SAT Question 083ef63a: Nonlinear Functions answer and explanation with Desmos solution

The value of $x$ for which $f\left(x\right)$ reaches its minimum can be found by rewriting the given equation in the form $f\left(x\right)={\left(x-h\right)}^2+k$, where $f\left(x\right)$ reaches its minimum, $k$, when the value of $x$ is $h$. The given equation, $f\left(x\right)=\left(x-2\right)\left(x+15\right)$, can be rewritten as $f\left(x\right)=x^2+13x-30$. By completing the square, this equation can be rewritten as $f\left(x\right)=\left(x^2+13x+{\left(\frac{13}{2}\right)}^2\right)-30-{\left(\frac{13}{2}\right)}^2$, which is equivalent to $f\left(x\right)={\left(x+\frac{13}{2}\right)}^2-\frac{289}{4}$, or $f\left(x\right)={\left(x-\left(-\frac{13}{2}\right)\right)}^2-\frac{289}{4}$. Therefore, $f\left(x\right)$ reaches its minimum when the value of $x$ is $-\frac{13}{2}$. Note that -13/2, -6.5, and -6.50 are examples of ways to enter a correct answer.