# How do you write y = -4x^2 + 8x + 60 into vertex form?

May 10, 2015

$y = - 4 {x}^{2} + 8 x + 60 = - 4 {\left(x - 1\right)}^{2} + 64$.

As I mentioned on another answer:

In general, for $y = a {x}^{2} + b x + c$, the $x$ coordinate of the vertex is $- \frac{b}{2 a}$, and the vertex form is:

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

May 10, 2015

Another way:
Vertex form:
$f \left(x\right) = a {\left(x - \frac{b}{2 a}\right)}^{3} + f \left(- \frac{b}{2 a}\right)$

-b/2a = -8/-8 = 1

f(1) = -4 + 8 + 60 = 64

Vertex form: $y = - 4. {\left(x - 1\right)}^{2} + 64$

Check:
Develop f(x) = -4(x - 1)^2 + 64 = -4(x^2 - 2x + 1) + 64 = -4x^2 + 8x - 4 + 64 = -4x^2 + 8x + 60 Correct