What is the vertex of y= -x^2-x-(3x+2)^2y=−x2−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 xx-intercept at around x = -0.5x=−0.5. This will be helpful in checking our answer.

1.1. First, we have to expand. So

-x^2 - x - 9x^2 - 4 - 12x−x2−x−9x2−4−12x

= -10x^2 - 13x -4=−10x2−13x−4

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

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

y = -10x^2 - 13x -4y=−10x2−13x−4

-y/10 = x^2 + 1.3x + 0.4−y10=x2+1.3x+0.4

-y/10 + 0.0225 = x^2 + 1.3y + 0.4225−y10+0.0225=x2+1.3y+0.4225

-y/10 + 0.0225 = (x+0.65)^2−y10+0.0225=(x+0.65)2

-y/10 =(x+0.65)^2 - 0.0225−y10=(x+0.65)2−0.0225

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

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

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