What is the vertex of # y= -x^2-x-(3x+2)^2#?

1 Answer
Elijah T. · Stefan V.
Jan 30, 2017

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Explanation:

graph{-x^2-x-(3x+2)^2 [-1.5, 0.5, -0.5, 1.5]}

It looks like this with an #x#-intercept at around #x = -0.5#. This will be helpful in checking our answer.

#1.# First, we have to expand. So

# -x^2 - x - 9x^2 - 4 - 12x#

#= -10x^2 - 13x -4#

Luckily, if we plug in #x = 0.5# for #x#, we get a value of zero in accordance to our #y# intercept, so we are on the right track.

#2.# Then we complete the square so that the equation is in vertex form (to find the vertex). So:

#y = -10x^2 - 13x -4#

#-y/10 = x^2 + 1.3x + 0.4#

#-y/10 + 0.0225 = x^2 + 1.3y + 0.4225#

#-y/10 + 0.0225 = (x+0.65)^2#

#-y/10 =(x+0.65)^2 - 0.0225#

#y = -10(x+0.65)^2 + 0.225#

So the vertex would be #(-0.65, 0.225)#

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