45. The Linear Fractional Function

ax + b
cx + d'

In order that

this function should really be fractional, it is essential that

a

b

c≤ 0, for if c = 0, it reduces to the linear function ≈ + x d d. The simplest linear fractional function is 1/x, obtained by setting bc, and a = d = 0, whose graph has been considered in Section 40. If bc, while a = d = 0, we have the function b/cx, or k/x, where k = b/c. Its graph may be obtained by multiplying by k the ordinates of points on the graph of 1/x (Theorem, page 89).

=

If we set y is seen that the graph is symmetrical

k/x, whence xy = k, it

with respect to the line y = x, since

-1 -3 -2 -1

·3+

2

xy-k

xy=1

10

B2 3

x

-1

-2

FIG. 66.

the equation is unchanged if x and y are interchanged. The graph is also symmetrical with respect to the origin.

The geometric significance of the constant k is obtained by solving the equations xy = k and y = x; the solutions are (√k, √k) and (− √k, Vk), which are the coördinates of the points of intersection of the graphs of the equations. The distance from the origin to the point of intersection A is found from the right triangle OAB to be √2k. Hence, for a small value of k, the graph lies close to the origin and axes, while if k is large, it lies at some distance from them. If k is negative, the graph will lie in the second and fourth quadrants.

=

The graph of xy k is called a rectangular, or equilateral, hyperbola. The former term is derived from the fact that the asymptotes are perpendicular.

In order to determine the form of the graph of the general linear fractional function, set

Multiplying both sides by cx + b, and subtracting ax + b from both sides,

cxy ax + dy - b = 0.

(2)

Let us now see if we can simplify this equation by translating the axes (Section 31, page 89).

Setting

we get

x = x' + h and

y = y' + k,

(3)

c(x' + h) (y' + k) − a(x' + h) + d(y' + k) − b = 0, (4)

or, removing the parentheses and collecting like terms, cx'y' + (ck - a)x' + (ch + d)y' + chk - ah + dk − b = 0. (5) Equating to zero the coefficients of x' and y', we get

Solving for h and k, which is always possible since c 0,

(6)

(7)

Simplifying, subtracting the constant term from both sides and dividing by c,

(8)

This equation, referred to the new axes, has the same graph as (1). It is of the form x'y' = k, where k = (bc — ad)/c2, or y' = k/x'. The graph is therefore a rectangular hyperbola. Hence we have the

Theorem. The graph of a linear fractional function (1), or of an equation of the form (2), is a rectangular hyperbola whose asymptotes are parallel to the axes of coördinates.

1. Plot the graph of y

=

(2x

EXERCISES

4)/(x + 3), (a) by translating the axes; (b) by finding the asymptotes (by the method on page 27), plotting one branch of the curve from a table of values, and the other by means of the symmetry with respect to the point of intersection of the asymptotes.

3. The linear fractional function may be written in the form a(x + b/a)/c(x + d/c). Prove that the linear factors cancel, and the function is a constant, if and only if bc ad 0. For this reason, it is always assumed that bc ad 0.

=

4. The relation between the radius of curvature of a concave mirror, R, the distance from the center of the mirror to the object, x, and the distance from the center to the reflected image, y, is

If R = 2 feet, construct the graph of the equation.

46. Integral Rational Functions. The general form of an integral rational function or polynomial is (see page 39)

We have already studied polynomials of the first and second degree in considering linear and quadratic functions.

The calculation of a table of values for a polynomial of higher degree than the second by direct substitution presents no new features, but because of the greater number of terms that may be present there is more labor involved. An alternative method which is simpler, in general, than that of direct substitution is developed in the next sections.

47. The Remainder Theorem.

EXAMPLE. Let us divide the polynomial 2x3 6x2+11 by x − 2, arranging the work as usual.

The result may be put in the form

4 is the quotient, denoted by q(x), and the remainder is

Equation (1) is an identity in accordance with the DEFINITION. An identity is an equation which is satisfied by

all values of the variable or variables.

Substituting 2 for x in the polynomial in the example,

Hence the remainder, R = 3, obtained by dividing the polynomial by x 2 is equal to the value of the polynomial when 2 is substituted for x.

This result is a verification of a theorem which is proved for any polynomial as follows:

Remainder Theorem. If a polynomial f(x) is divided by xa, the remainder is f(a).

If f(x) be divided by xa, the quotient q(x) is a polynomial of degree one less than that of f(x), and the remainder R is a constant.

The result of the division may be expressed in the form

Multiplying both members by xa, we obtain the identity

(4)

Since a a = 0, we have (aa)q(a) = 0.

48. Synthetic Division. The form of the division in the example of the preceding section is too cumbersome to be of practical use in calculating the table of values of a polynomial. The division may be simplified as follows:

We note that the operations involved in the division were performed on the coefficients of the dividend, divisor and quotient, the powers of x serving merely to determine the proper positions of these coefficients in carrying out the details of the division. If missing terms are supplied by zero coefficients, the division can be performed by the so-called method of detached coefficients as follows:

The first term in each partial product is canceled in the subtraction and hence may be omitted. The second term of each partial product may be written immediately under the term in the dividend from which it is subtracted. The first term in the divisor is not now needed, so that the process is further condensed as follows:

Noting that the coefficients of the quotient are identical with the numbers marked with an asterisk (*), it is unnecessary to write them under the divisor. As we shall use this process to find the value of f(2), it is convenient to replace 2 in the divisor by +2 and add throughout instead of subtracting. Also bringing down the first coefficient of the dividend, the work is now arranged in the form

The numbers in the last row are called partial remainders, the last being the remainder 3 (the value of the polynomial