DC CAPE Algebra I Interpreting Functions. Practice it free.

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 $y$-intercept is the value of $y$ when $x = 0\textsf{.}$ Substituting gives $-5y = -40\textsf{,}$ so $y = 8\textsf{.}$

Point-slope form is $y - y_1 = m(x - x_1)\textsf{.}$ The slope is the coefficient of $(x - x_1)\textsf{,}$ so $m = 5\textsf{.}$

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 3 > 0$, so the graph opens upward and the minimum occurs at the vertex. Here $h = 1$, so the vertex is at $x = 1$. The leading coefficient $3$ does not change the $x$-coordinate of the vertex. Therefore, $f(x)$ reaches its minimum when $x = 1$.

Because this parabola opens upward, its lowest point is 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, the lowest point has $x$-coordinate $2$.

At the $x$-intercept, $y = 0$, so $0 = -4x - 20$, $-4x = 20$, and $x = -5\textsf{.}$

Substitute $x = 2$: $f(2) = 3(2)^{2} + 2 = 3(4) + 2 = 12 + 2 = 14$.

Set $-3x + 20 = 5$. Then $-3x = -15$, so $x = 5$.