Chapter 4.1, Problem 29E

To determine

To sketch: the graph of the given polynomial function.

Explanation of Solution

Given information:

A polynomial function is given as

  $h\left(x\right)=x^4-2x^3+3x$

Concept used:

A polynomial function is of the form

  $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0$

Terms of a polynomial function should be arranged in descending order according to its degree to express it in a standard form and degree of each term should be a positive integer or whole number. The coefficients should be real numbers.

Leading coefficient of a polynomial function is the coefficient of the leading term.

Degree of the polynomial is the degree of leading term or the height degree in the polynomial function.

For the polynomial $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0$ the leading coefficient is $a_n$ and the degree is $n$ .

The end behavior can describe the graph of a polynomial function as  $x$  approaches $+\infty$ or as  $x$  approaches $-\infty$ .

The end behavior of a polynomial function can be determined by the leading coefficient and the degree of the polynomial.

If degree is even and leading coefficient is negative.

  $f\left(x\right)\to -\infty$ as $x\to -\infty$ and $f\left(x\right)\to -\infty$ as $x\to +\infty$ .

If degree is odd and leading coefficient is negative.

  $f\left(x\right)\to \infty$ as $x\to -\infty$ and $f\left(x\right)\to -\infty$ as $x\to +\infty$ .

If degree is odd and leading coefficient is positive.

  $f\left(x\right)\to -\infty$ as $x\to -\infty$ and $f\left(x\right)\to \infty$ as $x\to +\infty$ .

If degree is even and leading coefficient is positive.

  $f\left(x\right)\to \infty$ as $x\to -\infty$ and $f\left(x\right)\to \infty$ as $x\to +\infty$ .

Calculation:

Consider the given polynomial function.

  $h\left(x\right)=x^4-2x^3+3x$

Now, tabulate some points as shown below:

$x$	$h\left(x\right)=x^4-2x^3+3x$	$\left(x,h\left(x\right)\right)$
$0$	$h\left(0\right)={\left(0\right)}^4-2{\left(0\right)}^3+3\left(0\right)=0$	$\left(0,0\right)$
$1$	$h\left(1\right)={\left(1\right)}^4-2{\left(1\right)}^3+3\left(1\right)=2$	$\left(1,2\right)$
$-1$	$h\left(-1\right)={\left(1\right)}^4-2{\left(-1\right)}^3+3\left(1\right)=6$	$\left(-1,6\right)$
$2$	$h\left(2\right)={\left(2\right)}^4-2{\left(2\right)}^3+3\left(2\right)=6$	$\left(2,6\right)$
$-2$	$h\left(-2\right)={\left(-2\right)}^4-2{\left(-2\right)}^3+3\left(-2\right)=26$	$\left(-2,26\right)$

Graph:

Plot the points and draw a smooth curve to sketch the graph.

  

Interpretation:

Degree is even and leading coefficient is positive.

  $h\left(x\right)\to \infty$ as $x\to -\infty$ and $h\left(x\right)\to \infty$ as $x\to +\infty$ .