How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y = -x^2-2x+6y=−x2−2x+6?

1 Answer
George C.
Jun 6, 2015

y = -x^2-2x+6y=−x2−2x+6

= -(x^2+2x-6)=−(x2+2x−6)

= -((x+1)^2-7)=−((x+1)2−7)

This has vertex where (x+1)^2 = 0(x+1)2=0, that is where x = -1x=−1
and y = - (-7) = 7y=−(−7)=7. That is (-1, 7)(−1,7)

The axis of symmetry is the vertical line with equation x = -1x=−1

The intercept with the yy axis is where x=0x=0 and y=6y=6, that is at (0, 6)(0,6).

The intercepts with the xx axis are the roots of

-((x+1)^2-7) = 0−((x+1)2−7)=0

Hence (x+1)^2 = 7(x+1)2=7

So

x+1 = +-sqrt(7)x+1=±√7

x = -1+-sqrt(7)x=−1±√7

So the intercepts are:

(-1+sqrt(7), 0)(−1+√7,0) and (-1-sqrt(7), 0)(−1−√7,0)

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