Nonlinear functions SAT question

Choice D is correct. It's given that $f(x)=(x-14)(x+19)$, which can be rewritten as $f(x)=x^2+5x-266$. Since the coefficient of the $x^2$-term is positive, the graph of $y=f(x)$ 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 $f(x)$ reaches its minimum. For an equation in the form $f(x)=a{x}^2+bx+c$, where $a$, $b$, and $c$ are constants, the x-coordinate of the vertex is $-\frac{b}{2a}$. For the equation $f(x)=x^2+5x-266$, $a=1$, $b=5$, and $c=-266$. It follows that the x-coordinate of the vertex is $-\frac{5}{2(1)}$, or $-\frac{5}{2}$. Therefore, $f(x)$ reaches its minimum when the value of $x$ is $-\frac{5}{2}$.

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 $f(x)=(x-14)(x+19)$, it follows that the two x-intercepts of the graph of $y=f(x)$ in the xy-plane occur when $x=14$ and $x=-19$, or at the points $(14,0)$ and $(-19,0)$. The midpoint between two points, $(x_1,y_1)$ and $(x_2,y_2)$, is $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$. Therefore, the midpoint between $(14,0)$ and $(-19,0)$ is $(\frac{14+(-19)}{2},\frac{0+0}{2})$, or $(-\frac{5}{2},0)$. It follows that $f(x)$ reaches its minimum when the value of $x$ is $-\frac{5}{2}$.

Choice A is incorrect. This is the y-coordinate of the y-intercept of the graph of $y=f(x)$ in the xy-plane.

Choice B is incorrect. This is one of the x-coordinates of the x-intercepts of the graph of $y=f(x)$ in the xy-plane.

Choice C is incorrect and may result from conceptual or calculation errors.