SAT Question 6d9e01a2: Nonlinear Functions answer and explanation with Desmos solution

The given equation can be rewritten in the form $f\left(x\right)=a{\left(x-h\right)}^2+k$, where $a$, $h$, and $k$ are constants. When $a>0$, $h$ is the value of $x$ for which $f\left(x\right)$ reaches its minimum. The given equation can be rewritten as $f\left(x\right)=4\left(x^2-\frac{50}{4}x\right)+126$, which is equivalent to $f\left(x\right)=4\left(x^2-\frac{50}{4}x+{\left(\frac{50}{8}\right)}^2-{\left(\frac{50}{8}\right)}^2\right)+126$. This equation can be rewritten as $f\left(x\right)=4\left({\left(x-\frac{50}{8}\right)}^2-{\left(\frac{50}{8}\right)}^2\right)+126$, or $f\left(x\right)=4{\left(x-\frac{50}{8}\right)}^2-4{\left(\frac{50}{8}\right)}^2+126$, which is equivalent to $f\left(x\right)=4{\left(x-\frac{25}{4}\right)}^2-\frac{121}{4}$. Therefore, $h=\frac{25}{4}$, so the value of $x$ for which $f\left(x\right)$ reaches its minimum is $\frac{25}{4}$. Note that 25/4, 6.25, and 6.250 are examples of ways to enter a correct answer.