polynomial and rational functions

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Polynomial andRational Functions

Chapter 3


3.1 Polynomial Functions

and Modeling

Determine the behavior of the graph of a polynomial function using the leading-term test.

Factor polynomial functions and find the zeros of their multiplicities. Use a graphing calculator to graph a polynomial

function and find its real-number zeros, relative maximum and minimum values, and domain and range.

Solve applied problems using polynomial models; fit linear, quadratic, power, cubic, and quartic polynomial functions to data.


Polynomial Function

A polynomial function P is given by

where the coefficients an, an - 1, …, a1, a0 are real numbers and the exponents are whole numbers.

1 21 2 1 0( ) ... ,n n

n nP x a x a x a x a x a

f(x) = x4 3.2x3 + 0.1x4Quartic

f(x) = x3 +2x2 x + 113Cubic

f(x) = 4x2 x + 92Quadratic

f(x) = 3x + 11Linear

f(x) = 40Constant

ExampleDegreePolynomial Function


Quadratic Function


Cubic Function


Examples of Polynomial Functions


Examples of Nonpolynomial Functions


The Leading-Term Test


Example

Using the leading term-test, match each of the following functions with one of the graphs AD, which follow.

a) b)

c) d)

4 3( ) 3 2 3f x x x 3 2( ) 5 4 2f x x x x

5 14( ) 1f x x x 6 5 3( ) 4f x x x x


Graphs


Solution

CNegativeEvenx6

APositiveOddx5

BNegativeOdd5x3

DPositiveEven3x4

GraphSign of Leading Coeff.

Degree of Leading Term

Leading Term


Finding Zeros of Factored Polynomial Functions

If c is a real zero of a function (that is, f(c) = 0), then (c, 0) is an x-intercept of the graph of the function.

Example: Find the zeros of

3

( ) 5( 1)( 1)( 1)( 2)

5( 1) ( 2).

f x x x x x

x x


Finding Zeros of Factored Polynomial Functions continued

Solution:

To solve the equation f(x) = 0, we use the principle of zero products, solving x 1 = 0

and x + 2 = 0.

The zeros of f(x) are 1 and 2.

See graph on right.


Even and Odd Multiplicity

If (x c)k, k 1, is a factor of a polynomial function P(x) and:

k is odd, then the graph crosses the x-axis at (c, 0);

k is even, then the graph is tangent to the x-axis at (c, 0).


Finding Real Zeros on a Calculator

Find the real zeros of the function f given by f(x) = 0.2x3 1.1x2 0.3x + 3

Look for points where the graph crosses the x-axis. It appears that there are three zeros, one between 2 and 1, one near 2, and one near 5. We use the ZERO feature to find them.

The zeros are approximately

x = 1.56689, y = 0

x = 1.82693, y = 0

x = 5.2399, y = 0


3.2 Graphing

Polynomial Functions

Graph polynomial functions. Use the intermediate value theorem to determine

whether a function has a real zero between two given real numbers.


Graphing Polynomial Functions

If P(x) is a polynomial function of degree n, the graph of the function has:

at most n real zeros, and thus at most n x-intercepts;

at most n 1 turning points.

(Turning points on a graph, also called relative maxima and minima, occur when the function changes from decreasing to increasing or from increasing to decreasing.)


Steps to Graph a Polynomial Function

1. Use the leading-term test to determine the end behavior.2. Find the zeros of the function by solving f(x) = 0. Any real zeros

are the first coordinates of the x-intercepts.3. Use the x-intercepts (zeros) to divide the x-axis into intervals and

choose a test point in each interval to determine the sign of all function values in that interval.

4. Find f(0). This gives the y-intercept of the function.5. If necessary, find additional function values to determine the

general shape of the graph and then draw the graph.6. As a partial check, use the facts that the graph has at most n

x-intercepts and at most n 1 turning points. Multiplicity of zeros can also be considered in order to check where the graph crosses or is tangent to the x-axis.


Example

Graph the polynomial function f(x) = 2x3 + x2 8x 4.

Solution:

1. The leading term is 2x3. The degree, 3, is odd, the coefficient, 2, is positive. Thus the end behavior of the graph will appear as:

2. To find the zero, we solve f(x) = 0. Here we can use factoring by grouping.


Example continued

Factor:

The zeros are 1/2, 2, and 2. The x-intercepts are (2, 0), (1/2, 0), and (2, 0).

3. The zeros divide the x-axis into four intervals:

(, 2), (2, 1/2), (1/2, 2), and (2, ).

We choose a test value for x from each interval and find f(x).

3 2

2

2

2 8 4 0

(2 1) 4(2 1) 0

(2 1)( 4) 0

(2 1)( 2)( 2) 0

x x x

x x x

x x

x x x


Example continued

Above x-axis+353(2, )

Below x-axis91(1/2, 2)

Above x-axis+31(2, 1/2)

Below x-axis253(, 2)

Location of points on graph

Sign of f(x)Function value, f(x)

Test Value, x

Interval

4. To determine the y-intercept, we find f(0):

The y-intercept is (0, 4).

3 2

3 2

( ) 2 8 4

( ) 2( ) 8( )0 0 0 0 4 4

f x x x x

f


Example continued

5. We find a few additional points and complete the graph.

6. The degree of f is 3. The graph of f can have at most 3 x-intercepts and at most 2 turning points. It has 3 x-intercepts and 2 turning points. Each zero has a multiplicity of 1; thus the graph crosses the x-axis at 2, 1/2, and 2. The graph has the end behavior described in step (1). The graph appears to be correct.

71.5

3.51.5

92.5

f(x)x


Intermediate Value Theorem

For any polynomial function P(x) with real coefficients, suppose that for a b, P(a) and P(b) are of opposite signs. Then the function has a real zero between a and b.

Example: Using the intermediate value theorem, determine, if possible, whether the function has a real zero between a and b.

a) f(x) = x3 + x2 8x; a = 4 b = 1 b) f(x) = x3 + x2 8x; a = 1 b = 3


Solution

We find f(a) and f(b) and determine where they differ in sign. The graph of f(x) provides a visual check.

f(4) = (4)3 + (4)2 8(4) = 16

f(1) = (1)3 + (1)2 8(1) = 8

By the intermediate value theorem, since f(4) and f(1) have opposite signs, then f(x) has a zero between 4 and 1.

f(1) = (1)3 + (1)2 8(1) = 6

f(3) = (3)3 + (3)2 8(3) = 12

By the intermediate value theorem, since f(1) and f(3) have opposite signs, then f(x) has a zero between 1 and 3.


3.3 Polynomial Division; The

Remainder and Factor Theorems

Perform long division with polynomials and determine whether one polynomial is a factor of another. Use synthetic division to divide a polynomial by x c. Use the remainder theorem to find a function value f(c). Use the factor theorem to determine whether x c is a factor of f(x).


Division and Factors

When we divide one polynomial by another, we obtain a quotient and a remainder. If the remainder is 0, then the divisor is a factor of the dividend.

Example: Divide to determine whether x + 3 and x 1

are factors of 3 22 5 4.x x x


Division and Factors continued

Divide:

Since the remainder is –64, we know that x + 3 is not a factor.

2

3 2

3 2

2

2

5 20

3 2 5 4

3

5 5

5 15

2

rema

0 4

20 60

64 inder

x x

x x x x

x x

x x

x x

x

x


Division and Factors continued

Divide:

Since the remainder is 0, we know that x 1 is a factor.

2

3 2

3 2

2

2

4

1 2 5 4

5

4 4

4 4

0 remainder

x x

x x x x

x x

x x

x x

x

x


The Remainder Theorem

If a number c is substituted for x in a polynomial f(x), then the result f(c) is the remainder that would be obtained by dividing f(x) by x c. That is, if f(x) = (x c) • Q(x) + R, then f(c) = R.

Synthetic division is a “collapsed” version of long division; only the coefficients of the terms are written.


Example

Use synthetic division to find the quotient and remainder.

The quotient is – 4x4 – 7x3 – 8x2 – 14x – 28 and the remainder is –6.

5 4 3 24 6 2 50 ( 2)x x x x x

–6–28–14–8–7–4

–56–28–16–14–8

500261–42 Note: We must write a 0 for the missing term.


Example

Determine whether 4 is a zero of f(x), where f(x) = x3 6x2 + 11x 6.

We use synthetic division and the remainder theorem to find f(4).

Since f(4) 0, the number is not a zero of f(x).

63–21

12–84

–611–614


The Factor Theorem

For a polynomial f(x), if f(c) = 0, then x c is a factor of f(x).

Example: Let f(x) = x3 7x + 6. Factor f(x) and solve the equation f(x) = 0.

Solution: We look for linear factors of the form x c. Let’s try x 1.

Since f(1) = 0, we know that x 1 is one factor and the quotient x2 + x 6 is another.So, f(x) = (x 1)(x + 3)(x 2).For f(x) = 0, x = 3, 1, 2.

0–611

–611

6–701 1


3.4 Theorems about Zeros of

Polynomial Functions

Find a polynomial with specified zeros. For a polynomial function with integer coefficients, find

the rational zeros and the other zeros, if possible. Use Descartes’ rule of signs to find information about

the number of real zeros of a polynomial function with real coefficients.


The Fundamental Theorem of Algebra

Every polynomial function of degree n, with n 1, has at least one zero in the system of complex numbers.

Example: Find a polynomial function of degree 4 having zeros 1, 2, 4i, and 4i.

Solution: Such a polynomial has factors (x 1),(x 2), (x 4i), and (x + 4i), so we have:

Let an = 1.

( ) ( 1)( 2)( 4 )( 4 )nf x a x x x i x i

2 2

4 3 2 2

4 3 2

( ) ( 1)( 2)( 4 )( 4 )

( 3 2)( 16)

3 2 16 48 32

3 18 48 32

f x x x x i x i

x x x

x x x x x

x x x x


Zeros of Polynomial Functions with Real Coefficients

If a complex number a + bi, b 0, is a zero of a polynomial function f(x) with real coefficients, then its conjugate, a bi, is also a zero. (Nonreal zeros occur in conjugate pairs.)

Rational Coefficients If where a and b are rational and c is not a perfect square, is a zero of a polynomial function f(x) with rational coefficients, then is also a zero.

,a b c

a b c


Example

Suppose that a polynomial function of degree 6 with rational coefficients has 3 + 2i, 6i, and as three of its zeros. Find the other zeros.

Solution: The other zeros are the conjugates of the given zeros, 3 2i, 6i, and There are no other zeros because the polynomial of degree 6 can have at most 6 zeros.

1 2

1 2.


Rational Zeros Theorem

Let

where all the coefficients are integers. Consider a rational number denoted by p/q, where p and q are relatively prime (having no common factor besides 1 and 1). If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an.

1 21 2 1 0( ) ... ,n n

n nP x a x a x a x a x a


Example

Given f(x) = 2x3 3x2 11x + 6:

a) Find the rational zeros and then the other zeros.

b) Factor f(x) into linear factors.

Solution:

a) Because the degree of f(x) is 3, there are at most 3 distinct

zeros. The possibilities for p/q are:

3 31 12 2 2 2

1, 2, 3, 6:

1, 2

/ : 1, 1,2, 2,3, 3,6, 6, , , ,

Possibilities for p

Possibilities for q

Possibilities for p q


Example continued

Use synthetic division to help determine the zeros. It is easier to consider the integers before the fractions.

We try 1: We try 1:

Since f(1) = 6, 1 is not a zero. Since f(1) = 12, 1 is not zero.

–6–12–12

–12–12

6–11–321

12–6–52

65–2

6–11–32–1


Example continued

We try 3:

.

We can further factor 2x2 + 3x 2 as (2x 1)(x + 2).

0–232

–696

6–11–323

Since f(3) = 0, 3 is a zero. Thus x 3 is a factor. Using the results of the division above, we can express f(x) as

2( ) ( 3)(2 3 2)f x x x x


Example continued

The rational zeros are 2, 3 and

The complete factorization of f(x) is:

1.

2

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


Descartes’ Rule of Signs

Let P(x) be a polynomial function with real coefficients and a nonzeroconstant term. The number of positive real zeros of P(x) is either:

1. The same as the number of variations of sign in P(x), or2. Less than the number of variations of sign in P(x) by a positive even

integer.

The number of negative real zeros of P(x) is either:

3. The same as the number of variations of sign in P(x), or4. Less than the number of variations of sign in P(x) by a positive even

integer.

A zero of multiplicity m must be counted m times.


Example

What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros?

There are two variations of sign, so there are either two or zero positive real zeros to the equation.

4 3 2

4 3 2

( ) 3 6 7

3 2

2

6 7

P x x x x x

x x x x


Example continued

The number of negative real zeros is either two or zero.

Total Number of Zeros4

Positive 2 2 0 0

Negative 2 0 2 0

Nonreal 0 2 2 4

4 3 2

4 3 2

( ) 3( ) 6( ) ( ) 7( )

6 7 2

2

3

P x x x x x

x x xx


3.5 Rational Functions

For a rational function, find the domain and graph the function, identifying all of the asymptotes.

Solve applied problems involving rational functions.


Rational Function

A rational function is a function f that is a quotient of two polynomials, that is,

where p(x) and q(x) are polynomials and where q(x) is not the zero polynomial. The domain of f consists of all inputs x for which q(x) 0.

( )( ) ,

( )p x

f xq x


Example

Consider .

Find the domain and graph f.

Solution:

When the denominator x + 4 = 0, we have x = 4, so the only input that results in a denominator of 0 is 4. Thus the domain is {x|x 4} or (, 4) (4, ).

The graph of the function is the graph of y = 1/x translated to the left 4 units.

1( )

4f x

x


Vertical Asymptotes

The vertical asymptotes of a rational function f(x) = p(x)/q(x) are found by determining the zeros of q(x) that are not also zeros of p(x). If p(x) and q(x) are polynomials with no common factors other than constants, we need to determine only the zeros of the denominator q(x).

If a is a zero of the denominator, then the line x = a is a vertical asymptote for the graph of the function.


Example

Determine the vertical asymptotes of the function.

Factor to find the zeros of the denominator:x2 4 = (x + 2)(x 2)

Thus the vertical asymptotes are the lines x = 2 and x = 2.

2

2 3( )

4

xf x

x


Horizontal Asymptote

The line y = b is a horizontal asymptote.

When the numerator and the denominator of a rational function have the same degree, the line y = a/b is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively.

Example: Find the horizontal asymptote:

.

The numerator and denominator have the same degree. The ratio of the leading coefficients is 6/9, so the line y = 2/3 is the horizontal asymptote.

4 2

4

6 3 1( )

9 3 2

x xf x

x x


Determining a Horizontal Asymptote

When the numerator and the denominator of a rational function have the same degree, the line y = a/b is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively.

When the degree of the numerator of a rational function is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote.

When the degree of the numerator of a rational function is greater than the degree of the denominator, there is no horizontal asymptote.


True Statements

The graph of a rational function never crosses a vertical asymptote.

The graph of a rational function might cross a horizontal asymptote but does not necessarily do so.


Example

Graph .

Vertical asymptotes: x + 3 = 0, so x = 3.

The degree of the numerator and denominator is the same. Thus y = 2, is the horizontal asymptote.

1. Draw the asymptotes with dashed lines.

2. Compute and plot some ordered pairs and draw the curve.

2( )

3

xh x

x


Example continued

4/52

00

42

84

55

3.57

h(x)x


Oblique or Slant Asymptote

Find all the asymptotes of .

The line x = 2 is a vertical asymptote.

There is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator.

22 3 5( )

2

x xh x

x

22 3 5 3( ) (2 1)

2 2

x xh x x

x x


Oblique or Slant Asymptote continued

Divide to find an equivalent expression.

The line y = 2x 1 is the oblique asymptote.

22 3 5 3( ) (2 1)

2 2

x xh x x

x x

2

2

2 12 2 3 5

2 4

5

2

3

xx x x

x x

x

x


Occurrence of Lines as Asymptotes of Rational Functions

For a rational function f(x) = p(x)/q(x), where p(x) and q(x) have no common factors other than constants:

Vertical asymptotes occur at any x-values that make the denominator 0.

The x-axis is the horizontal asymptote when the degree of the numerator is less than the degree of the denominator.

A horizontal asymptote other than the x-axis occurs when the numerator and the denominator have the same degree.


Occurrence of Lines as Asymptotes of Rational Functions continued

An oblique asymptote occurs when the degree of the numerator is 1 greater than the degree of the denominator.

There can be only one horizontal asymptote or one oblique asymptote and never both.

An asymptote is not part of the graph of the function.


Graphing Rational Functions

1. Find the real zeros of the denominator. Determine the domain of the function and sketch any vertical asymptotes.

2. Find the horizontal or oblique asymptote, if there is one, and sketch it.

3. Find the zeros of the function. The zeros are found by determining the zeros of the numerator. These are the first coordinates of the x-intercepts of the graph.

4. Find f(0). This gives the y-intercept (0, f(0)), of the function.

5. Find other function values to determine the general shape. Then draw the graph.


Example

Graph .

1. Find the zeros by solving:

The zeros are 1/2 and 3, thus the domain excludes these values.

The graph has vertical asymptotes at x = 3 and x = 1/2. We sketch these with dashed lines.

2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, y = 0, is the horizontal asymptote.

2

3( )

2 5 3

xf x

x x

2

2

2 5 3 0

2 5 3 (2 1)( 3)

x x

x x x x


Example continued

3. To find the zeros of the numerator, we solve x + 3 = 0 and get x = 3. Thus, 3 is the zero of the function, and the pair (3, 0) is the x-intercept.

4. We find f(0):

Thus (0, 1) is the y-intercept.

2

00

0

3( )

2( ) 5( ) 3

31

3

0f


Example continued

5. We find other function values to determine the general shape of the graph and then draw the graph.

7/94

12

2/31

1/21

f(x)x


More Examples

Graph the following functions.

a)

b)

c)

3( )

2

xf x

x

2

2

8( )

9

xf x

x

2

( )1

xf x

x


Graph a:

Vertical Asymptote

x = 2 Horizontal Asymptote

y = 1 x-intercept

(3, 0) y-intercept

(0, 3/2)

3( )

2

xf x

x


Graph b:

Vertical Asymptote

x = 3, x = 3 Horizontal Asymptote

y = 1 x-intercepts

(2.828, 0) y-intercept

(0, 8/9)

2

2

8( )

9

xf x

x


Graph c:

Vertical Asymptote

x = 1 Oblique Asymptote

y = x 1 x-intercept

(0, 0) y-intercept

(0, 0)

2

( )1

xf x

x


3.6 Polynomial and

Rational Inequalities

Solve polynomial and rational inequalities.


Polynomial Inequalities

A quadratic inequality can be written in the form ax2 + bc + c > 0, where the symbol could be replaced with either <, , or .

A quadratic inequality is one type of polynomial inequality.

Examples:

4 2 3 232 2 4 6 0 8 2 6 5

4x x x x x x


Example

Solve: 4x3 7x2 15x.

We need to find all the zeros of the function so we solve the related equation.

The zeros are 0, 3 and 5/4. Thus the x-intercepts of the graph are (0, 0), (3, 0) and (5/4, 0).

3 2

3 2

2

4 7 15

4 7 15 0

(4 7 15) 0

(4 5)( 3) 0

x x x

x x x

x x x

x x x


Example continued

The zeros divide the x-axis into four intervals. For all x-values within a given interval, the sign of 4x3 7x2 15x 0 must be either positive or negative. To determine which, we choose a test value for x from each interval and find f(x).

Since we are solving 4x3 7x2 15x 0, the solution set consists of only two of the four intervals, those in which the sign of f(x) is negative. {x| < x < 5/4 or 0 < x < 3}.

Positivef(4) = 84(3, )

Negativef(1) = 18(0, 3)

Positivef(1) = 4(5/4, 0)

Negativef(2) = 30(, 5/4)

Sign of f(x)Test ValueInterval


To Solve a Polynomial Inequality

1. Find an equivalent inequality with 0 on one side.2. Solve the related polynomial equation.3. Use the solutions to divide the x-axis into intervals.

Then select a test value from each interval and determine the polynomial’s sign on the interval.

4. Determine the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is or .


To Solve a Rational Inequality

1. Find an equivalent inequality with 0 on one side.2. Change the inequality symbol to an equals sign and solve the

related equation.3. Find the values of the variable for which the related rational

function is not defined.3. The numbers found in steps (2) and (3) are called critical values.

Use the critical values to divide the x-axis into intervals. Then test an x-value from each interval to determine the function’s sign in that interval.

5. Select the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. If the inequality symbol is or , then the solutions to step (2) should be included in the solution set. The x-values found in step (3) are never included in the solution set.


Example

Solve .

The denominator tells us that f(x) is not defined for

x = 1 and x = 1.

Next, solve f(x) = 0.2

30

1

x

x

2

2

2 22

30

13

01

3( 1) 0( 1)

1

3 0

3

x

xx

xx

x xx

x

x


Example continued

The critical values are 3, 1, and 1. These values divide the x-axis into four intervals. We use a test value to determine the sign of f(x) in each interval.

+f(2) = 5/3(1, )

f(0) = 3(1, 1)

+f(2) = 1/3(3, 1)

f(4) = 1/15(, 3)

Sign of f(x)Test ValueInterval


Example continued

Function values are positive in the intervals (3, 1) and (1, ). Since neither 1 or 3 is in the domain of f, they cannot be part of the solution set.

The solution set is

(3, 1) (1, ).


3.7 Variation and Applications

Find equations of direct, inverse, and combined variation given values of the variables.

Solve applied problems involving variation.


Direct Variation

If a situation gives rise to a linear function f(x) = kx, or y = kx, where k is a positive constant, we say that we have direct variation, or that y varies directly as x, or that y is directly proportional to x. The number k is called the variation constant, or constant of proportionality.

The graph of y = kx, k > 0, always goes through the origin and rises from left to right. As x increases, y increases; that is, the function is increasing on the interval (0,). The constant k is also the slope of the line.


Direct Variation

Example: Find the variation constant and an equation of variation in which y varies directly as x, and y = 42 when x = 3.

Solution: We know that (3, 42) is a solution of y = kx. y = kx

42 = k 3

= k

14 = k The variation constant 14, is the rate of change of y with respect to

x. The equation of variation is y = 14x.

42

3


Application

Example: Wages. A cashier earns an hourly wage. If the cashier worked 18 hours and earned $168.30, how much will the cashier earn if she works 33 hours?

Solution: We can express the amount of money earned as a function of the amount of hours worked.

I(h) = kh I(18) = k 18 $168.30 = k 18 $9.35 = k The hourly wage is the variation constant.

Next, we use the equation to find how much the cashier will earn if she works 33 hours.

I(33) = $9.35(33) = $308.55


Inverse Variation

If a situation gives rise to a function f(x) = k/x, or y = k/x, where k is a positive constant, we say that we have inverse variation, or that y varies inversely as x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality.

For the graph y = k/x, k 0, as x increases, y decreases; that is, the function is decreasing on the interval (0, ).


Inverse Variation

Example: Find the variation constant and an equation of variation in which y varies inversely as x, and y = 22 when x = 0.4.

Solution:

The variation constant is 8.8. The equation of variation is y = 8.8/x.

220.4

(0.4)22

8.8

ky

xk

k

k


Application

Example: Road Construction. The time t required to do a job varies inversely as the number of people P who work on the job (assuming that all work at the same rate). If it takes 180 days for 12 workers to complete a job, how long will it take 15 workers to complete the same job?

Solution: We can express the amount of time required, in days, as a

function of the number of people working.

t varies inversely as P

This is the variation constant.

( )

(12)12

18012

2160

kt P

Pk

t

k

k


Application continued

The equation of variation is t(P) = 2160/P.

Next we compute t(15).

It would take 144 days for 15 people to complete the same job.

2160( )

2160(15)

15144

t PP

t

t


Combined Variation

Other kinds of variation: y varies directly as the nth power of x if there is some

positive constant k such that .

y varies inversely as the nth power of x if there is some positive constant k such that .

y varies jointly as x and z if there is some positive constant k such that y = kxz.

ny kx

n

ky

x


Example

The illuminance of a light (E) varies directly with the intensity (I) of the light and inversely with the square distance (D) from the light. At a distance of 10 feet, a light meter reads 3 units for a 50-cd lamp. Find the illuminance of a 27-cd lamp at a distance of 9 feet.

Solve for k.

Substitute the second set of data into the equation.

The lamp gives an illuminance reading of 2 units.

2

2

2

503

106

6 27

92

IE k

Dk

k

E

E

polynomial and rational functions

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