What is the vertex form of #y=5x^2 - 10x - 75 #?

1 Answer
Bill K.
Dec 19, 2015

#y=5(x-1)^2-80#, meaning the vertex is at the point #(x,y)=(1,-80)#.

Explanation:

First, factor out the coefficient of #x^2#, which is 5, out of the first two terms:

#y=5x^2-10x-75=5(x^2-2x)-75#.

Next, complete the square on the expression inside the parentheses. Take the coefficient of #x#, which is #-2#, divide it by 2 and square it to get #1#. Add this number inside the parentheses and compensate for this change by subtracting #5*1 = 5# outside of the parentheses as follows:

#y=5(x^2-2x+1)-75-5#.

This trick makes the expression inside the parentheses a perfect square to get the final answer:

#y=5(x-1)^2-80#.

The graph of this function is a parabola opening upward with a minimum at the vertex #(x,y)=(1,-80)#.

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