SOL Algebra II Functions and Relations. Practice it free.

Set $y = 0$: $(x - 15)(x + 7) = 0$. The zeros are $x = 15$ and $x = -7$. Of the choices, $15$ 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 $h = 9$, so the vertex is at $x = 9$. Therefore, $f(x)$ reaches its minimum when $x = 9$.

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 $a = 1$ and $b = 4$, so $x = -\dfrac{4}{2(1)} = -\dfrac{4}{2} = -2$. Therefore, $y$ is as small as possible when $x = -2$.

Substitute $x = 4$: $f(4) = \sqrt{4 + 21} = \sqrt{25} = 5$.

Substitute $x = 3$: $f(3) = -2(3) + 15 = -6 + 15 = 9$.

Each step multiplies by $1 + 0.50 = 1.5$. So $g(2) = 100 \cdot 1.5^2 = 100 \cdot 2.25 = 225$.

Substitute $s = 7\textsf{:}$ $r = 120 - 8(7) = 120 - 56 = 64\textsf{.}$ The estimated number remaining is $64\textsf{.}$

The $x$-intercept is the value of $x$ when $y = 0\textsf{.}$ Substituting gives $2x = -18\textsf{,}$ so $x = -9\textsf{.}$