Analyze polynomials in order to sketch their graph.

What you should be familiar with before taking this lesson

The end behavior of a function ff describes the behavior of its graph at the "ends" of the xx-axis. Algebraically, end behavior is determined by the following two questions:
  • As x+x\rightarrow +\infty, what does f(x)f(x) approach?
  • As xx\rightarrow -\infty, what does f(x)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 ff correspond to the xx-intercepts of its graph. If ff has a zero of odd multiplicity, its graph will cross the xx-axis at that xx value. If ff has a zero of even multiplicity, its graph will touch the xx-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)=(3x2)(x+2)2f(x)=(3x-2)(x+2)^2.

Finding the yy-intercept

To find the yy-intercept of the graph of ff, we can find f(0)f(0).
f(x)=(3x2)(x+2)2f(0)=(3(0)2)(0+2)2f(0)=(2)(4)f(0)=8\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 yy-intercept of the graph of y=f(x)y=f(x) is (0,8)(0,-8).

Finding the xx-intercepts

To find the xx-intercepts, we can solve the equation f(x)=0f(x)=0.
f(x)=(3x2)(x+2)20=(3x2)(x+2)2\begin{aligned} f(x)&=(3x-2)(x+2)^2 \\\\ \tealD 0&= (3x-2)(x+2)^2\\ \\ \end{aligned}
3x2=0orx+2=0Zero product propertyx=23orx=2\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 xx-intercepts of the graph of y=f(x)y=f(x) are (23,0)\left(\dfrac23,0\right) and (2,0)(-2,0).
Our work also shows that 23\dfrac 23 is a zero of multiplicity 11 and 2-2 is a zero of multiplicity 22. This means that the graph will cross the xx-axis at (23,0)\left (\dfrac 23, 0\right) and touch the xx-axis at (2,0)(-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.
f(x)=(3x2)(x+2)2f(x)=(3x2)(x2+4x+4)f(x)=3x3+12x2+12x2x28x8f(x)=3x3+10x2+4x8\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 xx.
For example, in f(x)=(3x2)(x+2)2f(x)=(\blueD{3x}-2)(\greenD{x}+2)^\greenD{2}, we can multiply 3x\blueD{3x} from the first factor by x2\greenD{x^2} in the second factor. This gives a leading term of 3x3\goldD{3x^3}.
The leading term of the polynomial is 3x3\goldD{3x^3}, and so the end behavior of function ff will be the same as the end behavior of 3x33x^3.
Since the degree is odd and the leading coefficient is positive, the end behavior will be: as x+x\rightarrow +\infty, f(x)+f(x)\rightarrow +\infty and as xx\rightarrow -\infty, f(x)f(x)\rightarrow -\infty.

Sketching a graph

We can use what we've found above to sketch a graph of y=f(x)y=f(x).
Let's start with end behavior:
  • As x+x\rightarrow +\infty, f(x)+f(x)\rightarrow +\infty.
  • As xx\rightarrow -\infty, f(x)f(x)\rightarrow -\infty.
This means that in the "ends," the graph will look like the graph of y=x3y=x^3.
Now we can add what we know about the xx-intercepts:
  • The graph touches the xx-axis at (2,0)(-2,0), since 2-2 is a zero of even multiplicity.
  • The graph crosses the xx-axis at (23,0)\left(\dfrac23,0\right), since 23\dfrac23 is a zero of odd multiplicity.
Finally, let's finish this process by plotting the yy-intercept (0,8)(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 ff's graph, it is easy to determine the intervals for which ff is positive, and those for which it is negative.
We see that ff is positive when x>23x>\dfrac{2}{3} and negative when x<2x<-2 or 2<x<23-2<x<\dfrac23.

Check your understanding

1) You will now work towards a sketch of g(x)=(x+1)(x2)(x+5)g(x)=(x+1)(x-2)(x+5) on your own.
a) What is the yy-intercept of the graph of g(x)=(x+1)(x2)(x+5)g(x)=(x+1)(x-2)(x+5)?
(0(0,
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
))

To find the yy-intercept we can find g(0)g(0).
g(x)=(x+1)(x2)(x+5)g(0)=(0+1)(02)(0+5)g(x)=(1)(2)(5)g(x)=10\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 yy-intercept of the graph of y=g(x)y=g(x) is (0,10)(0,-10).
b) What is the end behavior of the graph of g(x)=(x+1)(x2)(x+5)g(x)=(x+1)(x-2)(x+5)?
Choose 1 answer:
Choose 1 answer:

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.
g(x)=(x+1)(x2)(x+5)g(x)=(x+1)(x2+3x10)g(x)=x3+3x210x+x2+3x10g(x)=x3+4x27x10\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 x3\goldD{x^3}, and so the end behavior of function gg will be the same as the end behavior of x3x^3.
Since the degree is odd and the leading coefficient is positive, the end behavior will be: as x+x\rightarrow +\infty, g(x)+g(x)\rightarrow +\infty and as xx\rightarrow -\infty, g(x)g(x)\rightarrow -\infty.
c) What are the xx-intercepts of the graph of g(x)=(x+1)(x2)(x+5)g(x)=(x+1)(x-2)(x+5)?
Choose 1 answer:
Choose 1 answer:

To find the xx-intercepts, we can solve g(x)=0g(x)=0.
g(x)=(x+1)(x2)(x+5)0=(x+1)(x2)(x+5)\begin{aligned} g(x)&=(x+1)(x-2)(x+5)\\\\ \tealD 0&= (x+1)(x-2)(x+5) \end{aligned}
x+1=0x2=0orx+5=0Zero product propertyx=1x=2orx=5\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 xx-intercepts of the graph of y=g(x)y=g(x) are (1,0)\left(-1,0\right), (2,0)(2,0), and (5,0)(-5,0).
d) Which of the following graphs could be the graph of g(x)=(x+1)(x2)(x+5)g(x)=(x+1)(x-2)(x+5)?
Choose 1 answer:
Choose 1 answer:

We can pull together the above information to sketch a graph of y=g(x)y=g(x).
Sketching the yy-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 AA.
2) Which of the following could be the graph of y=(2x)(x+1)2y=(2-x)(x+1)^2
Choose 1 answer:
Choose 1 answer:

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

yy-intercept

To find the yy-intercept, we can substitute 00 in for xx and solve for yy.
y=(2x)(x+1)2y=(20)(0+1)2y=2\begin{aligned}y&=(2-x)(x+1)^2\\\\ y&=(2-0)(0+1)^2\\\\ y&=2 \end{aligned}
The yy-intercept is (0,2)(0,2).

xx-intercept

To find the xx-intercepts, we can substitute 00 in for yy and solve for xx.
y=(2x)(x+1)20=(2x)(x+1)22x=0orx+1=0x=2  orx=1\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 1-1 is a zero of even multiplicity, the graph will touch the xx-axis at (1,0)(-1,0), and since 22 is a zero of odd multiplicity, the graph will cross the xx-axis at (2,0)(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:
y=(2x)(x+1)2y=(2x)(x2+2x+1)y=2x2+4x+2x32x2xy=x3+3x+2\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=(2x)(x+1)2y=(2-x)(x+1)^2 will look like the graph of y=x3y=-x^3.
The end behavior is: as x+x \rightarrow +\infty, yy \rightarrow -\infty, and as xx \rightarrow -\infty, y+y \rightarrow +\infty.

Sketching a graph

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