Text_Section_2.1

# Text_Section_2.1 - Linear Quadratic Polynomial and Rational...

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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 3’2-3’1 = y4-y3 )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 3’2-3’1 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 xy-plane that has slope m and contains the point (x1, y1) ten used to denote is given by the equation the slope of a line. y-y1=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 xy-plane 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 xy-plane 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 y-plane 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 y-axis 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 y-axis 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 xy-plane 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 xy-plane 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 xy-plane 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 xy-plane 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 xy-plane 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 xy-plane 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, 3-1 ' 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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Text_Section_2.1 - Linear Quadratic Polynomial and Rational...

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