# How do you find the vertex of a parabola f(x)=x^2 +10x-11?

Mar 23, 2018

Vertex of the parabola is at :color(blue)((-5, -36)

#### Explanation:

The Standard form for the quadratic equation is :

$y = f \left(x\right) = a {x}^{2} + b x + c = 0$

Given:

$y = f \left(x\right) = {x}^{2} + 10 x - 11$

where a=1; b=10 and c= -11

Find the x-coordinate of the vertex using the formula $- \frac{b}{2 a}$

$x = - \frac{10}{2 \cdot 1}$

$\Rightarrow x = - 5$

To find the y-coordinate value of the vertex, set $x = - 15$ in $y = {x}^{2} + 10 x - 11$

$y = {\left(- 5\right)}^{2} + 10 \left(- 5\right) - 11$

Simplify the right-hand side.

$\Rightarrow 25 - 50 - 11$

$\Rightarrow 25 - 61$

$\Rightarrow - 36$

$y = - 36$

Hence,

Vertex of the parabola is at :color(blue)((-5, -36)

Also observe that, since the coefficient of ${x}^{2}$ term $\left(a\right)$ is less than zero, the parabola opens down and the graph has a minimum at the vertex.

Please refer to the image of the graph below: