How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=x22x+6?

1 Answer
Jun 6, 2015

y=x22x+6

=(x2+2x6)

=((x+1)27)

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)27)=0

Hence (x+1)2=7

So

x+1=±7

x=1±7

So the intercepts are:

(1+7,0) and (17,0)