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f ( t ) = 2 ( t 1 ) ( t + 2 ) ( t 3 )

y -intercept is ( 0 , 12 ) , t -intercepts are ( 1 , 0 ) ; ( 2 , 0 ) ; and  ( 3 , 0 ) .

g ( n ) = 2 ( 3 n 1 ) ( 2 n + 1 )

f ( x ) = x 4 16

y -intercept is ( 0 , 16 ) . x -intercepts are ( 2 , 0 ) and ( 2 , 0 ) .

f ( x ) = x 3 + 27

f ( x ) = x ( x 2 2 x 8 )

y -intercept is ( 0 , 0 ) . x -intercepts are ( 0 , 0 ) , ( 4 , 0 ) , and ( 2 ,   0 ) .

f ( x ) = ( x + 3 ) ( 4 x 2 1 )

Graphical

For the following exercises, determine the least possible degree of the polynomial function shown.

Graph of an odd-degree polynomial.

3

Graph of an even-degree polynomial.
Graph of an odd-degree polynomial.

5

Graph of an odd-degree polynomial.
Graph of an odd-degree polynomial.

3

Graph of an even-degree polynomial.
Graph of an odd-degree polynomial.

5

Graph of an even-degree polynomial.

For the following exercises, determine whether the graph of the function provided is a graph of a polynomial function. If so, determine the number of turning points and the least possible degree for the function.

Graph of an odd-degree polynomial.

Yes. Number of turning points is 2. Least possible degree is 3.

Graph of an equation.
Graph of an even-degree polynomial.

Yes. Number of turning points is 1. Least possible degree is 2.

Graph of an odd-degree polynomial.
Graph of an odd-degree polynomial.

Yes. Number of turning points is 0. Least possible degree is 1.

Graph of an equation.

No.

Graph of an odd-degree polynomial.

Yes. Number of turning points is 0. Least possible degree is 1.

Numeric

For the following exercises, make a table to confirm the end behavior of the function.

f ( x ) = x 3

f ( x ) = x 4 5 x 2

x f ( x )
10 9,500
100 99,950,000
–10 9,500
–100 99,950,000

as x , f ( x ) , as x , f ( x )

f ( x ) = x 2 ( 1 x ) 2

f ( x ) = ( x 1 ) ( x 2 ) ( 3 x )

x f ( x )
10 –504
100 –941,094
–10 1,716
–100 1,061,106

as x , f ( x ) , as x , f ( x )

f ( x ) = x 5 10 x 4

Technology

For the following exercises, graph the polynomial functions using a calculator. Based on the graph, determine the intercepts and the end behavior.

f ( x ) = x 3 ( x 2 )

Graph of f(x)=x^3(x-2).

The y - intercept is ( 0 ,   0 ) . The x - intercepts are ( 0 ,   0 ) ,   ( 2 ,   0 ) . As x , f ( x ) , as x , f ( x )

f ( x ) = x ( x 3 ) ( x + 3 )

f ( x ) = x ( 14 2 x ) ( 10 2 x )

Graph of f(x)=x(14-2x)(10-2x).

The y - intercept is ( 0 , 0 ) . The x - intercepts are ( 0 ,   0 ) ,   ( 5 ,   0 ) ,   ( 7 ,   0 ) . As x , f ( x ) , as x , f ( x )

f ( x ) = x ( 14 2 x ) ( 10 2 x ) 2

f ( x ) = x 3 16 x

The y - intercept is ( 0 ,   0 ) . The x - intercept is ( 4 ,   0 ) ,   ( 0 ,   0 ) ,   ( 4 ,   0 ) . A s x , f ( x ) , as x , f ( x )

f ( x ) = x 3 27

f ( x ) = x 4 81

Graph of f(x)=x^3-27.

The y - intercept is ( 0 ,   81 ) . The x - intercept are ( 3 ,   0 ) ,   ( 3 ,   0 ) . As x , f ( x ) , as x , f ( x )

f ( x ) = x 3 + x 2 + 2 x

f ( x ) = x 3 2 x 2 15 x

Graph of f(x)=-x^3+x^2+2x.

The y - intercept is ( 0 ,   0 ) . The x - intercepts are ( 3 ,   0 ) ,   ( 0 ,   0 ) ,   ( 5 ,   0 ) . As x , f ( x ) , as x , f ( x )

f ( x ) = x 3 0.01 x

Extensions

For the following exercises, use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is 1 or –1. There may be more than one correct answer.

The y - intercept is ( 0 , 4 ) . The x - intercepts are ( 2 , 0 ) , ( 2 , 0 ) . Degree is 2.

End behavior: as x , f ( x ) , as x , f ( x ) .

f ( x ) = x 2 4

The y - intercept is ( 0 , 9 ) . The x - intercepts are ( 3 , 0 ) , ( 3 , 0 ) . Degree is 2.

End behavior: as x , f ( x ) , as x , f ( x ) .

The y - intercept is ( 0 , 0 ) . The x - intercepts are ( 0 , 0 ) , ( 2 , 0 ) . Degree is 3.

End behavior: as x , f ( x ) , as x , f ( x ) .

f ( x ) = x 3 4 x 2 + 4 x

The y - intercept is ( 0 , 1 ) . The x - intercept is ( 1 , 0 ) . Degree is 3.

End behavior: as x , f ( x ) , as x , f ( x ) .

The y - intercept is ( 0 , 1 ) . There is no x - intercept. Degree is 4.

End behavior: as x , f ( x ) , as x , f ( x ) .

f ( x ) = x 4 + 1

Real-world applications

For the following exercises, use the written statements to construct a polynomial function that represents the required information.

An oil slick is expanding as a circle. The radius of the circle is increasing at the rate of 20 meters per day. Express the area of the circle as a function of d , the number of days elapsed.

A cube has an edge of 3 feet. The edge is increasing at the rate of 2 feet per minute. Express the volume of the cube as a function of m , the number of minutes elapsed.

V ( m ) = 8 m 3 + 36 m 2 + 54 m + 27

A rectangle has a length of 10 inches and a width of 6 inches. If the length is increased by x inches and the width increased by twice that amount, express the area of the rectangle as a function of x .

An open box is to be constructed by cutting out square corners of x - inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of x .

V ( x ) = 4 x 3 32 x 2 + 64 x

A rectangle is twice as long as it is wide. Squares of side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width ( x ).

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Source:  OpenStax, Essential precalculus, part 1. OpenStax CNX. Aug 26, 2015 Download for free at http://legacy.cnx.org/content/col11871/1.1
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