SOL Algebra I Relations and Functions. Practice it free.

Set $y = 0$: $-2(x + 15)(x - 8)(x + 3) = 0$. The constant $-2 \neq 0$, so the zeros are $x = -15$, $x = 8$, and $x = -3$. Among the choices, only $8$ is an x-coordinate of an x-intercept.

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$, so the graph opens upward and the minimum occurs at the vertex. Here $h = 8$, so the vertex is at $x = 8$. Therefore, $f(x)$ reaches its minimum when $x = 8$.

The parabola opens upward, so the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$, the $x$-value at the vertex is $x = -\dfrac{b}{2a}$. Here $a = 3$ and $b = -30$, so $x = -\dfrac{-30}{2(3)} = \dfrac{30}{6} = 5$. Therefore, $y$ is smallest when $x = 5$.

In $y = mx + b$, the slope is $m = \dfrac{1}{2}$.

Substitute $x = 3$: $8(3) + c = 10$, so $24 + c = 10$ and $c = -14$. The rule is $g(x) = 8x - 14$, so $g(5) = 8(5) - 14 = 40 - 14 = 26$.

The shop starts with $64$ posters (intercept) and adds $3.5$ posters each hour (slope), so $y = 3.5x + 64\textsf{.}$

Slope $= \dfrac{1 - (-1)}{2 - (-2)} = \dfrac{2}{4} = \dfrac{1}{2}\textsf{.}$

Substitute $x = 16$: $f(16) = \sqrt{2(16) + 17} = \sqrt{32 + 17} = \sqrt{49} = 7$.