What is the vertex form #y=x^2+2x-8#?

1 Answer
Lil Del · EZ as pi
Oct 30, 2016

The vertex form of the equation is #y = (x + 1)^2 - 9#

Explanation:

Changing a quadratic function from standard form to vertex form actually requires that we go through the process of completing the square. To do this, we need the #x^2# and #x# terms only on the right side of the equation.

#y = x^2 + 2x - 8#
#y + 8 = x^2 + 2x - 8 + 8#
#y + 8 = x^2 + 2x - 8 + 8#
#y + 8 = x^2 + 2x#

Now, the right side has the #ax^2 + bx# terms, and we need to find #c#, using the formula #c = (b/2)^2#.

In our prepared equation, #b = 2#, so
#c = (2/2)^2 = 1^2 = 1#

Now, we add #c# to both sides of our equation, simplify the left side, and factor the right side.

#y + 8 + 1 = x^2 + 2x + 1#
#y + 9 = (x +1)^2#

To finish putting the equation in vertex form, subtract #9# from both sides, thus isolating the #y#:

#y + 9 - 9 = (x + 1)^2 - 9#
#y = (x + 1)^2 - 9#

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