Graphing Polynomial Functions Practice

1. A polynomial $p$ is graphed.

   
   

   What could be the equation of $p$?

   Choose 1 answer:

   1. $p(x)=x^2(2x-5)(x+2)$
   2. $p(x)=x(2x-5)(x+2)^2$
   3. $p(x)=x^2(2x-5)^2(x+2)$

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

2. A polynomial $p$ is graphed.

   
   

   What could be the equation of $p$?

   Choose 1 answer:

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

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

3. A polynomial $p$ is graphed.

   
   

   What could be the equation of $p$?

   Choose 1 answer:

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

   4. $p(x)=2x^2(x-2)^2$

4. A polynomial $p$ is graphed.

   
   

   What could be the equation of $p$?

   Choose 1 answer:

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

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-graphs/x2ec2f6f830c9fb89:poly-intervals/e/poly-zeros-mult
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. 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:
   

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

   ---

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:

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

   ---

3. 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:

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

   ---

4. 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:

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