How do you find the vertex and axis of symmetry for a quadratic equation #y = -x^2 + 6x -9#?

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Answer 1 · 2015-07-11T18:13:12

The vertex is at (#3,0#).
The axis of symmetry is #x = 3#.

The standard form for the equation of a parabola is

#y= a^2 + bx +c#

Your equation is

#y = -x^2 + 6x - 9#

So

#a = -1#, #b = 6#, and #c = -9#.

Vertex

Since #a < 0#, the parabola opens downwards.

The #x#-coordinate of the vertex is at

#x = –b/(2a) = -6/(2(-1)) = -6/-2 = 3#.

Insert this value of #x# back into the equation.

#y = -x^2 + 6x - 9 = -3^2 + 6×3 - 9 = -9 + 18 – 9 = 0#

The vertex is at (#3,0#).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is #x = 3#.