How is the behavior of the graph at a zero different when the factor appears once (like $(x-1)$) versus when it appears squared (like $(x-1)^2$)? Which kind of behavior corresponds to being tangent to the x-axis?

Using your answer about multiplicities, write a general factored form $f(x)=a\cdot(\text{factors in }x)$ with a single unknown constant $a$ in front.

Plug $x=0$ and $f(0)=24$ into your general factored form. What equation do you get for $a$, and what value of $a$ does it give?

Type each option into Desmos as a separate function, for example f(x)=2(x+3)(x-4)(x-1)^2, then g(x)=-2(x+3)(x-4)^2(x-1), etc. Turn them on and off so you can see one graph at a time clearly.

For each graph, look at the x-intercepts. Verify that the graph has zeros at $x=-3$, $x=1$, and $x=4$. Then zoom in near each zero to see whether the graph crosses the x-axis or just touches and turns around there. The correct function will cross at $x=-3$ and $x=4$ and be tangent at $x=1$.

For any remaining candidate that matches the x-intercept behavior, either:

• Click where $x=0$ on the graph, or
• Type the expression with 0 substituted for x (like f(0)) into Desmos.

Read off the y-value. The correct function is the one whose value at $x=0$ is 24 (so the graph passes through the point $(0,24)$).

The graph crosses the x-axis at $x=-3$ and $x=4$, so $x=-3$ and $x=4$ are zeros of $f(x)$.

• That means $(x+3)$ and $(x-4)$ must be factors of $f(x)$.

Because the graph is tangent to the x-axis at $x=1$, $x=1$ is also a zero, so $(x-1)$ is a factor as well.

For polynomials:

• If the graph crosses the x-axis at a zero, that zero has odd multiplicity, usually 1 in the least-degree case.
• If the graph is tangent to the x-axis at a zero, that zero has even multiplicity, usually 2 in the least-degree case.

So, to get least possible degree:

• Use multiplicity 1 for the zeros where the graph crosses, at $x=-3$ and $x=4$.
• Use multiplicity 2 for the zero where the graph is tangent, at $x=1$.

That gives a general form

where $a$ is some nonzero integer constant (to keep integer coefficients).

The y-intercept is $(0,24)$, which means $f(0)=24$.

Substitute $x=0$ into the general form:

We are told $f(0)=24$, so

Solve for $a$ to get $a=-2$.

Substitute $a=-2$ into the factored form:

This polynomial has integer coefficients, least possible degree, zeros at $-3$ and $4$ where it crosses, a double zero at $1$ where it is tangent, and y-intercept $(0,24)$. Therefore, the correct choice is $-2(x+3)(x-4)(x-1)^2$.