Question: Determine an equation of the graph below in the following forms:
a) In factored form, $y = a(x - r)(x - s)$
b) in vertex form, $y = a(x - h)^2 + k$
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Points given: $(-4, 0)$, $(2, 6)$, $(8, 0)$
Graph shape and math: A downward-opening parabola with x-intercepts at $(-4, 0)$ and $(8, 0)$ and vertex at $(2, 6)$, centered at $x = 2$.
1. **State the problem:**
We need to find the equation of a parabola that passes through the points $(-4, 0)$, $(2, 6)$, and $(8, 0)$.
2. **Factored form:**
The factored form of a quadratic is $$y = a(x - r)(x - s)$$ where $r$ and $s$ are the roots (x-intercepts).
Given roots are $r = -4$ and $s = 8$, so:
$$y = a(x + 4)(x - 8)$$
3. **Find $a$ using the vertex point $(2, 6)$:**
Substitute $x = 2$, $y = 6$:
$$6 = a(2 + 4)(2 - 8)$$
$$6 = a(6)(-6)$$
$$6 = -36a$$
Divide both sides by $-36$:
$$a = \frac{6}{-36}$$
$$a = -\frac{1}{6}$$
4. **Final factored form:**
$$y = -\frac{1}{6}(x + 4)(x - 8)$$
5. **Vertex form:**
The vertex form is $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex.
Given vertex is $(2, 6)$ and $a = -\frac{1}{6}$, so:
$$y = -\frac{1}{6}(x - 2)^2 + 6$$
6. **Summary:**
- Factored form: $$y = -\frac{1}{6}(x + 4)(x - 8)$$
- Vertex form: $$y = -\frac{1}{6}(x - 2)^2 + 6$$