# Graphs of polynomials

CCSS Math: HSF.IF.C.7c
Analyze polynomials in order to sketch their graph.

#### What you should be familiar with before taking this lesson

The end behavior of a function $f$ describes the behavior of its graph at the "ends" of the $x$-axis. Algebraically, end behavior is determined by the following two questions:
• As $x\rightarrow +\infty$, what does $f(x)$ approach?
• As $x\rightarrow -\infty$, what does $f(x)$ approach?
If this is new to you, we recommend that you check out our end behavior of polynomials article.
The zeros of a function $f$ correspond to the $x$-intercepts of its graph. If $f$ has a zero of odd multiplicity, its graph will cross the $x$-axis at that $x$ value. If $f$ has a zero of even multiplicity, its graph will touch the $x$-axis at that point.
If this is new to you, we recommend that you check out our zeros of polynomials article.

#### What you will learn in this lesson

In this lesson, we will use the above features in order to analyze and sketch graphs of polynomials. We will then use the sketch to find the polynomial's positive and negative intervals.

## Analyzing polynomial functions

We will now analyze several features of the graph of the polynomial $f(x)=(3x-2)(x+2)^2$.

### Finding the $y$-intercept

To find the $y$-intercept of the graph of $f$, we can find $f(0)$.
\begin{aligned} f(x)&=(3x-2)(x+2)^2 \\\\ f(\tealD0)&= (3(\tealD 0)-2)(\tealD0+2)^2\\ \\ f(0)&= (-2)(4)\\\\ f(0)&=-8 \end{aligned}
The $y$-intercept of the graph of $y=f(x)$ is $(0,-8)$.

### Finding the $x$-intercepts

To find the $x$-intercepts, we can solve the equation $f(x)=0$.
\begin{aligned} f(x)&=(3x-2)(x+2)^2 \\\\ \tealD 0&= (3x-2)(x+2)^2\\ \\ \end{aligned}
\begin{aligned}&\swarrow&\searrow\\\\ 3x-2&=0&\text{or}\quad x+2&=0&\small{\gray{\text{Zero product property}}}\\\\ x&=\dfrac{2}{3}&\text{or}\qquad x&=-2\end{aligned}
The $x$-intercepts of the graph of $y=f(x)$ are $\left(\dfrac23,0\right)$ and $(-2,0)$.
Our work also shows that $\dfrac 23$ is a zero of multiplicity $1$ and $-2$ is a zero of multiplicity $2$. This means that the graph will cross the $x$-axis at $\left (\dfrac 23, 0\right)$ and touch the $x$-axis at $(-2,0)$.

### Finding the end behavior

To find the end behavior of a function, we can examine the leading term when the function is written in standard form.
Let's write the equation in standard form.
\begin{aligned}f(x)&=(3x-2)(x+2)^2\\ \\ f(x)&=(3x-2)(x^2+4x+4)\\ \\ f(x)&=3x^3+12x^2+12x-2x^2-8x-8\\ \\ f(x)&=\goldD{3x^3}+10x^2+4x-8 \end{aligned}
The leading term of a polynomial is the first term when a polynomial is written in standard form. In other words, it is the term with the highest degree.
Since we only need to know the leading term (and not the remainder of the polynomial), we can find this by finding the product of the terms with the largest powers of $x$.
For example, in $f(x)=(\blueD{3x}-2)(\greenD{x}+2)^\greenD{2}$, we can multiply $\blueD{3x}$ from the first factor by $\greenD{x^2}$ in the second factor. This gives a leading term of $\goldD{3x^3}$.
The leading term of the polynomial is $\goldD{3x^3}$, and so the end behavior of function $f$ will be the same as the end behavior of $3x^3$.
Since the degree is odd and the leading coefficient is positive, the end behavior will be: as $x\rightarrow +\infty$, $f(x)\rightarrow +\infty$ and as $x\rightarrow -\infty$, $f(x)\rightarrow -\infty$.

### Sketching a graph

We can use what we've found above to sketch a graph of $y=f(x)$.
• As $x\rightarrow +\infty$, $f(x)\rightarrow +\infty$.
• As $x\rightarrow -\infty$, $f(x)\rightarrow -\infty$.
This means that in the "ends," the graph will look like the graph of $y=x^3$.
Now we can add what we know about the $x$-intercepts:
• The graph touches the $x$-axis at $(-2,0)$, since $-2$ is a zero of even multiplicity.
• The graph crosses the $x$-axis at $\left(\dfrac23,0\right)$, since $\dfrac23$ is a zero of odd multiplicity.
Finally, let's finish this process by plotting the $y$-intercept $(0,-8)$ and filling in the gaps with a smooth, continuous curve.
While we don't know exactly where the turning points are, we still have a good idea of the overall shape of the function's graph!

### Positive and negative intervals

Now that we have a sketch of $f$'s graph, it is easy to determine the intervals for which $f$ is positive, and those for which it is negative.
We see that $f$ is positive when $x>\dfrac{2}{3}$ and negative when $x<-2$ or $-2.

1) You will now work towards a sketch of $g(x)=(x+1)(x-2)(x+5)$ on your own.
a) What is the $y$-intercept of the graph of $g(x)=(x+1)(x-2)(x+5)$?
$(0$,
$)$

To find the $y$-intercept we can find $g(0)$.
\begin{aligned} g(x)&=(x+1)(x-2)(x+5) \\\\ g(\tealD 0)&= (\tealD 0+1)(\tealD0-2)(\tealD0+5)\\ \\ g(x) &= (1)(-2)(5)\\\\ g(x)&=-10 \end{aligned}
The $y$-intercept of the graph of $y=g(x)$ is $(0,-10)$.
b) What is the end behavior of the graph of $g(x)=(x+1)(x-2)(x+5)$?

To find the end behavior of a function, we can examine the leading term when the function is written in standard form.
Let's write the equation in standard form.
\begin{aligned}g(x)&=(x+1)(x-2)(x+5)\\ \\ g(x)&=(x+1)(x^2+3x-10)\\ \\ g(x)&=x^3+3x^2-10x+x^2+3x-10\\ \\ g(x)&=\goldD{x^3}+4x^2-7x-10 \end{aligned}
The leading term of the polynomial is $\goldD{x^3}$, and so the end behavior of function $g$ will be the same as the end behavior of $x^3$.
Since the degree is odd and the leading coefficient is positive, the end behavior will be: as $x\rightarrow +\infty$, $g(x)\rightarrow +\infty$ and as $x\rightarrow -\infty$, $g(x)\rightarrow -\infty$.
c) What are the $x$-intercepts of the graph of $g(x)=(x+1)(x-2)(x+5)$?

To find the $x$-intercepts, we can solve $g(x)=0$.
\begin{aligned} g(x)&=(x+1)(x-2)(x+5)\\\\ \tealD 0&= (x+1)(x-2)(x+5) \end{aligned}
\begin{aligned} \swarrow &\qquad \downarrow & \searrow\\\\ x+1=0 &\quad x-2=0 & \text{or}\quad & x+5=0 &\small{\gray{\text{Zero product property}}}\\\\ x=-1 &\quad x=2 & \text{or}\quad & x=-5 \end{aligned}
The $x$-intercepts of the graph of $y=g(x)$ are $\left(-1,0\right)$, $(2,0)$, and $(-5,0)$.
d) Which of the following graphs could be the graph of $g(x)=(x+1)(x-2)(x+5)$?

We can pull together the above information to sketch a graph of $y=g(x)$.
Sketching the $y$-intercept, end behavior of the polynomial, and what we know about the zeros would give us:
We could then fill in the gaps with a smooth, continuous curve to graph the polynomial.
This corresponds with graph $A$.
2) Which of the following could be the graph of $y=(2-x)(x+1)^2$

Let's find the end behavior and any intercepts of $y=(2-x)(x+1)^2$. Then, we can use this to sketch its graph.

#### $y$-intercept

To find the $y$-intercept, we can substitute $0$ in for $x$ and solve for $y$.
\begin{aligned}y&=(2-x)(x+1)^2\\\\ y&=(2-0)(0+1)^2\\\\ y&=2 \end{aligned}
The $y$-intercept is $(0,2)$.

#### $x$-intercept

To find the $x$-intercepts, we can substitute $0$ in for $y$ and solve for $x$.
\begin{aligned}y&=(2-x)(x+1)^2\\\\ 0&=(2-x)(x+1)^2\\\\ 2&-x=0 \qquad\text{or} \qquad x+1=0\\\\ x&=2~~\quad\qquad \text{or} \qquad x=-1 \end{aligned}
Since $-1$ is a zero of even multiplicity, the graph will touch the $x$-axis at $(-1,0)$, and since $2$ is a zero of odd multiplicity, the graph will cross the $x$-axis at $(2,0)$.

#### End behavior

To determine the end behavior, we can examine the leading term of the polynomial. If we write the polynomial in standard form, we get:
\begin{aligned}y&=(2-x)(x+1)^2\\\\ y&=(2-x)(x^2+2x+1)\\\\ y&=2x^2+4x+2-x^3-2x^2-x\\\\ y&=\goldD{-x^3}+3x+2 \end{aligned}
So in the end, the graph of $y=(2-x)(x+1)^2$ will look like the graph of $y=-x^3$.
The end behavior is: as $x \rightarrow +\infty$, $y \rightarrow -\infty$, and as $x \rightarrow -\infty$, $y \rightarrow +\infty$.

#### Sketching a graph

Putting this all together, we see that the only possible graph could be graph $D$.