - 4 x + 4$ 
B)  $x ^ { 2 } = - 4 y + 4$ 
C)  $y ^ { 2 } = 4 x + 4$ 
D)  $x ^ { 2 } = 4 y + 4$

Question 9

Graph the equation.
-$$\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 10

Graph the hyperbola.
-$(y+1)^{2}-9(x-2)^{2}=9$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 11

Find a polar equation for the conic. A focus is at the pole.
-$e = 1$; directrix is parallel to the polar axis 1 above the pole

Accepted Answer

A)  $r = \frac { 3 } { 1 + \cos \theta }$ 
B)  $r = \frac { 3 } { 1 + \sin \theta }$ 
C)  $r = \frac { 3 } { 1 - \cos \theta }$ 
D)  $r = \frac { 3 } { 1 - \sin \theta }$ 
A)  $r = \frac { 3 } { 1 + \cos \theta }$ 
B)  $r = \frac { 3 } { 1 + \sin \theta }$ 
C)  $r = \frac { 3 } { 1 - \cos \theta }$ 
D)  $r = \frac { 3 } { 1 - \sin \theta }$

Question 12

Match the equation to the graph.
-$$( x - 2 ) ^ { 2 } = 7 ( y - 1 )$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 13

Find an equation for the ellipse described. Graph the equation.
-Foci at $( 3 , - 1 )$ and $( 3 , - 7 )$; length of major axis is 10

Accepted Answer

A) 
 
$\frac{(y+4)^{2}}{25}+\frac{(x+3)^{2}}{16}=1$ 
B)

$\frac{(x+4)^{2}}{16}+\frac{(y+3)^{2}}{25}=1$ 
C) 
  
D) 
 
 $$\frac { ( x + 4 ) ^ { 2 } } { 25 } + \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 \quad \frac { ( y + 4 ) ^ { 2 } } { 25 } + \frac { ( x - 3 ) ^ { 2 } } { 16 } = 1$$ 
A) 
 
$\frac{(y+4)^{2}}{25}+\frac{(x+3)^{2}}{16}=1$ 
B)

$\frac{(x+4)^{2}}{16}+\frac{(y+3)^{2}}{25}=1$ 
C) 
  
D) 
 
 $$\frac { ( x + 4 ) ^ { 2 } } { 25 } + \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 \quad \frac { ( y + 4 ) ^ { 2 } } { 25 } + \frac { ( x - 3 ) ^ { 2 } } { 16 } = 1$$

Question 14

Solve the problem.
-A satellite following the hyperbolic path shown in the picture turns rapidly at $( 0,4 )$ and then moves closer and closer to the line $y = \frac { 8 } { 5 } x$ as it gets farther from the tracking station at the origin. Find the equation that describes the path of the rocket if the center of the hyperbola is at $( 0,0 )$.

Accepted Answer

A)  $\frac { y ^ { 2 } } { \frac { 25 } { 4 } } - \frac { x ^ { 2 } } { 16 } = 1$ 
B)  $\frac { x ^ { 2 } } { \left( \frac { 12 } { 5 } \right) ^ { 2 } } - \frac { y ^ { 2 } } { 9 } = 1$ 
C)  $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { \left( \frac { 16 } { 5 } \right) ^ { 2 } } = 1$ 
D)  $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { \frac { 25 } { 4 } } = 1$ 
A)  $\frac { y ^ { 2 } } { \frac { 25 } { 4 } } - \frac { x ^ { 2 } } { 16 } = 1$ 
B)  $\frac { x ^ { 2 } } { \left( \frac { 12 } { 5 } \right) ^ { 2 } } - \frac { y ^ { 2 } } { 9 } = 1$ 
C)  $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { \left( \frac { 16 } { 5 } \right) ^ { 2 } } = 1$ 
D)  $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { \frac { 25 } { 4 } } = 1$

Question 15

Match the equation to its graph.
-$$x ^ { 2 } = 7 y$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 16

Find a rectangular equation for the plane curve defined by the parametric equations.
-$$x = 2 t - 1 , y = t ^ { 2 } + 4 ; - 4 \leq t \leq 4$$

Accepted Answer

A)  $y = - \frac { 1 } { 2 } x + 30$; for $x$ in $- 6 \leq x \leq 4$ 
B)  $y = \frac { 1 } { 2 } x ^ { 2 } + 1$; for $x$ in $- 6 \leq x \leq 4$ 
C)  $y = x ^ { 2 } + 1 ;$ for $x$ in $- 2 \leq x \leq 2$ 
D)  $y = \frac { 1 } { 4 } x ^ { 2 } + \frac { 1 } { 2 } x + \frac { 17 } { 4 } ;$ for $x$ in $- 9 \leq x \leq 7$ 
A)  $y = - \frac { 1 } { 2 } x + 30$; for $x$ in $- 6 \leq x \leq 4$ 
B)  $y = \frac { 1 } { 2 } x ^ { 2 } + 1$; for $x$ in $- 6 \leq x \leq 4$ 
C)  $y = x ^ { 2 } + 1 ;$ for $x$ in $- 2 \leq x \leq 2$ 
D)  $y = \frac { 1 } { 4 } x ^ { 2 } + \frac { 1 } { 2 } x + \frac { 17 } { 4 } ;$ for $x$ in $- 9 \leq x \leq 7$

Question 17

Graph the curve whose parametric equations are given.
-$x=2 t-1, y=t^{2}+5 ;-4 \leq t \leq 4$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Question 18

Find the center, foci, and vertices of the ellipse.
-$$4 x ^ { 2 } + 64 y ^ { 2 } = 256$$

Accepted Answer

A) center at $ (0,0) $
foci at $ (0,-2) $ and $ (0,2) $
vertices at $ (0,-4),(0,4) $ 
B) center at $ (0,0) $
foci at $ (0,-2 \sqrt{15}) $ and $ (0,2 \sqrt{15}) $
vertices at $ (0,-8),(0,8) $ 
C)  center at $ (0,0) $
foci at $ (-2 \sqrt{15}, 0) $ and $ (2 \sqrt{15}, 0) $
vertices at $ (-8,0),(8,0) $ 
D) center at $ (0,0) $
foci at $ (-8,0) $ and $ (8,0) $
vertices at $ (-64,0),(64,0) $ 
A) center at $ (0,0) $
foci at $ (0,-2) $ and $ (0,2) $
vertices at $ (0,-4),(0,4) $ 
B) center at $ (0,0) $
foci at $ (0,-2 \sqrt{15}) $ and $ (0,2 \sqrt{15}) $
vertices at $ (0,-8),(0,8) $ 
C)  center at $ (0,0) $
foci at $ (-2 \sqrt{15}, 0) $ and $ (2 \sqrt{15}, 0) $
vertices at $ (-8,0),(8,0) $ 
D) center at $ (0,0) $
foci at $ (-8,0) $ and $ (8,0) $
vertices at $ (-64,0),(64,0) $

Question 19

Rotate the axes so that the new equation contains no xy-term. Graph the new equation.
-$17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0$

Accepted Answer

A) 
  
B) 
  
A) 
  
B)

Question 20

Choose the one alternative that best completes the statement or answers the question.
Match the equation to the graph.
-$$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
A) 
  
B) 
  
C) 
  
D)

Find an equation for the ellipse described. Graph the equation. -Vertices at $( 5 , - 4 )$ and $( 5,8 )$ ; length of minor axis is 6

Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - $31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$ $Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$

Find the vertex, focus, and directrix of the parabola. Graph the equation. - $(x-3)^{2}=(y-3)$ $Find the vertex, focus, and directrix of the parabola. Graph the equation. - (x-3)^{2}=(y-3)$

Find the asymptotes of the hyperbola. - $\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$

Solve the problem. -A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 24 inches across at its opening and is 4 feet deep, where will the light be concentrated?

Discuss the equation and graph it. - $\mathrm { r } = \frac { 3 } { 2 + 4 \sin \theta }$ $Discuss the equation and graph it. - \mathrm { r } = \frac { 3 } { 2 + 4 \sin \theta }$

Determine the appropriate rotation formulas to use so that the new equation contains no xy-term. - $3 x ^ { 2 } + 5 x y + 3 y ^ { 2 } - 8 x + 8 y = 0$

Convert the polar equation to a rectangular equation. - $\mathrm { r } = \frac { 2 } { 2 - 2 \cos \theta }$

Graph the equation. - $\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$

Graph the hyperbola. - $(y+1)^{2}-9(x-2)^{2}=9$ $Graph the hyperbola. - (y+1)^{2}-9(x-2)^{2}=9$

Find a polar equation for the conic. A focus is at the pole. - $e = 1$ ; directrix is parallel to the polar axis 1 above the pole

Match the equation to the graph. - $( x - 2 ) ^ { 2 } = 7 ( y - 1 )$

Find an equation for the ellipse described. Graph the equation. -Foci at $( 3 , - 1 )$ and $( 3 , - 7 )$ ; length of major axis is 10

Match the equation to its graph. - $x ^ { 2 } = 7 y$

Find a rectangular equation for the plane curve defined by the parametric equations. - $x = 2 t - 1 , y = t ^ { 2 } + 4 ; - 4 \leq t \leq 4$

Graph the curve whose parametric equations are given. - $x=2 t-1, y=t^{2}+5 ;-4 \leq t \leq 4$ $Graph the curve whose parametric equations are given. - x=2 t-1, y=t^{2}+5 ;-4 \leq t \leq 4$

Find the center, foci, and vertices of the ellipse. - $4 x ^ { 2 } + 64 y ^ { 2 } = 256$

Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - $17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0$ $Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0$

Choose the one alternative that best completes the statement or answers the question. Match the equation to the graph. - $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$

Functions and Their Graphs

Linear and Quadratic Functions

Polynomial and Rational Functions

Exponential and Logarithmic Functions

Trigonometric Functions

Analytic Trigonometry

Applications of Trigonometric Functions

Polar Coordinates; Vectors

Systems of Equations and Inequalities

Sequences; Induction; the Binomial Theorem

Counting and Probability

A Preview of Calculus: the Limit, Derivative, and Integral of a Function

Foundations: a Prelude to Functions

Graphing Utilities

Filters

Exam 9: Analytic Geometry

Find an equation for the ellipse described. Graph the equation. -Vertices at (5,−4)( 5 , - 4 )(5,−4) and (5,8)( 5,8 )(5,8) ; length of minor axis is 6

Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 31x2+103xy+21y2−144=031 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=031x2+103​xy+21y2−144=0

Find the vertex, focus, and directrix of the parabola. Graph the equation. - (x−3)2=(y−3)(x-3)^{2}=(y-3)(x−3)2=(y−3)

Find the asymptotes of the hyperbola. - (x−2)24−(y−1)225=1\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 25 } = 14(x−2)2​−25(y−1)2​=1

Solve the problem. -A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 24 inches across at its opening and is 4 feet deep, where will the light be concentrated?

Discuss the equation and graph it. - r=32+4sin⁡θ\mathrm { r } = \frac { 3 } { 2 + 4 \sin \theta }r=2+4sinθ3​

Determine the appropriate rotation formulas to use so that the new equation contains no xy-term. - 3x2+5xy+3y2−8x+8y=03 x ^ { 2 } + 5 x y + 3 y ^ { 2 } - 8 x + 8 y = 03x2+5xy+3y2−8x+8y=0

Convert the polar equation to a rectangular equation. - r=22−2cos⁡θ\mathrm { r } = \frac { 2 } { 2 - 2 \cos \theta }r=2−2cosθ2​

Graph the equation. - (x+1)29+(y−2)216=1\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 19(x+1)2​+16(y−2)2​=1

Graph the hyperbola. - (y+1)2−9(x−2)2=9(y+1)^{2}-9(x-2)^{2}=9(y+1)2−9(x−2)2=9

Find a polar equation for the conic. A focus is at the pole. - e=1e = 1e=1 ; directrix is parallel to the polar axis 1 above the pole

Match the equation to the graph. - (x−2)2=7(y−1)( x - 2 ) ^ { 2 } = 7 ( y - 1 )(x−2)2=7(y−1)

Find an equation for the ellipse described. Graph the equation. -Foci at (3,−1)( 3 , - 1 )(3,−1) and (3,−7)( 3 , - 7 )(3,−7) ; length of major axis is 10

Match the equation to its graph. - x2=7yx ^ { 2 } = 7 yx2=7y

Find a rectangular equation for the plane curve defined by the parametric equations. - x=2t−1,y=t2+4;−4≤t≤4x = 2 t - 1 , y = t ^ { 2 } + 4 ; - 4 \leq t \leq 4x=2t−1,y=t2+4;−4≤t≤4

Graph the curve whose parametric equations are given. - x=2t−1,y=t2+5;−4≤t≤4x=2 t-1, y=t^{2}+5 ;-4 \leq t \leq 4x=2t−1,y=t2+5;−4≤t≤4

Find the center, foci, and vertices of the ellipse. - 4x2+64y2=2564 x ^ { 2 } + 64 y ^ { 2 } = 2564x2+64y2=256

Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17x2−12xy+8y2−68x+24y−12=017 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=017x2−12xy+8y2−68x+24y−12=0

Choose the one alternative that best completes the statement or answers the question. Match the equation to the graph. - x216−y24=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 116x2​−4y2​=1

Functions and Their Graphs

Linear and Quadratic Functions

Polynomial and Rational Functions

Exponential and Logarithmic Functions

Trigonometric Functions

Analytic Trigonometry

Applications of Trigonometric Functions

Polar Coordinates; Vectors

Systems of Equations and Inequalities

Sequences; Induction; the Binomial Theorem

Counting and Probability

A Preview of Calculus: the Limit, Derivative, and Integral of a Function

Foundations: a Prelude to Functions

Graphing Utilities

Filters

Find an equation for the ellipse described. Graph the equation. -Vertices at $( 5 , - 4 )$ and $( 5,8 )$ ; length of minor axis is 6

Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - $31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$ $Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 31 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0$

Find the vertex, focus, and directrix of the parabola. Graph the equation. - $(x-3)^{2}=(y-3)$ $Find the vertex, focus, and directrix of the parabola. Graph the equation. - (x-3)^{2}=(y-3)$

Find the asymptotes of the hyperbola. - $\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$

Discuss the equation and graph it. - $\mathrm { r } = \frac { 3 } { 2 + 4 \sin \theta }$ $Discuss the equation and graph it. - \mathrm { r } = \frac { 3 } { 2 + 4 \sin \theta }$

Determine the appropriate rotation formulas to use so that the new equation contains no xy-term. - $3 x ^ { 2 } + 5 x y + 3 y ^ { 2 } - 8 x + 8 y = 0$

Convert the polar equation to a rectangular equation. - $\mathrm { r } = \frac { 2 } { 2 - 2 \cos \theta }$

Graph the equation. - $\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 2 ) ^ { 2 } } { 16 } = 1$

Graph the hyperbola. - $(y+1)^{2}-9(x-2)^{2}=9$ $Graph the hyperbola. - (y+1)^{2}-9(x-2)^{2}=9$

Find a polar equation for the conic. A focus is at the pole. - $e = 1$ ; directrix is parallel to the polar axis 1 above the pole

Match the equation to the graph. - $( x - 2 ) ^ { 2 } = 7 ( y - 1 )$

Find an equation for the ellipse described. Graph the equation. -Foci at $( 3 , - 1 )$ and $( 3 , - 7 )$ ; length of major axis is 10

Match the equation to its graph. - $x ^ { 2 } = 7 y$

Find a rectangular equation for the plane curve defined by the parametric equations. - $x = 2 t - 1 , y = t ^ { 2 } + 4 ; - 4 \leq t \leq 4$

Graph the curve whose parametric equations are given. - $x=2 t-1, y=t^{2}+5 ;-4 \leq t \leq 4$ $Graph the curve whose parametric equations are given. - x=2 t-1, y=t^{2}+5 ;-4 \leq t \leq 4$

Find the center, foci, and vertices of the ellipse. - $4 x ^ { 2 } + 64 y ^ { 2 } = 256$

Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - $17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0$ $Rotate the axes so that the new equation contains no xy-term. Graph the new equation. - 17 x^{2}-12 x y+8 y^{2}-68 x+24 y-12=0$

Choose the one alternative that best completes the statement or answers the question. Match the equation to the graph. - $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$