Question

How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of `y = -x^2 + 6x -9`?

Answer 1

This one is easy to factor to find the vertex to be the point `(x,y)=(3,0)` so that `x=3` is the axis of symmetry. The `y`-intercept is `-9`, the `x`-intercept is `x=3`, the domain is all of `RR`, and the range is `(-infty,0]`.

Explanation:

You can factor `y=f(x)=-x^2+6x-9` as the opposite of a perfect square (there's no need to "complete the square" here):

`y=-x^2+6x-9=-(x^2-6x+9)=-(x-3)^2`.

This means the parabola opens downward with a vertex (high point in this case) at `(x,y)=(3,0)` (all other values of the function are negative). This also gives `x=3` as the axis of symmetry and the (one) `x`-intercept.

For the `y`-intercept, compute `f(0)=-0^2+6*0-9=-9`.

The domain is the entire real number system `RR` because you can plug in anything you want for `x`. The range is the set of non-positive numbers `(-infty,0]` because that's the set of possible outputs.

The graph is shown below. Make sure you think about all these answers in relation to the graph.

graph{-x^2+6x-9 [-40, 40, -20, 20]}