Question

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

Answer 1

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)`.