Exam 7: Conic Sections

Use the center, vertices, and asymptotes to graph the hyperbola. - $( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4$ $Use the center, vertices, and asymptotes to graph the hyperbola. - ( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4$

(Multiple Choice)

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Question 21

Graph the parabola. - $x ^ { 2 } = 6 y$ $Graph the parabola. - x ^ { 2 } = 6 y$

(Multiple Choice)

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Question 22

Find the solution set for the system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. - $\left\{ \begin{array} { l } \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1 \\y = 3\end{array} \right.$ $Find the solution set for the system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. - \left\{ \begin{array} { l } \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1 \\ y = 3 \end{array} \right.$

(Multiple Choice)

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Question 23

Convert the equation to the standard form for a hyperbola by completing the square on x and y. - $x ^ { 2 } - y ^ { 2 } - 4 x + 4 y - 1 = 0$

(Multiple Choice)

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Question 24

Graph the ellipse and locate the foci. - $4 x ^ { 2 } = 36 - 9 y ^ { 2 }$ $Graph the ellipse and locate the foci. - 4 x ^ { 2 } = 36 - 9 y ^ { 2 }$

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Question 25

Graph Hyperbolas Centered at the Origin Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes. - $4 x^{2}-9 y^{2}=36$ $Graph Hyperbolas Centered at the Origin Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes. - 4 x^{2}-9 y^{2}=36$

(Multiple Choice)

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Question 26

Additional Concepts Use the relation's graph to determine its domain and range. - $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$ $Additional Concepts Use the relation's graph to determine its domain and range. - \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$

(Multiple Choice)

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Question 27

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - $( y - 4 ) ^ { 2 } = - 16 ( x - 2 )$

(Multiple Choice)

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Question 28

Additional Concepts Use the relation's graph to determine its domain and range. - $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$ $Additional Concepts Use the relation's graph to determine its domain and range. - \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$

(Multiple Choice)

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Question 29

Write Equations of Ellipses in Standard Form - Center at $( - 1,1 )$

(Multiple Choice)

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Question 30

Find the standard form of the equation of the hyperbola. -

(Multiple Choice)

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Question 31

Graph Ellipses Not Centered at the Origin - $16 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 64$ $Graph Ellipses Not Centered at the Origin - 16 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 64$

(Multiple Choice)

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Question 32

Graph Parabolas with Vertices at the Origin Find the focus and directrix of the parabola with the given equation. - $x ^ { 2 } = - 8 y$

(Multiple Choice)

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Question 33

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - $( x + 4 ) ^ { 2 } = - 4 ( y + 1 )$

(Multiple Choice)

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Question 34

Graph Ellipses Not Centered at the Origin - $\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$ $Graph Ellipses Not Centered at the Origin - \frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$

(Multiple Choice)

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Question 35

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. - $x = - ( y + 1 ) ^ { 2 } + 8$

(Multiple Choice)

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Question 36

Find the standard form of the equation of the ellipse satisfying the given conditions. -Foci: $( - 3,0 ) , ( 3,0 )$ ; vertices: $( - 4,0 ) , ( 4,0 )$

(Multiple Choice)

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Question 37

Find the foci of the ellipse whose equation is given. - $36 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 900$

(Multiple Choice)

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Question 38

Graph the parabola. - $y ^ { 2 } = - 6 x$ $Graph the parabola. - y ^ { 2 } = - 6 x$

(Multiple Choice)

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Question 39

Graph the parabola. - $y ^ { 2 } + 12 x = 0$ $Graph the parabola. - y ^ { 2 } + 12 x = 0$

(Multiple Choice)

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Question 40

Showing 21 - 40 of 120

Use the center, vertices, and asymptotes to graph the hyperbola. - $( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4$ $Use the center, vertices, and asymptotes to graph the hyperbola. - ( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4$

Graph the parabola. - $x ^ { 2 } = 6 y$ $Graph the parabola. - x ^ { 2 } = 6 y$

Convert the equation to the standard form for a hyperbola by completing the square on x and y. - $x ^ { 2 } - y ^ { 2 } - 4 x + 4 y - 1 = 0$

Graph the ellipse and locate the foci. - $4 x ^ { 2 } = 36 - 9 y ^ { 2 }$ $Graph the ellipse and locate the foci. - 4 x ^ { 2 } = 36 - 9 y ^ { 2 }$

Additional Concepts Use the relation's graph to determine its domain and range. - $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$ $Additional Concepts Use the relation's graph to determine its domain and range. - \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - $( y - 4 ) ^ { 2 } = - 16 ( x - 2 )$

Additional Concepts Use the relation's graph to determine its domain and range. - $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$ $Additional Concepts Use the relation's graph to determine its domain and range. - \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$

Write Equations of Ellipses in Standard Form - Center at $( - 1,1 )$

Find the standard form of the equation of the hyperbola. -

Graph Ellipses Not Centered at the Origin - $16 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 64$ $Graph Ellipses Not Centered at the Origin - 16 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 64$

Graph Parabolas with Vertices at the Origin Find the focus and directrix of the parabola with the given equation. - $x ^ { 2 } = - 8 y$

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - $( x + 4 ) ^ { 2 } = - 4 ( y + 1 )$

Graph Ellipses Not Centered at the Origin - $\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$ $Graph Ellipses Not Centered at the Origin - \frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. - $x = - ( y + 1 ) ^ { 2 } + 8$

Find the standard form of the equation of the ellipse satisfying the given conditions. -Foci: $( - 3,0 ) , ( 3,0 )$ ; vertices: $( - 4,0 ) , ( 4,0 )$

Find the foci of the ellipse whose equation is given. - $36 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 900$

Graph the parabola. - $y ^ { 2 } = - 6 x$ $Graph the parabola. - y ^ { 2 } = - 6 x$

Graph the parabola. - $y ^ { 2 } + 12 x = 0$ $Graph the parabola. - y ^ { 2 } + 12 x = 0$

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Exam 7: Conic Sections

Use the center, vertices, and asymptotes to graph the hyperbola. - (x+2)2−4(y+2)2=4( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4(x+2)2−4(y+2)2=4

Graph the parabola. - x2=6yx ^ { 2 } = 6 yx2=6y

Convert the equation to the standard form for a hyperbola by completing the square on x and y. - x2−y2−4x+4y−1=0x ^ { 2 } - y ^ { 2 } - 4 x + 4 y - 1 = 0x2−y2−4x+4y−1=0

Graph the ellipse and locate the foci. - 4x2=36−9y24 x ^ { 2 } = 36 - 9 y ^ { 2 }4x2=36−9y2

Graph Hyperbolas Centered at the Origin Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes. - 4x2−9y2=364 x^{2}-9 y^{2}=364x2−9y2=36

Additional Concepts Use the relation's graph to determine its domain and range. - y24−x225=1\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 14y2​−25x2​=1

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - (y−4)2=−16(x−2)( y - 4 ) ^ { 2 } = - 16 ( x - 2 )(y−4)2=−16(x−2)

Additional Concepts Use the relation's graph to determine its domain and range. - x24−y216=1\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 14x2​−16y2​=1

Write Equations of Ellipses in Standard Form - Center at (−1,1)( - 1,1 )(−1,1)

Find the standard form of the equation of the hyperbola. -

Graph Ellipses Not Centered at the Origin - 16(x−1)2+4(y−2)2=6416 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 6416(x−1)2+4(y−2)2=64

Graph Parabolas with Vertices at the Origin Find the focus and directrix of the parabola with the given equation. - x2=−8yx ^ { 2 } = - 8 yx2=−8y

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - (x+4)2=−4(y+1)( x + 4 ) ^ { 2 } = - 4 ( y + 1 )(x+4)2=−4(y+1)

Graph Ellipses Not Centered at the Origin - (x+1)29+(y−1)216=1\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 19(x+1)2​+16(y−1)2​=1

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. - x=−(y+1)2+8x = - ( y + 1 ) ^ { 2 } + 8x=−(y+1)2+8

Find the standard form of the equation of the ellipse satisfying the given conditions. -Foci: (−3,0),(3,0)( - 3,0 ) , ( 3,0 )(−3,0),(3,0) ; vertices: (−4,0),(4,0)( - 4,0 ) , ( 4,0 )(−4,0),(4,0)

Find the foci of the ellipse whose equation is given. - 36(x−1)2+25(y−3)2=90036 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 90036(x−1)2+25(y−3)2=900

Graph the parabola. - y2=−6xy ^ { 2 } = - 6 xy2=−6x

Graph the parabola. - y2+12x=0y ^ { 2 } + 12 x = 0y2+12x=0

Equations and Inequalities

Functions and Graphs

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Use the center, vertices, and asymptotes to graph the hyperbola. - $( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4$ $Use the center, vertices, and asymptotes to graph the hyperbola. - ( x + 2 ) ^ { 2 } - 4 ( y + 2 ) ^ { 2 } = 4$

Graph the parabola. - $x ^ { 2 } = 6 y$ $Graph the parabola. - x ^ { 2 } = 6 y$

Convert the equation to the standard form for a hyperbola by completing the square on x and y. - $x ^ { 2 } - y ^ { 2 } - 4 x + 4 y - 1 = 0$

Graph the ellipse and locate the foci. - $4 x ^ { 2 } = 36 - 9 y ^ { 2 }$ $Graph the ellipse and locate the foci. - 4 x ^ { 2 } = 36 - 9 y ^ { 2 }$

Additional Concepts Use the relation's graph to determine its domain and range. - $\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$ $Additional Concepts Use the relation's graph to determine its domain and range. - \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 25 } = 1$

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - $( y - 4 ) ^ { 2 } = - 16 ( x - 2 )$

Additional Concepts Use the relation's graph to determine its domain and range. - $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$ $Additional Concepts Use the relation's graph to determine its domain and range. - \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 16 } = 1$

Write Equations of Ellipses in Standard Form - Center at $( - 1,1 )$

Graph Ellipses Not Centered at the Origin - $16 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 64$ $Graph Ellipses Not Centered at the Origin - 16 ( x - 1 ) ^ { 2 } + 4 ( y - 2 ) ^ { 2 } = 64$

Graph Parabolas with Vertices at the Origin Find the focus and directrix of the parabola with the given equation. - $x ^ { 2 } = - 8 y$

Graph Parabolas with Vertices Not at the Origin Find the vertex, focus, and directrix of the parabola with the given equation. - $( x + 4 ) ^ { 2 } = - 4 ( y + 1 )$

Graph Ellipses Not Centered at the Origin - $\frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$ $Graph Ellipses Not Centered at the Origin - \frac { ( x + 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 16 } = 1$

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. - $x = - ( y + 1 ) ^ { 2 } + 8$

Find the standard form of the equation of the ellipse satisfying the given conditions. -Foci: $( - 3,0 ) , ( 3,0 )$ ; vertices: $( - 4,0 ) , ( 4,0 )$

Find the foci of the ellipse whose equation is given. - $36 ( \mathrm { x } - 1 ) ^ { 2 } + 25 ( \mathrm { y } - 3 ) ^ { 2 } = 900$

Graph the parabola. - $y ^ { 2 } = - 6 x$ $Graph the parabola. - y ^ { 2 } = - 6 x$

Graph the parabola. - $y ^ { 2 } + 12 x = 0$ $Graph the parabola. - y ^ { 2 } + 12 x = 0$