SAT Question 4d7064a6: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. It's given that $f\left(x\right)=\left(x-10\right)\left(x+13\right)$, which can be rewritten as $f\left(x\right)=x^2+3x-130$. Since the coefficient of the $x^2$ -term is positive, the graph of $y=f\left(x\right)$ in the xy-plane opens upward and reaches its minimum value at its vertex. The x-coordinate of the vertex is the value of $x$ such that $f\left(x\right)$ reaches its minimum. For an equation in the form $f\left(x\right)=a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants, the x-coordinate of the vertex is $-\frac{b}{2a}$. For the equation $f\left(x\right)=x^2+3x-130$, $a=1$, $b=3$, and $c=-130$. It follows that the x-coordinate of the vertex is $-\frac{3}{2\left(1\right)}$, or $-\frac{3}{2}$. Therefore, $f\left(x\right)$ reaches its minimum when the value of $x$ is $-\frac{3}{2}$.

Alternate approach: The value of $x$ for the vertex of a parabola is the x-value of the midpoint between the two x-intercepts of the parabola. Since it's given that $f\left(x\right)=\left(x-10\right)\left(x+13\right)$, it follows that the two x-intercepts of the graph of $y=f\left(x\right)$ in the xy-plane occur when $x=10$ and $x=-13$, or at the points $\left(10,0\right)$ and $\left(-13,0\right)$. The midpoint between two points, $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$, is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. Therefore, the midpoint between $\left(10,0\right)$ and $\left(-13,0\right)$ is $\left(\frac{10+\left(-13\right)}{2},\frac{0+0}{2}\right)$, or $\left(-\frac{3}{2},0\right)$. It follows that $f\left(x\right)$ reaches its minimum when the value of $x$ is $-\frac{3}{2}$.