# How do you write the equation of the quadratic function with roots 6 and 10 and a vertex at (8, 2)?

Jun 16, 2015

In vertex form:

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

In standard form:

$y = - \frac{1}{2} {x}^{2} + 8 x - 30$

#### Explanation:

Since we're given the vertex the equation may be written

$y = a {\left(x - 8\right)}^{2} + 2$ for some constant $a$.

Putting $x = 10$, $y = 0$ into this equation, we find:

$0 = a {\left(10 - 8\right)}^{2} + 2 = 4 a + 2$

Hence $a = - \frac{1}{2}$

So in vertex form, the equation is:

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

Expand to get standard form:

$y = - \frac{1}{2} {x}^{2} + 8 x - 30$