Question 61·Hard·Advanced Math

A zero is an input where $f(x)=0$. The graph crosses the $x$-axis at $-3$ and $4$, so it needs factors $x+3$ and $x-4$: each factor becomes $0$ at its matching input. One copy of each is enough for the least possible degree.

At $x=1$, the graph is tangent to the $x$-axis: it touches there instead of crossing. The factor $x-1$ must occur an even number of times, and two is the fewest. So the least-degree form is $f(x)=a(x+3)(x-4)(x-1)^2$, where $a$ is a nonzero integer multiplier.

The $y$-intercept $(0,24)$ says that input $0$ gives output $24$. Put $x=0$ into the factored form: $a(0+3)(0-4)(0-1)^2=24$. The zeros determined the factors; this equation determines the multiplier.

Type $a(0+3)(0-4)(0-1)^2\sim24$. The $\sim$ tells Desmos to fit $a$; it shows $a=-2$ under PARAMETERS. So $f(x)=-2(x+3)(x-4)(x-1)^2$ has the required crossings, touch, and $y$-intercept. Choice D.