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

1 Answer
Trevor Ryan.
Jun 10, 2016

#(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]}

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