# What are the local extema of f(x)=x^2-4x-5?

Apr 27, 2016

At $\left(2 , - 9\right)$ There is a minima.

#### Explanation:

Given -

$y = {x}^{2} - 4 x - 5$
Find the first two derivatives
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 4$

Maxima and Minima is to be determined by the second derivative.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 > 0$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 2 x - 4 = 0$
$2 x = 4$
$x = \frac{4}{2} = 2$

At x=2 ; y= 2^2-4(2)-5

$y = 4 - 8 - 5$
$y = 4 - 13 = - 9$

Since the second derivative is greater than one.
At $\left(2 , - 9\right)$ There is a minima.