How do you find the vertex and axis of symmetry for a quadratic equation #y = x^2-5#?

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Answer 1 · 2015-07-11T17:58:12

The vertex is at (#0,-5#).
The axis of symmetry is #x = 0#.

The standard form for the equation of a parabola is

#y= a^2 + bx +c#

Your equation is

#y = x^2 - 5#

So

#a = 1#, #b = 0#, and #c = -5#.

Vertex

Since #a > 0#, the parabola opens upwards.

The #x#-coordinate of the vertex is at

#x = –b/(2a) = -0/(2×1) = -0/2 = 0#.

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

#y = x^2 - 5 = 0^2 - 5 = 0 - 5 = -5#

The vertex is at (#0,-5#).

Axis of symmetry

The axis of symmetry must pass through the vertex, so

The axis of symmetry is #x = 0#.