SAT Question bd4d0e0c: Nonlinear Functions answer and explanation with Desmos solution

Choice A is correct. The value of $f\left(0\right)$ is the value of $f\left(x\right)$ when $x=0$. The function $f$ is defined by $f\left(x\right)=10x^2-32x-152$. Substituting $0$ for $x$ in this equation yields $f\left(0\right)=10{\left(0\right)}^2-32\left(0\right)-152$. This equation can be rewritten as $f\left(0\right)=10\left(0\right)-0-152$, or $f\left(0\right)=-152$. Therefore, the value of $f\left(0\right)$ is $-152$.