Question

given the function
$f(x) = \

$$\begin{cases} -5x^2 + 8 & x < 1 \\ -10x^2 + 8 & x \\geq 1 \\end{cases}$$

$
evaluate the following:
$f(-2) = \square$
$f(1) = \square$
$f(2) = \square$
question help: \video

Explanation:

Step1: Evaluate $f(-2)$

Since $-2 < 1$, use $f(x) = -5x^2 + 8$.

$$\begin{align*} f(-2) &= -5(-2)^2 + 8 \\ &= -5(4) + 8 \\ &= -20 + 8 \\ &= -12 \end{align*}$$

Step2: Evaluate $f(1)$

Since $1 \geq 1$, use $f(x) = -10x^2 + 8$.

$$\begin{align*} f(1) &= -10(1)^2 + 8 \\ &= -10(1) + 8 \\ &= -10 + 8 \\ &= -2 \end{align*}$$

Step3: Evaluate $f(2)$

Since $2 \geq 1$, use $f(x) = -10x^2 + 8$.

$$\begin{align*} f(2) &= -10(2)^2 + 8 \\ &= -10(4) + 8 \\ &= -40 + 8 \\ &= -32 \end{align*}$$

Answer:

$f(-2) = -12$
$f(1) = -2$
$f(2) = -32$