Question

How do you tell whether the graph opens up or down, find the vertex and the axis of symmetry of `y=-4(x-3)^2`?

Answer 1

Please see below.

Explanation:

As the equation is already in vertex form and is `y=-4(x-3)^2` i.e. `(y-0)=-4(x-3)^2`

the vertex is `(3,0)` and axis of symmetry is `x=3`. Further, as coefficient of `x^2` is negative, parabola opens downwards.

graph{ -4(x-3)^2 [ -20, 20, -35, 5] }