Question

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

Answer 1

`(1/2,-81/4)`

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 `dy/dx=0`

ie when `2x-1=0` which implies `x=1/2`.

The corresponding y values is then `y(1/2)=(1/2)^2-1/2-20=-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 `(d^2y)/(dx^2)|_(x=1/2)=2>0`.

The corresponding graph is given for completeness.

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