# What is the vertex of  y=1/2x^2 +2x - 8 ?

Nov 14, 2015

The vertex of a quadratic curve is the point where the slope of the curve is zero.

#### Explanation:

$y = {x}^{2} / 2 + 2 x - 8$
=> $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot 2 \cdot x + 2$ (Differentiating both sides with respect to x)
=>$\frac{\mathrm{dy}}{\mathrm{dx}} = x + 2$

Now the slope of the quadratic curve is given by $\frac{\mathrm{dy}}{\mathrm{dx}}$

Thus, at the vertex ( as mentioned earlier ), $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Therefore $x + 2 = 0$
Or $x = - 2$

The corresponding y coordinate can be obtained by substituting $x = - 2$ in the original equation.

$y = {x}^{2} / 2 + 2 x - 8$
=> $y = {2}^{2} / 2 + 2 \cdot 2 - 8$
=>$y = 2 + 4 - 8$
=>$y = - 2$

This the required vertex is: $\left(x , y\right) = \left(- 2 , - 2\right)$