Graphing Techniques & Graphing a Polynomial Function

Comprehensive Graph

We have discussed several characteristics of the graphs of polynomial functions that are useful for graphing the function by hand. A comprehensive graph of a polynomial function will show the following characteristics.

1.all x-intercepts (zeros) and the behavior of the graph at these zeros
the y-intercept
the sign of f(x) within the intervals formed by the x-intercepts
enough of the domain to show the end behavior

Graphing a Polynomial Function

Let $LaTeX: f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0,\:a_n\ne0$ $f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0,\:a_n\ne0$ , be a polynomial function of degree n. To sketch its graph, follow these steps.

Find the real zeros of f. Plot them as x-intercepts.
Find f(0) = a0. Plot this as the y-intercept.
Use end behavior, whether the graph crosses, bounces on, or wiggles through the x-axis at the x-intercepts, and selected points as necessary to complete the graph.

Example

Graph $LaTeX: f\left(x\right)=2x^3+5x^2-x-6$ $f\left(x\right)=2x^3+5x^2-x-6$

Solution

The possible rational zeros are $LaTeX: \pm1,\:\pm2,\:\pm3,\:\pm6,\:\pm\frac{1}{2},\:$ $\pm1,\:\pm2,\:\pm3,\:\pm6,\:\pm\frac{1}{2},\:$ and $LaTeX: \pm\frac{3}{2}$ $\pm\frac{3}{2}$ . Use synthetic division to show that 1 is a zero.

Showing 1 is a zero because there is a 0 remainder with synthetic division.PNG

Thus,

$LaTeX: f\left(x\right)=\left(x-1\right)\left(2x^2+7x+6\right)$ $f\left(x\right)=\left(x-1\right)\left(2x^2+7x+6\right)$

Factoring:

$LaTeX: f\left(x\right)=\left(x-1\right)\left(2x+3\right)\left(x+2\right)$ $f\left(x\right)=\left(x-1\right)\left(2x+3\right)\left(x+2\right)$

Set each factor equal to 0, then solve for x to find zeros. The three zeros are 1, $LaTeX: -\frac{3}{2}$ $-\frac{3}{2}$ , and LaTeX: -2. $-2.$

Since the polynomial $LaTeX: f\left(x\right)=2x^3+5x^2-x-6$ $f\left(x\right)=2x^3+5x^2-x-6$ has a positive leading coefficient and an odd degree, the end behavior is

An incomplete graph just showing the zeros and the end behavior.PNG

Each zero of f(x) occurs with multiplicity 1, meaning that the graph of f(x) will cross the x-axis at each of its zeros. Because the graph of a polynomial function has no breaks, gaps, or sudden jumps, we now have sufficient information to sketch the graph of f(x).

Begin sketching at either end of the graph with the appropriate end behavior, and draw a smooth curve that crosses the x-axis at each zero, has a turning point between successive zeros, and passes through the
y-intercept as shown.

Now a complete graph.PNG

Additional points may be used to verify whether the graph is above or below the x-axis between the zeros and to add detail to the sketch of the graph. The zeros divide the x-axis into four intervals: $LaTeX: \left(-\infty,\:-2\right),\left(-2,\:-\frac{3}{2}\right),\left(-\frac{3}{2},1\right),\left(1,\infty\right)$ $\left(-\infty,\:-2\right),\left(-2,\:-\frac{3}{2}\right),\left(-\frac{3}{2},1\right),\left(1,\infty\right)$ .

Select an x-value as a test point in each interval, and substitute it into the equation for f(x) to determine additional points on the graph. A typical selection of test points and results of the tests are shown in this table:

Interval	Test Point	Value of $LaTeX: f\left(x\right)$ $f\left(x\right)$	Sign of $LaTeX: f\left(x\right)$ $f\left(x\right)$	Graph Above or Below x-Axis
$LaTeX: \left(-\infty,\:-2\right)$ $\left(-\infty,\:-2\right)$	-3	-12	Negative	Below
$LaTeX: \left(-2,-\frac{3}{2}\right)$ $\left(-2,-\frac{3}{2}\right)$	$LaTeX: -\frac{7}{4}$ $-\frac{7}{4}$	$LaTeX: \frac{11}{32}$ $\frac{11}{32}$	Positive	Above
$LaTeX: \left(-\frac{3}{2},1\right)$ $\left(-\frac{3}{2},1\right)$	0	-6	Negative	Below
$LaTeX: \left(1,\infty\right)$ $\left(1,\infty\right)$	2	28	Positive	Above

The sketch could be improved by plotting the points found in each interval in the table.

Now a complete graph.PNG

Example

Graph $LaTeX: f\left(x\right)=-\left(x-1\right)\left(x-3\right)\left(x+2\right)^2$ $f\left(x\right)=-\left(x-1\right)\left(x-3\right)\left(x+2\right)^2$

Solution

Since the polynomial is given in factored form, the zeros can be determined by inspection. They are 1, 3, and −2. Plot these as x-intercepts of the graph of f(x).

To find the y-intercept, find f(0).

$LaTeX: f\left(0\right)=-\left(0-1\right)\left(0-3\right)\left(0+2\right)^2=-\left(-1\right)\left(-3\right)\left(4\right)=-12$ $f\left(0\right)=-\left(0-1\right)\left(0-3\right)\left(0+2\right)^2=-\left(-1\right)\left(-3\right)\left(4\right)=-12$

Plot $LaTeX: \left(0,-12\right)$ $\left(0,-12\right)$ on the x-axis.

The dominating term of f(x) can be found by multiplying the factors and identifying the term of greatest degree. Here it is $LaTeX: -\left(x\right)\left(x\right)\left(x\right)^2=-x^4$ $-\left(x\right)\left(x\right)\left(x\right)^2=-x^4$ , indicating that the end behavior of the graph is

Because 1 and 3 are zeros of multiplicity 1, the graph will cross the x-axis at these zeros. The graph of f(x) will touch the x-axis at LaTeX: -2 $-2$ and then change direction because it is a zero of even multiplicity. (The multiplicity is the exponent on the associated factor in f(x))

Begin at either end of the graph with the appropriate end behavior and draw a smooth curve that crosses the x-axis at 1 and 3 and that touches the x-axis at −2, then turns and changes direction. The graph will also pass through the y-intercept −12.

Using test points within intervals formed by the x-intercepts is a good way to add detail to the graph and verify the accuracy of the sketch. A typical selection of test points is (−3, −24), (−1, −8), (2, 16), and (4, −108).

A comprehensive graph of the function.PNG