Which is the vertex of x^2+10x=-17x2+10x=−17?

1 Answer
Anjali G
Nov 21, 2016

(-5,-8)(−5,−8)

Explanation:

x^2+10x=-17x2+10x=−17
0=-x^2-10x-170=−x2−10x−17
0=-[x^2+10x+17]0=−[x2+10x+17]
0=-[(x+5)^2-8]0=−[(x+5)2−8]
0=-(x+5)^2+80=−(x+5)2+8

x^2+10x+17=0x2+10x+17=0
(x+5)^2-8=0(x+5)2−8=0

Vertex is at x = -5x=−5. It is unclear whether the coefficient of the highest degree is positive or negative. If the parabola is negative, then the vertex is at (-5,8)(−5,8). If the parabola is positive, the vertex is at (-5,-8)(−5,−8)

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