# How do you find the vertex of F(x)=x^2+2x-8?

Apr 10, 2018

-1

#### Explanation:

For a quadratic function like this, the vertices are points of symmetry.

We can use a shortcut such that:
$x = \frac{- b}{2 a}$

In your case, $b$ is 2 and $a$ is 1. So:
$- \frac{2}{2 \cdot 1} = - 1$

Confirm it on a graph.
graph{x^2+2x-8 [-25.8, 25.8, -18.1, 7.7]}

Apr 10, 2018

The vertex is when the gradient is 0.

#### Explanation:

Differentiate and find $x$ where $F ' \left(x\right) = 0$

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

$F ' \left(x\right) = 2 x + 2$

$0 = 2 x + 2$

$x = - 1$

$F \left(- 1\right) = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 8$

$F \left(- 1\right) = - 9$

Therefore the vertex is: $\left(- 1 , - 9\right)$

Apr 10, 2018

Vertex is (-1, -9)

#### Explanation:

To find the x-value of the vertex, we use the formula $x = \frac{- b}{2 a}$ (This is considered the axis of symmetry).

$a = 1$ and $b = 2$ with $c = - 8$ (These are the numbers in order)

By substituting we get $x = \frac{- \left(2\right)}{2 \cdot 1}$, which simplifies to $x = - 1$.

Once you find x, substitute back into the equation to find y.
$y = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 8$
$y = 1 - 2 - 8$
$y = - 9$

Putting the two numbers together you get the coordinates $\left(- 1 , - 9\right)$.