Question

find the range of the quadratic function.
y = -3x² + 30x - 71
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: Identify parabola direction

The coefficient of $x^2$ is $-3 < 0$, so the parabola opens downward, meaning it has a maximum value (no minimum, extends to $-\infty$).

Step2: Find vertex x-coordinate

For $y=ax^2+bx+c$, vertex $x = -\frac{b}{2a}$.
$x = -\frac{30}{2(-3)} = \frac{30}{6} = 5$

Step3: Find vertex y-value (max y)

Substitute $x=5$ into the function:
$y = -3(5)^2 + 30(5) - 71 = -3(25) + 150 - 71 = -75 + 150 - 71 = 4$

Step4: Define the range

Since the parabola opens downward, $y$ can be at most 4, and all values less than 4 are included.

Answer:

$y \leq 4$