# How do you find all local maximum and minimum points given y=x^2-98x+4?

Aug 30, 2017

The minimum point is at $\left(49 , - 2397\right)$ [Ans]

#### Explanation:

$y = {x}^{2} - 98 x + 4$ . At turning point , $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 98 \therefore 2 x - 98 = 0 \mathmr{and} 2 x = 98 \mathmr{and} x = 49$

At x=49 ; y = 49^2 -98*49+4 or y= -2397

So turrning point is at $\left(49 , - 2397\right)$.

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2$ . To distinguish maximum ar minimum point

we know if $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 > 0$ then the point must be a minimum.

Here $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 > 0$ , so the point $\left(49 , - 2397\right)$ is minimum.

The minimum point is at $\left(49 , - 2397\right)$ [Ans]