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. For parabolas of the form: f(x) = ax2 + bx + c, the vertex can be found by completing the square, resulting in the vertex form: f(x) = a(x - h)2 + k, where the vertex is (h, k). Alternatively, you can use the vertex formula to find the vertex: (-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
#1:
Instructions: Find the vertex form and state the vertex.
$$a)\hspace{.2em}f(x)=3x^2 - 24x + 38$$
$$b)\hspace{.2em}f(x)=x^2 - 18x + 91$$
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#2:
Instructions: Find the vertex form and state the vertex.
$$a)\hspace{.2em}f(x)=2x^2 - 8x + 10$$
$$b)\hspace{.2em}f(x)=-x^2 + 4x + 4$$
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#3:
Instructions: Find the vertex form and state the vertex.
$$a)\hspace{.2em}f(x)=\frac{1}{3}x^2 - 4x + 21$$
$$b)\hspace{.2em}f(x)=-18x^2 - 144x - 279$$
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#4:
Instructions: Find the vertex form and state the vertex.
$$a)\hspace{.2em}f(x)=-2x^2 + 9$$
$$b)\hspace{.2em}f(x)=-\frac{1}{3}x^2 + 6x - 32$$
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#5:
Instructions: Find the vertex form and state the vertex.
$$a)\hspace{.2em}f(x)=10x^2 + 6$$
$$b)\hspace{.2em}f(x)=-8x^2 + 80x - 194$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}f(x)=3(x - 4)^2 - 10$$ $$\text{Vertex:}\hspace{.25em}(4, -10)$$
$$b)\hspace{.2em}f(x)=(x - 9)^2 + 10$$ $$\text{Vertex:}\hspace{.25em}(9, 10)$$
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#2:
Solutions:
$$a)\hspace{.2em}f(x)=2(x - 2)^2 + 2$$ $$\text{Vertex:}\hspace{.25em}(2, 2)$$
$$b)\hspace{.2em}f(x)=-(x - 2)^2 + 8$$ $$\text{Vertex:}\hspace{.25em}(2, 8)$$
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#3:
Solutions:
$$a)\hspace{.2em}f(x)=\frac{1}{3}(x - 6)^2 + 9$$ $$\text{Vertex:}\hspace{.25em}(6, 9)$$
$$b)\hspace{.2em}f(x)=-18(x + 4)^2 + 9$$ $$\text{Vertex:}\hspace{.25em}(-4, 9)$$
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#4:
Solutions:
$$a)\hspace{.2em}f(x)=-2x^2 + 9$$ $$\text{Vertex:}\hspace{.25em}(0, 9)$$
$$b)\hspace{.2em}f(x)=-\frac{1}{3}(x - 9)^2 - 5$$ $$\text{Vertex:}\hspace{.25em}(9, -5)$$
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#5:
Solutions:
$$a)\hspace{.2em}f(x)=10x^2 + 6$$ $$\text{Vertex:}\hspace{.25em}(0, 6)$$
$$b)\hspace{.2em}f(x)=-8(x - 5)^2 + 6$$ $$\text{Vertex:}\hspace{.25em}(5, 6)$$
