SAT Question a1e65979: Nonlinear Functions answer and explanation with Desmos solution

Choice C is correct. It’s given that function $f$ is a quadratic function, and in the xy-plane, the graph of $y=f(x)$ has a vertex at $(1,9)$. In the xy-plane, the graph of a quadratic function with a vertex at $(h,k)$ has an equation of the form $y=a(x-h{)}^2+k$, where $a$ is a constant. Therefore, function $f$ can be written as $f\left(x\right)=a(x-1{)}^2+9$. It’s also given that the graph of $y=f(x)$ passes through the point $(2,33)$. Substituting $2$ for $x$ and $33$ for $f(x)$ in the equation $f\left(x\right)=a(x-1{)}^2+9$ yields $33=a(2-1{)}^2+9$, or $33=a+9$. Subtracting $9$ from both sides of this equation yields $24=a$. Substituting $24$ for $a$ in the equation $f\left(x\right)=a(x-1{)}^2+9$ yields $f(x)=24(x-1{)}^2+9$. The value of $f(-2)$ can be found by substituting $-2$ for $x$ in function $f$. Therefore, $f\left(-2\right)=24(-2-1{)}^2+9$, which yields $f(-2)=24(9)+9$, or $f(-2)=225$. The value of $f(0)$ can be found by substituting $0$ for $x$ in function $f$. Therefore, $f\left(0\right)=24(0-1{)}^2+9$, which yields $f(0)=24+9$, or $f(0)=33$. It follows that $f(-2)-f(0)=225-33$, or $f(-2)-f(0)=192$.