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)#.