SAT Question f5e8ccf1: Nonlinear Functions answer and explanation with Desmos solution

Choice A is incorrect because this point is an $x$-intercept of the graph of the function $f$ in the $xy$-plane. By definition, an $x$-intercept is a point on the graph of the form $(k,0)$, where $k$ is a constant. Substituting $-4$ for $x$ and $0$ for $f(x)$ in the given equation yields $0=(-4+4)(-4-1)(2(-4)-3)$, or $0=0$. Since this is a true statement, the point $(-4,0)$ lies on the graph of the function $f$ and is an $x$-intercept of the graph.

Choice C is incorrect because this point is an $x$-intercept of the graph of the function $f$ in the $xy$-plane. By definition, an $x$-intercept is a point on the graph of the form $(k,0)$, where $k$ is a constant. Substituting $1$ for $x$ and $0$ for $f(x)$ in the given equation yields $0=(1+4)(1-1)(2\times 1-3)$, or $0=0$. Since this is a true statement, the point $(1,0)$ lies on the graph of the function $f$ and is an $x$-intercept of the graph.

Choice D is incorrect because this point is an $x$-intercept of the graph of the function $f$ in the $xy$-plane. By definition, an $x$-intercept is a point on the graph of the form $(k,0)$, where $k$ is a constant. Substituting $\frac{3}{2}$ for $x$ and $0$ for $f(x)$ in the given equation yields $0=\left(\frac{3}{2}+4\right)\left(\frac{3}{2}-1\right)\left(2\times \frac{3}{2}-3\right)$, or $0=0$. Since this is a true statement, the point $\left(\frac{3}{2},0\right)$ lies on the graph of the function $f$ and is an $x$-intercept of the graph.