About Vertex Form of a Parabola:

The vertex of a parabola is the lowest point for an upward-facing parabola or the highest point for a downward-facing parabola. When working with parabolas of the form: f(x) = ax2 + bx + c, we can find the vertex by completing the square and obtaining the vertex form or from the vertex formula. When we complete the square, we are able to change the form to: f(x) = a(x - h)2 + k, where the vertex is given as: (h, k). Alternatively, we can just find the vertex as: (-b/2a, f(-b/2a)).


Test Objectives
  • Demonstrate the ability to write a quadratic equation in vertex form
  • Demonstrate the ability to find the vertex for a parabola
Vertex Form of a Parabola Practice Test:

#1:

Instructions: Find the vertex form and state the vertex.

$$a)\hspace{.2em}f(x)=3x^2 - 24x + 38$$

$$b)\hspace{.2em}f(x)=x^2 - 18x + 91$$


#2:

Instructions: Find the vertex form and state the vertex.

$$a)\hspace{.2em}f(x)=2x^2 - 8x + 10$$

$$b)\hspace{.2em}f(x)=-x^2 + 4x + 4$$


#3:

Instructions: Find the vertex form and state the vertex.

$$a)\hspace{.2em}f(x)=\frac{1}{3}x^2 - 4x + 21$$

$$b)\hspace{.2em}f(x)=-18x^2 - 144x - 279$$


#4:

Instructions: Find the vertex form and state the vertex.

$$a)\hspace{.2em}f(x)=-2x^2 + 9$$

$$b)\hspace{.2em}f(x)=-\frac{1}{3}x^2 + 6x - 32$$


#5:

Instructions: Find the vertex form and state the vertex.

$$a)\hspace{.2em}f(x)=10x^2 + 6$$

$$b)\hspace{.2em}f(x)=-8x^2 + 80x - 194$$


Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}f(x)=3(x - 4)^2 - 10$$ $$\text{Vertex:}\hspace{.25em}(4, -10)$$

$$b)\hspace{.2em}f(x)=(x - 9)^2 + 10$$ $$\text{Vertex:}\hspace{.25em}(9, 10)$$


#2:

Solutions:

$$a)\hspace{.2em}f(x)=2(x - 2)^2 + 2$$ $$\text{Vertex:}\hspace{.25em}(2, 2)$$

$$b)\hspace{.2em}f(x)=-(x - 2)^2 + 8$$ $$\text{Vertex:}\hspace{.25em}(2, 8)$$


#3:

Solutions:

$$a)\hspace{.2em}f(x)=\frac{1}{3}(x - 6)^2 + 9$$ $$\text{Vertex:}\hspace{.25em}(6, 9)$$

$$b)\hspace{.2em}f(x)=-18(x + 4)^2 + 9$$ $$\text{Vertex:}\hspace{.25em}(-4, 9)$$


#4:

Solutions:

$$a)\hspace{.2em}f(x)=-2x^2 + 9$$ $$\text{Vertex:}\hspace{.25em}(0, 9)$$

$$b)\hspace{.2em}f(x)=-\frac{1}{3}(x - 9)^2 - 5$$ $$\text{Vertex:}\hspace{.25em}(9, -5)$$


#5:

Solutions:

$$a)\hspace{.2em}f(x)=10x^2 + 6$$ $$\text{Vertex:}\hspace{.25em}(0, 6)$$

$$b)\hspace{.2em}f(x)=-8(x - 5)^2 + 6$$ $$\text{Vertex:}\hspace{.25em}(5, 6)$$