SAT Question 8e1da169: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. It's given that $f\left(x\right)=\left(x-44\right)\left(x-46\right)$, which can be rewritten as $f\left(x\right)=x^2-90x+2,024$. 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. 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 $f\left(x\right)=x^2-90x+2,024$, $a=1$, $b=-90$, and $c=2,024$. It follows that the x-coordinate of the vertex is $-\frac{\left(-90\right)}{2\left(1\right)}$, or $45$. Therefore, $f\left(x\right)$ reaches its minimum when the value of $x$ is $45$.