OSTP Grade 11 CCRA Functions. Practice it free.

When $w = 2$, the exponent is $2/2 = 1$. So $L(2) = 180(0.88)^1 = 158.4$ lumens.

$10 \div 5 = 2$ half-lives. Remaining: $\left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}$.

Slope-intercept form is $f(x) = mx + b\textsf{.}$ When $x = 0\textsf{,}$ the cost is $f(0) = 10\textsf{,}$ so $b = 10\textsf{.}$

The temperature decreases by $3$ degrees each hour, and at noon the temperature is $72\textsf{.}$ So $w(x) = -3x + 72\textsf{.}$

First write slope-intercept form: $5 = -4(1) + b$ gives $b = 9$, so $y = -4x + 9$. Rearranging into standard form gives $4x + y = 9\textsf{.}$

Set $y = 0$: $-(x + 8)(x - 2) = 0$. The factor $-1$ is nonzero, so the zeros come from $x + 8 = 0 \Rightarrow x = -8$ and $x - 2 = 0 \Rightarrow x = 2$. Of the choices, $-8$ is an x-coordinate of an x-intercept.

Set $y=\dfrac{3}{x-1}$. Swap: $x=\dfrac{3}{y-1}$. Multiply: $x(y-1)=3$. Divide by $x$: $y-1=\dfrac{3}{x}$. Add $1$: $y=\dfrac{3}{x}+1$. So $f^{-1}(x)=\dfrac{3}{x}+1\textsf{.}$

Multiplying the outputs by $-1$ replaces each $y$-value with its opposite, which reflects the graph over the $x$-axis.