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

1 Answer
George C.
Jun 6, 2015

y=−x2−2x+6

=−(x2+2x−6)

=−((x+1)2−7)

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

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

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

The intercepts with the x axis are the roots of

−((x+1)2−7)=0

Hence (x+1)2=7

So

x+1=±√7

x=−1±√7

So the intercepts are:

(−1+√7,0) and (−1−√7,0)

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