Question 670·Hard·Math

At $(-2,0)$, the curve touches and turns. The factor $x+2$ is $0$ there, so square it: $(x+2)^2$ doesn't change sign across $-2$. At $(1,0)$, the curve crosses, so use $x-1$ once. Touching calls for an even-powered factor; crossing calls for an odd-powered factor.

Multiplying by a coefficient, a number out front, changes the curve's heights without moving those zeros. So among the choices, the matching factor pattern has the form $f(x)=a(x-1)(x+2)^2$. The unknown multiplier $a$ still matters: the factors alone don't tell you the height at $x=0$.

The graph passes through $(0,-4)$. A $y$-intercept has input $0$, so $f(0)=-4$. Substitute $0$ for $x$ in the factored form: $-4=a(0+2)^2(0-1)$.

Type $-4\sim x_{1}(0+2)^2(0-1)$ in Desmos. Use $x_{1}$ for the unknown multiplier and $\sim$ to have Desmos fit its value. Under PARAMETERS, Desmos shows $x_{1}=1$. So $f(x)=(x+2)^2(x-1)$ has the required intercept and factor pattern. Choice A.