# How do you find and classify all the critical points and then use the second derivative to check your results given y=x^2+10x-11?

Jul 30, 2016

Vertex $\left(- 5 , - 36\right)$
Y-intercept $\left(0 , - 11\right)$
X-intercepts $\left(- 11 , 0\right)$and $\left(1 , 0\right)$

#### Explanation:

Given -

$y = {x}^{2} + 10 x - 11$

It is a quadratic equation .
It has only one critical point.
It is the vertex.

$x = \frac{- b}{2 a} = \frac{- 10}{2 \times 1} = - 5$

At x=-5; y= (-5)^2+10(-5)-11

$y = 25 - 50 - 11 = 25 - 61 = - 36$

Vertex is $\left(- 5 , - 36\right)$

Derivatives of the function are

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 10$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 > 0$

Its second derivative is greater than zero. The curve is concave upwards.

Its other important points are

Y-intercept

At x=0; y=0^2+10(0)-11=-11

At (0, -11 the curve cuts the Y-axis

X- intercepts

At y=0; x^2+10x-11=0

${x}^{2} + 11 x - x - 11 = 0$
$x \left(x + 11\right) - 1 \left(x + 11\right) = 0$
$\left(x + 11\right) \left(x - 1\right) = 0$
$x + 11 = 0$
$x = - 11$

$\left(- 11 , 0\right)$ is one of the x- intercept

$x - 1 = 0$
$x = 1$

$\left(1 , 0\right)$ is another x-intercept.

At points $\left(- 11 , 0\right)$and $\left(1 , 0\right)$ , the curve cuts the x-axis