SOL Algebra II Equations and Inequalities. Practice it free.

Set $f(x) = 0$: $6(x + 9)(x - 13) = 0$. The constant $6 \neq 0$, so the zeros are $x = -9$ and $x = 13$. Of the choices, $-9$ 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 $(x + 6)^2$ as $(x - (-6))^2$: the vertex is at $x = -6$. Therefore, $f(x)$ reaches its minimum when $x = -6$.

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

Subtract the third-year base from the first: $\text{\char36}35{,}200 - \text{\char36}31{,}000 = \text{\char36}4{,}200$. Over three years there are two equal increases, so divide by $3 - 1 = 2$ to get $x = \text{\char36}2{,}100$. Add $\text{\char36}2{,}100$ each year after the first: $\text{\char36}31{,}000$, $\text{\char36}33{,}100$, $\text{\char36}35{,}200$. Sum all three years: $\text{\char36}99{,}300$.

Above $32$ and below $40$ are both strict, so $32 < t < 40$.

Factor out $x$: $p(x)=x(x^2+4x+4)=x(x+2)^2$. Set each factor equal to zero: $x=0 \Rightarrow x=0$; $(x+2)^2=0$, so $(x+2)=0 \Rightarrow x=-2$. The zero $-2$ has multiplicity $2$, so the zeros are $0$, $-2$, and $-2$.

Set $y = 0$: $2x = 12$, so $x = 6$.

The slope is $m = \dfrac{5 - (-1)}{2 - (-4)} = 1$. Substitute $(-4, -1)$: $-1 = 1(-4) + b$, so $b = 3$. The equation is $y = x + 3$.