# How do you find the vertex and intercepts for y = - 2x^2 + 8x + 4?

Mar 16, 2016

The x-intercepts are simply the roots of the quadratic, $x = 0.58 , x = 3.4$.
The y-intercept is simply the constant term, $y = 4$.
The vertex is the point at which the slope (gradient) is zero, so differentiate and then solve to find the point $\left(2 , 12\right)$.

#### Explanation:

I'll use the method that requires calculus. Another answerer may be able to answer using only algebra.

First visualise the curve: the ${x}^{2}$ term is negative, so this is an 'upside-down' parabola, with its vertex at the top and the opening toward the bottom.

The x-intercepts occur when $y = 0$, so:

$- 2 {x}^{2} + 8 x + 4 = 0$

Solve using the quadratic formula or otherwise:

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot \left(- 2\right) \cdot 4}}{2 \left(- 2\right)} = \frac{- 8 \pm \sqrt{32}}{-} 4$
$x = 3.4 \mathmr{and} 0.58$

To find the vertex, we know that it is a point where the gradient = $0$. The first derivative of the expression gives the slope:

$y = - 2 {x}^{2} + 8 x + 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 4 x + 8$

Set this equal to zero:

$0 = - 4 x + 8$

Solving, $x = 2$. To find the $y$ value of the vertex, substitute this in the original expression:

$y = - 2 {\left(2\right)}^{2} + 8 \left(2\right) + 4 = 12$

Therefore the coordinates of the vertex are $\left(2 , 12\right)$.