6.3.1 REVIEW A Test Review

Authored by Kyleigh O'Boyle

Mathematics

12th Grade

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9 questions

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DROPDOWN QUESTION

2 mins • 2 pts

Identify the vertices and co-vertices of the ellipse.
vertices: (a)
co-vertices: (b)

$\left(-5,\ 0\right);\ \left(5,\ 0\right)$

$\left(0,\ -3\right);\ \left(0,\ 3\right)$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Use the following equation of a parabola to determine the coordinates of the vertex, the focus, and the equation of the directrix. $y=\frac{1}{2}\left(x+10\right)^2+5$

vertex (5, -10);

focus (5, -8);

directrix y=-12

vertex (-10, 5);

focus (-10, 5.5);

directrix y=4.5

vertex (-5, 10);

focus (-5, 12);

directrix y=8

vertex (10, -5);

focus (-10, -4.5);

directrix y=-5.5

MATCH QUESTION

2 mins • 4 pts

Match each graph with the correct equation.

$\frac{y^2}{9}-\frac{x^2}{16}=1$

$\frac{x^2}{16}+\frac{y^2}{9}=1$

$\frac{x^2}{16}-\frac{y^2}{9}=1$

$\frac{x^2}{9}+\frac{y^2}{16}=1$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Given the two questions below, find all points of intersection.
$x^2+y^2=16$
$2x^2-3y^2=12$

$\left(2,4\right);\ \left(2,\ -4\right);\$

$\left(-2,\ 4\right);\ \left(-2,\ -4\right)$

$\left(2\ \sqrt[]{3},\ 2\right);\ \left(2\ \sqrt[]{3},\ -2\right)$

$\left(-2\ \sqrt[]{3},\ 2\right);\ \left(-2\ \sqrt[]{3},\ -2\right)$

$\left(2\ \sqrt[]{5},\ 2\right);\ \left(2\ \sqrt[]{5},\ -2\right)$

$\left(-2\ \sqrt[]{5},\ 2\right);\ \left(-2\ \sqrt[]{5},\ -2\right)$

$\left(1,\ 4\right);\ \left(1,\ -4\right)$

$\left(-1,\ 4\right);\ \left(-1,\ -4\right)$

CATEGORIZE QUESTION

3 mins • 6 pts

Determine if the equation is a circle, ellipse, or a hyperbola.

Groups:

(a) circle

(b) ellipse

$-2x^2-3y^2+3x-6y+1=0$

$4x^2+y^2+2x-3y-5=0$

$4x^2+4y^2-2x+8y-4=0$

$4x^2-3y^2+2x+y-5=0$

$y^2-2x^2+2y-4x+8=0$

$x^2+y^2-2x-5y-9=0$

DRAW QUESTION

3 mins • 1 pt

Use the given graph paper to graph the following equation:
$\left(x+2\right)^2+\left(y-1\right)^2=16$

DRAW QUESTION

3 mins • 1 pt

Use the given graph paper to graph the following equation:
$x-4=\left(y+1\right)^2$

DRAW QUESTION

3 mins • 1 pt

Use the given graph paper to graph the following equation:
$\frac{\left(x+7\right)^2}{25}+\frac{\left(y-2\right)^2}{4}=1$

DRAW QUESTION

3 mins • 1 pt

Use the given graph paper to graph the following equation:
$\frac{\left(x+3\right)^2}{36}-\frac{\left(y+7\right)^2}{9}=1$

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6.3.1 REVIEW A Test Review

Identify the vertices and co-vertices of the ellipse.vertices: ​ (a) co-vertices: ​ (b)

Use the following equation of a parabola to determine the coordinates of the vertex, the focus, and the equation of the directrix. y=12(x+10)2+5y=\frac{1}{2}\left(x+10\right)^2+5y=21​(x+10)2+5

Match each graph with the correct equation.

Given the two questions below, find all points of intersection. x2+y2=16x^2+y^2=16x2+y2=16 2x2−3y2=122x^2-3y^2=122x2−3y2=12

Determine if the equation is a circle, ellipse, or a hyperbola.

Use the given graph paper to graph the following equation: (x+2)2+(y−1)2=16\left(x+2\right)^2+\left(y-1\right)^2=16(x+2)2+(y−1)2=16

Use the given graph paper to graph the following equation: x−4=(y+1)2x-4=\left(y+1\right)^2x−4=(y+1)2

Use the given graph paper to graph the following equation: (x+7)225+(y−2)24=1\frac{\left(x+7\right)^2}{25}+\frac{\left(y-2\right)^2}{4}=125(x+7)2​+4(y−2)2​=1

Use the given graph paper to graph the following equation: (x+3)236−(y+7)29=1\frac{\left(x+3\right)^2}{36}-\frac{\left(y+7\right)^2}{9}=136(x+3)2​−9(y+7)2​=1

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Discover more resources for Mathematics

Identify the vertices and co-vertices of the ellipse.
vertices: (a)
co-vertices: (b)

Use the following equation of a parabola to determine the coordinates of the vertex, the focus, and the equation of the directrix. $y=\frac{1}{2}\left(x+10\right)^2+5$

Given the two questions below, find all points of intersection.
$x^2+y^2=16$
$2x^2-3y^2=12$

Use the given graph paper to graph the following equation:
$\left(x+2\right)^2+\left(y-1\right)^2=16$

Use the given graph paper to graph the following equation:
$x-4=\left(y+1\right)^2$

Use the given graph paper to graph the following equation:
$\frac{\left(x+7\right)^2}{25}+\frac{\left(y-2\right)^2}{4}=1$

Use the given graph paper to graph the following equation:
$\frac{\left(x+3\right)^2}{36}-\frac{\left(y+7\right)^2}{9}=1$