SAT Question d139cf4b: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. It’s given that function $g$ is defined by $g(t)=f(t)+3$ and that $f(t)=55t-2t^2$. Substituting $55t-2t^2$ for $f(t)$ in $g(t)=f(t)+3$ yields $g(t)=55t-2t^2+3$, or $g(t)=-2t^2+55t+3$. The maximum value of $g(t)$ can be found by completing the square to rewrite the equation defining $g$ in the form $g\left(t\right)=a(t-h{)}^2+k$, where the maximum value of the function is $k$, which occurs when $t=h$, and $a$ is a negative constant. The equation $g\left(t\right)=-2t^2+55t+3$ is equivalent to $g(t)=-2(t^2-\frac{55}{2}t)+3$, which can be rewritten as $g\left(t\right)=-2(t^2-\frac{55}{2}t+(\frac{55}{4}{)}^2)+3+2(\frac{55}{4}{)}^2$, or $g\left(t\right)=-2(t-\frac{55}{4}{)}^2+3+2(\frac{55}{4}{)}^2$. This equation is in the form $g\left(t\right)=a(t-h{)}^2+k$, where $a=-2$, $h=\frac{55}{4}$, and $k=3+2(\frac{55}{4}{)}^2$. Thus, the maximum value of $g(t)$ is $3+2(\frac{55}{4}{)}^2$.

Alternate approach: Since the function $f$ is a quadratic function, the maximum value of $f(t)$ occurs at the value of $t$ that’s halfway between the two zeros of the function. The zeros of function $f$ can be found by substituting $0$ for $f(t)$ in the equation defining $f$, which yields $0=55t-2t^2$. This equation can be rewritten as $0=t(55-2t)$. By the zero product property, it follows that $t=0$ or $55-2t=0$. Subtracting $55$ from each side of the equation $55-2t=0$ yields $-2t=-55$. Dividing each side of this equation by $-2$ yields $t=\frac{55}{2}$. Therefore, the zeros of function $f$ are $0$ and $\frac{55}{2}$. The value that’s halfway between $0$ and $\frac{55}{2}$ can be found by calculating the average of $0$ and $\frac{55}{2}$, which is $\frac{0+\frac{55}{2}}{2}$, or $\frac{55}{4}$. It follows that the maximum of function $f$ occurs when $t=\frac{55}{4}$. Substituting $\frac{55}{4}$ for $t$ in the equation defining function $f$ yields $f(\frac{55}{4})=55(\frac{55}{4})-2(\frac{55}{4}{)}^2$, which is equivalent to $f(\frac{55}{4})=\frac{55^2}{4}-2(\frac{55^2}{4^2})$. Multiplying $\frac{55^2}{4}$ by $\frac{4}{4}$ in this equation to get a common denominator yields $f(\frac{55}{4})=4(\frac{55^2}{4^2})-2(\frac{55^2}{4^2})$, or $f(\frac{55}{4})=2(\frac{55^2}{4^2})$, which is equivalent to $f(\frac{55}{4})=2(\frac{55}{4}{)}^2$. Thus, the maximum value of $f(t)$ is $2(\frac{55}{4}{)}^2$. Since the equation defining $g(t)$ is $g(t)=f(t)+3$, the maximum value of $g(t)$ is $3$ greater than the maximum value of $f(t)$. It follows that the maximum value of $g(t)$ is $3+2(\frac{55}{4}{)}^2$.