HSF.IF.C.7 math practice. Learn by doing.

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.

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 $a = 1$ and $b = -6$, so $x = -\dfrac{-6}{2(1)} = \dfrac{6}{2} = 3$. Therefore, $y$ is smallest when $x = 3$.

Set $y = 0$: $4(x + 12)(x - 3)(x + 2) = 0$. The constant $4 \neq 0$, so the zeros are $x = -12$, $x = 3$, and $x = -2$. Among the choices, only $3$ 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. Rewrite $(x + 3)^2$ as $(x - (-3))^2$: the vertex is at $x = -3$. Therefore, $f(x)$ reaches its minimum when $x = -3$.

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 $b = 8$, so $x = -\dfrac{8}{2(1)} = -\dfrac{8}{2} = -4$. Therefore, $y$ is smallest when $x = -4$.

Set $y = 0$: $-5(x - 11)(x + 4) = 0$. The constant $-5 \neq 0$, so the zeros are $x = 11$ and $x = -4$. Of the choices, $-4$ 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 $h = 2$, so the vertex is at $x = 2$. Therefore, $f(x)$ reaches its minimum when $x = 2$.