# How do you graph y=-2x^2+8x?

Aug 13, 2016

A parabola with zeros at $x = 0$ and $x = 4$; a maximum value of $8$ at $x = 2$ (See graph below).

#### Explanation:

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

Therefore $y = 0$ at $x = 0 \mathmr{and} 4$

$y ' = - 4 x + 8$

$y ' = 0$ at $x = 2$

Since the coefficient of x^2 is negative, $y$ has a maximum value at $x = 2$

Therefore, ${y}_{\text{max}} = - 2 \cdot {2}^{2} + 8 \cdot 2 = - 8 + 16 = 8$

graph{-2x^2+8x [-12.6, 15.89, -4.86, 9.38]}