What is the vertex form of y= x^2-x-20 ?

Jun 10, 2016

$\left(\frac{1}{2} , - \frac{81}{4}\right)$

Explanation:

The vertex or turning point is the relative extreme point of the function and occurs at the point where the derivative of the function is zero.

That is, when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

ie when $2 x - 1 = 0$ which implies $x = \frac{1}{2}$.

The corresponding y values is then $y \left(\frac{1}{2}\right) = {\left(\frac{1}{2}\right)}^{2} - \frac{1}{2} - 20 = - \frac{81}{4}$.

Since the coefficient of ${x}^{2}$ is $1 > 0$, it implies the arms of the corresponding parabola graph of this quadratic function go up and hence the relative extremum is a relative (and in fact an absolute) minimum. One could also check this by showing that the second derivative $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} {|}_{x = \frac{1}{2}} = 2 > 0$.

The corresponding graph is given for completeness.

graph{x^2-x-20 [-11.95, 39.39, -22.35, 3.28]}