# What is the vertex form of y=-32x^2+80x+2?

##### 2 Answers
Jan 9, 2017

Vertex form of equation is $y = - 32 {\left({x}^{2} - \frac{5}{4}\right)}^{2} + 52$

#### Explanation:

Vertex form of equation is $y = a {\left(x - h\right)}^{2} + k$

As we have $y = - 32 {x}^{2} + 80 x + 2$

or $y = - 32 \left({x}^{2} - \frac{80}{32} x\right) + 2$

or $y = - 32 \left({x}^{2} - \frac{5}{2} x\right) + 2$

or $y = - 32 \left({x}^{2} - 2 \times \frac{5}{4} x + {\left(\frac{5}{4}\right)}^{2}\right) + 2 - \left(- 32\right) \times {\left(\frac{5}{4}\right)}^{2}$

or $y = - 32 {\left({x}^{2} - \frac{5}{4}\right)}^{2} + 2 + 32 \times \frac{25}{16}$

or $y = - 32 {\left({x}^{2} - \frac{5}{4}\right)}^{2} + 2 + 50$

or $y = - 32 {\left({x}^{2} - \frac{5}{4}\right)}^{2} + 52$, where vertex is $\left(- \frac{5}{4} , - 48\right)$
graph{-32x^2+80x+2 [-10, 10, -60, 60]}

Jan 9, 2017

y = - 32(x - 5/4)^2 + 52

#### Explanation:

$y = - 32 {x}^{2} + 80 x + 2$
x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{80}{64} = \frac{5}{4}$
y-coordinate of vertex:
$y \left(\frac{5}{4}\right) = - 32 \left(\frac{25}{16}\right) + 80 \left(\frac{5}{4}\right) + 2 = - 50 + 100 + 2 = 52$
Vertex form of y:
$y = - 32 {\left(x - \frac{5}{4}\right)}^{2} + 52$