Question

What is the opening, vertex, and the axis of symmetry of `f(x)=-3(x-2)^2+5`?

What is the domain and range? How do you graph it?

Answer 1

The parabola opens down, has a vertex of `(2,5)`, and an axis of symmetry of `x=2`.

Explanation:

`f(x)=color(red)(-3)(x-color(blue)2)^2+color(blue)5`

This function is written in the "vertex form" of a parabola, which is
`f(x)=color(red)a(x-h)^2+k` where `a` is a constant and `(h,k)` is the vertex.

If `a` is positive, the parabola opens up.

If `a` is negative, the parabola opens down.

In our example, `color(red)(a)=color(red)(-3)`, so the parabola opens down.

The vertex `(color(blue)h, color(blue)k)= (color(blue)2, color(blue)5)`. Note that because `color(blue)h` is subtracted in vertex form, the `x` coordinate of the vertex is `color(blue)2`, not `-2`.

The axis of symmetry goes through the vertex and is `x=2`.

The graph of the `color(red)("parabola")` and the `color(blue)("axis of symmetry")` is shown below.