Graphing Polynomial Functions Practice

All the options have the same linear factors: $x$,
$(2x-5)$, and $(x+2)$. This makes sense, as the graph has zeros at $x=0$,  $x=\dfrac{5}2$, and $x=-2$.

The difference between the options is in the  of the zeros. To determine which option is correct, we need to think about multiplicity:

• If the multiplicity of a zero is an odd number, the graph will cross the $x$-axis at that zero.
• If the multiplicity of a zero is an even number, the graph will only touch the $x$-axis at that zero.

We see that the graph crosses the $x$-axis at $x=-2$ and $x=\dfrac{5}2$. So their multiplicities must be odd numbers.

We see that the graph touches the $x$-axis at $x=0$. So its multiplicity must be an even number.

In conclusion, this is the correct answer:

$p(x)=x^2(2x-5)(x+2)$

All the options have the same linear factors: $(x+3)$, $(x+1)$, and $(2x-1)$. This makes sense, as the graph has zeros at $x=-3$, $x=-1$, and $x=\dfrac{1}2$.

The difference between the options is in the  of the zeros. To determine which option is correct, we need to think about multiplicity:

• If the multiplicity of a zero is an odd number, the graph will cross the $x$-axis at that zero.
• If the multiplicity of a zero is an even number, the graph will only touch the $x$-axis at that zero.

We see that the graph crosses the $x$-axis at $x=-3$ and  $x=\dfrac{1}2$. So their multiplicities must be odd numbers.

We see that the graph touches the $x$-axis at $x=-1$. So its multiplicity must be an even number.

In conclusion, this is the correct answer:

$p(x)=(x+3)(x+1)^2(2x-1)$

All the options have the same linear factors: $2x$ and $(x-2)$. This makes sense, as the graph has zeros at $x=0$ and $x=2$.

The difference between the options is in the  of the zeros. To determine which option is correct, we need to think about multiplicity:

• If the multiplicity of a zero is an odd number, the graph will cross the $x$-axis at that zero.
• If the multiplicity of a zero is an even number, the graph will only touch the $x$-axis at that zero.

We see that the graph crosses the $x$-axis at $x=2$. So its multiplicity must be an odd number.

We see that the graph touches the $x$-axis at $x=0$. So its multiplicity must be an even number.

In conclusion, this is the correct answer:

$p(x)=2x^2(x-2)^3$

All the options have the same linear factors: $(x-1)$,
$(2x+7)$, and $(x+1)$. This makes sense, as the graph has zeros at $x=1$, $x=-\dfrac{7}2$, and $x=-1$.

The difference between the options is in the  of the zeros. To determine which option is correct, we need to think about multiplicity:

• If the multiplicity of a zero is an odd number, the graph will cross the $x$-axis at that zero.
• If the multiplicity of a zero is an even number, the graph will only touch the $x$-axis at that zero.

We see that the graph crosses the $x$-axis at $x=1$. So its multiplicity must be an odd number.

We see that the graph touches the $x$-axis at $x=-\dfrac{7}2$ and $x=-1$. So their multiplicities must be even numbers.

In conclusion, this is the correct answer:

$p(x)=(x-1)(2x+7)^2(x+1)^2$