Question

find the range of the quadratic function.
$y = -x^2 + 6x - 11$
write your answer as an inequality using x or y as appropriate.
or, you may instead click on \empty set\ or \all reals\ as the answer.

Explanation:

Step1: Rewrite in vertex form

Complete the square for $y=-x^2+6x-11$:
First factor out -1 from the first two terms:
$y=-(x^2-6x)-11$
Add and subtract $(\frac{-6}{2})^2=9$ inside the parentheses:
$y=-(x^2-6x+9-9)-11$
Rewrite as:
$y=-(x-3)^2 + 9 - 11$
Simplify:
$y=-(x-3)^2 - 2$

Step2: Analyze the vertex form

The quadratic is in the form $y=a(x-h)^2+k$, where $a=-1$, $h=3$, $k=-2$.
Since $a<0$, the parabola opens downward, so the vertex $(3,-2)$ is the maximum point.
The $y$-value of the vertex is the highest possible value of $y$, and $y$ can decrease infinitely.

Answer:

$y \leq -2$