What is the vertex form of y=5x^2 - 10x - 75 y=5x2−10x−75?

1 Answer
Bill K.
Dec 19, 2015

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

Explanation:

First, factor out the coefficient of x^2x2, which is 5, out of the first two terms:

y=5x^2-10x-75=5(x^2-2x)-75y=5x2−10x−75=5(x2−2x)−75.

Next, complete the square on the expression inside the parentheses. Take the coefficient of xx, which is -2−2, divide it by 2 and square it to get 11. Add this number inside the parentheses and compensate for this change by subtracting 5*1 = 55⋅1=5 outside of the parentheses as follows:

y=5(x^2-2x+1)-75-5y=5(x2−2x+1)−75−5.

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

y=5(x-1)^2-80y=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)(x,y)=(1,−80).

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