What is the axis of symmetry and vertex for the graph f(x)=-4x^2 ?

1 Answer
Jun 2, 2018

See below

Explanation:

The axis of symmetry can be calculated for a quadratic in standard form (ax^2+bx+c) by the equation x=-b/(2a)
In the equation in your question, a=-4, b=0, and c=0. Thus, the axis of symmetry is at x=0:
x=-b/(2a)=-0/(2*-4)=0/-8=0

To find the vertex, substitute the x-coordinate of the axis of symmetry for x in the original equation to find its y-coordinate:
y=-4x^2 = -4*0^2=-4*0=0

So the axis of symmetry is x=0 and the vertex is at (0,0).