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

Oct 15, 2015

This one is easy to factor to find the vertex to be the point $\left(x , y\right) = \left(3 , 0\right)$ 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 $\mathbb{R}$, and the range is $\left(- \infty , 0\right]$.

#### Explanation:

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

$y = - {x}^{2} + 6 x - 9 = - \left({x}^{2} - 6 x + 9\right) = - {\left(x - 3\right)}^{2}$.

This means the parabola opens downward with a vertex (high point in this case) at $\left(x , y\right) = \left(3 , 0\right)$ (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 $f \left(0\right) = - {0}^{2} + 6 \cdot 0 - 9 = - 9$.

The domain is the entire real number system $\mathbb{R}$ because you can plug in anything you want for $x$. The range is the set of non-positive numbers $\left(- \infty , 0\right]$ 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]}