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Unformatted text preview: Linear, Quadratic, Polynomial,
and Rational Functions In this chapter we focus on four important special classes of functions. The
main themes of the last chapteretransformations of functions, composition
of functions, and inverses of functions—appear in this chapter in the context
of these special classes of functions. Linear functions form our ﬁrst special class of functions. Lines and their
slopes, although simple concepts, have immense importance. After dealing with linear functions and lines, we will turn to the quadratic
functions, our second class of special functions. The graph of a quadratic
function is a parabola. We will see how to ﬁnd the vertex of a parabola, how
to perform the useful algebraic operation of completing the square, and how
to solve quadratic equations. Then we will take a brief diversion to review the algebra of integer expo
nents in preparation for dealing with the polynomial functions, which form
our third special class of functions. From polynomials we will move on to
rational functions, our fourth special class of functions. This chapter concludes with two optional sections—one discusses com
plex numbers (needed to solve arbitrary quadratic equations) and one dis
cusses systems of equations and matrices. H7 J ‘ ‘ .l
33 E CHAPT R 1
Tmmsm‘mm This detail from The
School of Athens
(painted by Raphael
around 1 51 0) depicts
Euclid explaining geometry. 118 CHAPTER 2 Linear, Quadratic, Polynomial, and Rational Functions In this ﬁgure, each
side of the larger tri—
angle has twice the
length of the cor
responding side of
the smaller triangle. 3 Linear Functions and Lines SECTION OBJECTIVES
By the end of this section you should I understand the concept of the slope of a line; I be able to ﬁnd the equation of a line given its slope and a point on it;
I be able to ﬁnd the equation of a line given two points on it; I understand why parallel lines have the same slope; I be able to find the equation of a line perpendicular to a given line and
containing a given point. Slope Consider a line in the coordinate plane. along with four points (x1, 311),
(X2, 3/2), (x3, )3), and (X4, 324) on the line. Draw two right triangles with
horizontal and vertical edges as in the ﬁgure below: Similar triangles. The two right triangles in the ﬁgure above are similar because their angles
are equal. Thus the ratios of the corresponding sides of the two triangles
above are equal. Speciﬁcally, taking the ratio of the vertical side and hori
zontal side for each triangle, we have 3’23’1 = y4y3
)Cz—xl X4#X3I The equation above states that for any pair of points (x1, 3/1) and (X2. 3/2)
on the line, the ratio JéLjé—i does not depend on the particular pair of points
chosen on the line. If we choose another pair of points on the line, say
(963,313) and (901,324) instead of (xhyl) and (X2012), then the difference of
second coordinates divided by the difference of ﬁrst coordinates remains the
same, as shown by the equation above. Thus the ratio 33% is a constant depending only on the line and not on
the particular points (x1, 3/1) and (X2, ya) chosen on the line. This constant
is called the slope of the line. SECTION 2.1 Linear Functions and Lines 1 19 Slope If (xhyl) and (362,312) are any two points on a line, with x1 94 x;, then
the slope of the line is 3’23’1
x2"x1. Find the slope of the line containing the points (2,1) and (5. 3). SOLUTION The line containing (2, 1) and (5, 3) is shown here. The slope of this line
is which equals slope = 2
Aline with positive slope slants
up from left to right; a line with neg
. slope = l
ative slope slants down from left to
right. Lines whose slopes have lar er 51°“ = 1/2
absolute value are stee er than lines
WWW slope = —1/2
value; This figure shows some lines slope : ,1
and their slopes; the same scale has
been used on both axes.
slope = —2 In the ﬁgure above, the horizontal axis has e 0 a does every hori
zontal line. Vertical lines, including the vertical axis, do not have a slope, beciu:e a vertlcal line does not contain two p01nts (X1. 321) and (xz, yg) With
X1 2 . The Equation of a Line Consider a line with slope m, and suppose (361,311) is a point on this line.
Let (x, 3/) denote a typical point on the line, as shown here.
Because this line has slope m, we have y’yl
X—X1 Multi 1 ' ' ation above b x — x we get the following formula: 21%. EXAMPLE ], 3’ /y} (x1, 3’1)
X A line with slope m. V 120 CHAPTERZ Linear, Quadratic, Polynomial, and Rational Functions The equation of a line, given its slope and one point on it The symbol m is of The line in the xyplane that has slope m and contains the point (x1, y1) ten used to denote is given by the equation
the slope of a line. yy1=m(x—x1). 5'3; The equation above can be solved for y to get an equation for the line in
the form 3/ 2 mx + h, where m and h are constants. EXAMPLE 2 Find the equation of the line in the xyplane that has slope 4 and contains the point
(2, 3). SOLUTION In this'case the equation displayed above becomes 32 — 3 : 4(x i 2).
Adding 3 to both sides and simplifying, we get y=4x—S. Always perform this As a check, if we take x = 2 in the equation above. we get y : 3. Thus the point
kind of check to (2,3) is indeed on this line. determine if an er WWWWMWMMWMWMWW
ror has been made. Suppose we want to ﬁnd the equation of the line containing two Speciﬁc
points. We can reduce this problem to a problem we have already solved
by computing the slope of the line and then using the formula in the box
above. Speciﬁcally, suppose we want to find the equation of the line con
taining the points {3:1, yl) and (X2, 3/2), where x1 9E x2. This line has Slope
(y; — 3/1)] (x; — x1). Thus the formula above gives the following result: The equation of a line, given two points on it The line in the xyplane that contains the points (am, 321) and (x2, ya),
where x; 49 X2, is given by the equation yvy1=(i%)(x—x1). 32W EXAMPLE g Find the equation of the line in the x yplane that contains the points (2, 1) and (5. 3)
(this line is shown with Example 1). SOLUTION In this case the equation above becomes y—l = (gt—Dove). SECTION 2.1 Linear Functions and Lines 121 Solving this equation for y, we get
32 = §x — As a check, if we take x : 2 in the equation above, we get 3/ = 1, and if we take x : 5 in the equation above, we get 32 = 3; thus the points (2, 1) and (5,3) are indeed on
this line. Suppose we want to find the equation of the line in the xy—plane with
slope m that intersects the yaxis at the point (0,19). Because (0.19) is a The point Where a line
point on the line, we can use the formula for the equation of a line given its intersects the y'EXI'S
slope and one point on it. In this case, that equation becomes 1'5 Often 031180? the J" intercept.
y — b = m(x * 0). 3 Solving this equation for y, we have the following result: The equation of a line, given its slope and vertical axis
intersection The line in the xy—plane with slepe m that intersects the yaxis at (0.19)
is given by the equation
3! = mx + is. If a line contains the origin, then b = 0 in the equation above. For example,
the line in the xyplane that has slope 2 and contains the origin is given
by the equation 32 = 2x. The figure below Example 1 shows several lines
containing the origin.
We have seen that a line in the xyplane with slope m is characterized
by the equation y = mx + la, where b is some constant. To restate this
conclusion in terms of functions, let f be the function defined by f (x) =
mx + b, where m and b are constants. Then the graph of f is a line with
slope m. Functions of this form are so important that they have a name—
linear functions: Differential calculus
focuses on approxi
mating an arbitrary Linear functions
function on a small A linear function is a function f of the form part ofits domain by a
linear function. Thus
f(x) 2 mx + b, you will frequently
encounter linear func
where m and b are constants. tions in future mathe matics courses. Find the function f such that ﬁx) equals the temperature on the Fahrenheit scale [ng ,\M pl E 4
corresponding to temperature x on the Celsius scale. ' ' 122 CHAPTER 2 Linear, Quadratic, Polynomial, and Rational Functions illllllﬁlillltﬁlllll I D!
lllllllll u I? J
3
l
'I
:I This thermometer
shows Celsius degrees on the left, Fahrenheit
degrees on the right. 3" 212 32 I X
100 The graph of
for) = 2x + 32 on the
interval [—10, 110]. SOLUTION Changing from one system of units to another system of units is modeled
by a linear function. Thus f has the form f (x) = mx + b for some constants in and h.
To ﬁnd m and b. we recall that the freezing temperature of water equals 0 degrees Celsius and 32 degrees Fahrenheit; also, the boiling point of water equals 100 degrees
Celsius and 212 degrees Fahrenheit. Thus f(0)=32 and f(100)=212. But f(0) : h, and thus I) = 32. Now we know that f(x) = mx + 32. Hence f(100) = 100m + 32. Setting this last quantity equal to 212 and then solving for in shows that m = %. Thus A special type of linear function is obtained when considering functions
of the form f(x) = mx + h withm = 0: Constant functions A constant function is a function f of the form
f(x) = b.
where b is a constant.
The graph of a constant function is a horizontal line.
y 2 The graph of the constant function f deﬁned by
1 f(x) = 2 on the interval [1,4]. Parallel Lines Consider two parallel lines in the coordinate plane, as shown in the ﬁgure
below: 3’ Parallel lines. SECTION 2.1 Linear Functions and Lines 123 Because the two lines are parallel, the corresponding angles in the two trian
gles above are equal (as shown by the arcs in the ﬁgure above), and thus the
two right triangles are similar. This implies that
r_______________ Because 3 is the slope of the top line and g is the slope of the bottom line, we conclude that these parallel lines have the same 310 e. The logic used in the paragraph above is reversible. Speciﬁcally, suppose
instead of starting with the assumption that the two lines in the ﬁgure above
are parallel, we start with the assumption that the two lines have the same
slope. This implies that g = g, which implies that the two right triangles in
the ﬁgure above are similar. This then implies that the two lines make equal
angles with the horizontal axis, as shown by the arcs in the ﬁgure, which
implies that the two lines are parallel. The ﬁgure and reasoning given above do not work if both lines are hor
izontal or both lines are vertical. But horizontal lines all have slope 0, and
the slope is not deﬁned for vertical lines. Thus we can summarize our char
acterization of parallel lines as follows: Parallel lines Two nonvertical lines in the coordinate plane are parallel if and only if
they have the same slope. For example, the lines in the xyplane given by the equations
y=4x—S and y=4x+18 are parallel because they have the same slope (which equals 4). As another
example, the lines in the xy—plane given by the equations y=6x+5 and y:7x+5 are not parallel because their slopes are not equalithe first line has slope 6
and the second line has slope 7. Sometimes seeing a second explanation for an important result can help
lead to better understanding. We used geometry to derive the conclusion in
the box above. The same result can also be derived algebraically. To do that,
ﬁrst we need to examine the procedure for ﬁnding the intersection of two
lines, which is illustrated by the following example. Find the intersection of the lines in the xyplane given by the equations y=4x—5 and y=—3x+7. 4? The phrase “if and
only if”, when con
necting two state—
ments, means that
the two statements
are either both true or
both false. For exam
ple,x+1 >6ifand
only if x > 5. EXAMPLE 5 124 CHAPTER 2 Linear, Quadratic, Polynomial, and Rational Functions The lines
v : 4x — 5 (blue) and
y = —3x + 7 (red). The neXt several para
graphs give an alge
braic explanation for
the characterization
of parallel lines, as
compared to the ge
ometric explanation
that we saw earlier. SOLUTION The figure here is good enough to give an estimate of the coordinates of
the point of intersection of the two lines, but to ﬁnd the exact coordinates we need to use algebra.
To ﬁnd the intersection of the two lines, we need to ﬁnd a point (x, y) that satisﬁes both the equations above. The simplest way to solve this system of two
simultaneous equations is to notice that the left side of both equations equals 3/;
thus the right sides must be equal. Setting the two right sides equal to each other, we get the equation
4x — 5 = ~3x + 7. Solving this equation for x gives x = Then this value of x can be substituted in
the first equation y = 4x — 5, giving 3/ = As a check that we have not made an
error, we can substitute x = 172 into our second equation 3/ = —3x + 7, again getting
y = g (getting a different value of y would indicate that an error has been made somewhere).
Thus the two lines intersect at the point (1—72, g), which is consistent with the ﬁgure above. Suppose we have two distinct lines in the xyplane given by the equations y=mix+b1 and y=m2x+h2. Two distinct lines in the coordinate plane are parallel if and only if they do
not intersect. Thus to determine whether or not these lines are parallel, we
can determine algebraically whether or not they intersect. To do this, we set
the two right sides of the equations above equal to each other, getting mlx + 191: max + 192.
This equation is equivalent to the equation
(m1 — m2)x = 102 — b1. First suppose m1 2 1712. Then the equation above becomes 0x = in — hz.
Because 191 qé 192 (otherwise the two lines would not be distinct), this equation
has no solutions. Thus the two lines do not intersect and hence they are
parallel. Now suppose m1 74 1712. Then we can divide both sides of the displayed
equation above by m1 — m2, getting a solution to the equation, which leads
to a point of intersection of the two lines. Thus if m1 aé 1112, then the lines
are not parallel. We have shOWn algebraically that the two lines are parallel if and only if
m1 = 1412. In other words, the two lines are parallel if and only if they have
the same slope. Perpendicular Lines Our next goal is to show that two lines with slopes m1 and 7112 are perpen
dicular if and only if mlmg : —1. If mlmg = —1, then either ml or 1412 is negative. Thus before beginning our treatment of perpendicular lines, we
take a brief detour to make clear the geometry of a line with negative slope. A line with negative slope slants down from left to right. The ﬁgure below
shows a line with negative slope; to avoid clutter the coordinate axes are not shown: In the ﬁgure above, a. is the length of the horizontal line segment and c
is the length of the vertical line segment. Of course a and c are positive
numbers, because lengths are positive. In terms of the coordinates as shown
in the ﬁgure above, we have a : x2 7 x1 and c = y1 ,— yg. The slope of this A line with slope —§. line equals (yg — yﬂ/(xz — x1), which equals 7cm. Now consider two perpendicular lines, as
shown in blue in the ﬁgure here, where again to avoid clutter the coordinate axes shown. In addition to the two perpendicular
lines in blue, the ﬁgure shows the horizontal
line segment PS and the vertical line segment QT, which intersect at S. We assume that the angle PQT is 6 degrees.
To check that the other three labeled angles
in the ﬁgure are labeled correctly, ﬁrst note
that the triangle QPT is a right triangle, one of
whose angles is 6 degrees. Thus the angle PTQ
is 90 — 6 degrees, as shown in the ﬁgure. Con sideration of the right triangle PST that angle TPS is 6 degrees, as labeled. This then implies that angle QPS is 90 —
as shown in the ﬁgure. The line containing the points P and Q has slope b for, as can be seen from
the ﬁgure. Furthermore, as can be seen from our brief discussion of lines
with negative slope, the line containing the points P and T has slope —c for.
To ﬁnd a relationship between the slopes of these two lines, we will ﬁnd a formula connecting a, la, and c. Consider the right triangles PSQ and TSP in the ﬁgure. These triangles
have the same angles, and thus they are similar. Thus the ratios of corre are not now shows 6 degrees, spending sides are equal. Speciﬁcally, we have h (1 Multiplying both sides of this equation by —C for, we get SECTION 2.1 Linear Functions and lines 125 This ﬁgure contains
three righ t triangles:
QPT, PSQ, and TSP. Let a denote the
length of the line seg
ment PS, let l9 denote
the length of the line
segment Q5, and let :2 denote the length of
the line segment ST,
as shown in the ﬁgure. 126 CHAPTER 2 Linear, Quadratic, Polynomial, and Rational Functions Numbers 1111 and 1112
such that mlmz = —1
are sometimes called
negative reciprocals
of each other. EXAMPLE 6 EXAMPLE 7 (2)  (7%) As we have already seen, the ﬁrst quantity on the left above is the slope of
the line containing the points P and Q, and the second quantity is the slope
of the line containing the points P and T. Thus we can conclude that the
product of the slopes of these two perpendicular lines equals 71. The logic used above is reversible. Speciﬁcally, supposa instead of starting
with the assumption that the two lines in blue are perpendicular, we start
with the assumption that the product of their slopes equals w 1. This implies
that g : %, which implies that the two right triangles PSQ and TSP are
similar; thus these two triangles have the same angles. This implies that the
angles are labeled correctly in the ﬁgure above (assuming that we start by
declaring that angle PQS measures 6 degrees). This then implies that angle
QPT measures 90". Thus the two lines in blue are perpendicular. In conclusion, we have derived the following characterization of perpen
dicular lines: Perpendicular lines Two nonvertical lines are perpendicular if and only if the product of their
slopes equals —1. Show that the lines in the xyplane given by the equations
y=4x—5 and y=—%x+18
are perpendicular. SOLUTION The ﬁrst line has slope 4; the second line has slope —%. The product of
these slopes is 4  [—41], which equals —1. Because the product of the two slopes
equals *1, the two lines are perpendicular. To show that two lines are perpendicular, we only need to know the slopes
of the lines, not their full equations. The following example illustrates this
point. Show that the line containing the points (1, 2} and (3, 7) is perpendicular to the line
containing the points (9, 3) and {4, 5). B which equals Also, 31 '
which equals 73. Because the J) SOLUTION The line containing (1,2) and (3,7) has slope the line containing (9,3) and (4,5) has Slope E, product  (—§) equals —1, the two lines are perpendicular. ...
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