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Writing Quadratic Functions

Writing Quadratic Functions

Assessment

Presentation

•

Mathematics

•

9th - 12th Grade

•

Hard

Created by

Joseph Anderson

FREE Resource

5 Slides • 10 Questions

1

Interactive Lesson

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2

Writing Quadratic Equations in Vertex Form

Have your notebook nearby to show your work!

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3

Forms of Quadratic Functions
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4

Multiple Choice

What form of a quadratic is given?

f(x)=2x2+3x−4f\left(x\right)=2x^2+3x-4  

1

Vertex Form

2

Standard Form

3

Root Form

5

Multiple Choice

What form of a quadratic is given?

f(x)=3(x−2)2+4f\left(x\right)=3\left(x-2\right)^2+4  

1

Vertex Form

2

Standard Form

3

Root Form

6

7

Fill in the Blanks

Given the equation

f(x)=−5(x−2)2−4f\left(x\right)=-5\left(x-2\right)^2-4 , what is the vertex for the parabola? 
Write your answer as a coordinate point. (x, y)



Type answer...

8

Multiple Choice

Lets do another problem just to make sure we've got this.



Find the equation of a parabola that has a vertex at (2, -7) and goes through the point (3, -4).

1

y=325(x+2)2−7y=\frac{3}{25}\left(x+2\right)^2-7

2

y=3(x−2)2−7y=3\left(x-2\right)^2-7

3

y=3(x+2)2−7y=3\left(x+2\right)^2-7

4

y=(x−2)2−7y=\left(x-2\right)^2-7

9

Multiple Choice

Find the equation for a parabola that has a vertex at (3, 4) and goes through the point (2, 2).

1

y=(x−3)2+4y=\left(x-3\right)^2+4

2

y=−225(x+3)2+4y=-\frac{2}{25}\left(x+3\right)^2+4

3

y=−2(x−3)2+4y=-2\left(x-3\right)^2+4

4

y=−2(x+3)2−4y=-2\left(x+3\right)^2-4

10

Multiple Choice

Find the equation for a parabola that has a vertex at (2,3) and goes through the point (4,5).

1

y=(x−4)2+5y=\left(x-4\right)^2+5

2

y=(x−2)2+3y=\left(x-2\right)^2+3

3

y=(x−4)2−5y=\left(x-4\right)^2-5

4

y=(x−2)2−3y=\left(x-2\right)^2-3

11

Multiple Choice

Find the equation for a parabola that has a vertex at (-1,2) and goes through the point (3,4).

1

y=18(x−3)2+4y=\frac{1}{8}\left(x-3\right)^2+4

2

y=18(x+1)2+2y=\frac{1}{8}\left(x+1\right)^2+2

3

y=14(x+1)2+2y=\frac{1}{4}\left(x+1\right)^2+2

4

y=18(x−1)2+2y=\frac{1}{8}\left(x-1\right)^2+2

12

13

Multiple Choice

Lets do another problem just to make sure we've got this.



Find the equation of a parabola that has x-intercepts at -4 and 3 and passes through the point (2,7)

1

y=−76(x−4)(x−3)y=-\frac{7}{6}\left(x-4\right)\left(x-3\right)

2

y=−76(x+4)(x−3)y=-\frac{7}{6}\left(x+4\right)\left(x-3\right)

3

y=72(x−4)(x−3)y=\frac{7}{2}\left(x-4\right)\left(x-3\right)

4

y=76(x+4)(x−3)y=\frac{7}{6}\left(x+4\right)\left(x-3\right)

14

Multiple Choice


Find the equation of a parabola that has x-intercepts at -3 and 5 and passes through the point (3,-6)

1

y=−12(x+3)(x−5)y=-\frac{1}{2}\left(x+3\right)\left(x-5\right)

2

y=12(x+3)(x−5)y=\frac{1}{2}\left(x+3\right)\left(x-5\right)

3

y=12(x−3)(x−5)y=\frac{1}{2}\left(x-3\right)\left(x-5\right)

4

y=2(x+3)(x−5)y=2\left(x+3\right)\left(x-5\right)

15

Multiple Choice


Find the equation of a parabola that has x-intercepts at 3 and 5 and passes through the point (2,-3)

1

y=(x−3)(x−5)y=\left(x-3\right)\left(x-5\right)

2

y=−(x−3)(x−5)y=-\left(x-3\right)\left(x-5\right)

3

y=−(x+3)(x+5)y=-\left(x+3\right)\left(x+5\right)

4

y=13(x−3)(x−5)y=\frac{1}{3}\left(x-3\right)\left(x-5\right)

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