parabolas

What is the vertex of this quadratic equation?

y = 2(x + 2)² + 5

• (-2,5)

• (2,5)

• (5,-2)

• (-5,-2)

Where is the axis of symmetry for this parabola?

• x = −1
• x = 0
• y = 3
• x = −3

Identify the vertex and whether the graph opens up or down.

• (5, 2); opens up
• (5, 2); opens down
• (-5, 2); opens up
• (-5, 2); opens down

Convert to vertex form.

• y=2(x+3)² - 9
• y=2(x-3)² - 9
• y=2(x+6)² - 9
• y=2(x-6)² - 9

The graph of y = x² + 1 could be?

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Tags

CCSS.HSF-IF.C.7A

What is the equation of this graph?

• f(x) = (x -5)² + 1
• f(x) = (x - 1)² - 5
• f(x) = (x + 1)² - 5
• f(x) = (x + 5)² +1

Tags

CCSS.HSF-IF.C.7A

Which equation has a vertex of (-3, 4)?

• y = 2(x+3)² + 4
• y = (x+3)² - 4
• y = (x-3)² + 4
• y = 2(x - 3)² + 4

Write the following equation is (h,k) form: 
 $y\ =\ -x^2\ -6x\ -5$  

•  $-1\left(x+3\right)^2\ +4$  

•  $-1\left(x\ +\ 6\right)^2\ -14$  

•  $-1\left(x\ +\ 3\right)^2\ -14$  

•  $-1\left(x\ +\ 6\right)^2\ +4$  

 $y\ =\ \left(x-1\right)^2\ -9$  Find the y-intercept:

• (0, -10)

• (0 ,-8)

• (-8, 0)

• (-10, 0)

 $Y=\ -\frac{1}{8\ }\left(x\ -\ 4\right)^2\ +\ 5$  Find the focus:

• (4, 5)

• (4, 7)

• (4, -2)

• (4, 3)