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Example 2.3Find the distance between the points ( 5, 8) and ( 10, 14).Solution.Applying the distance formula we findp d (14 8)2 ( 10 ( 5))2 36 25 61.The Midpoint FormulaThe point halfway between the endpoints of a line segment is called themidpoint. Thus, a midpoint divides a line segment into two equal parts.Let M (a, b) be the midpoint of the line segment with endpoints A(x1 , y1 ) andB(x2 , y2 ). See Figure 2.3.Figure 2.3The triangles M AN and BM P are similar so that we can write M A AN . BM M P But M A BM so that AN M P . Also, M P N C so that AN N C . Thus, N is the midpoint of the line segment with endpoints A2and C. It follows that a x1 x2 a or a x1 x. A similar argument shows2y1 y2that b 2 . Thus, the midpoint M is given by the midpoint formula x1 x2 y1 y2M,.22Example 2.4Find the midpoint of the line segment with endpoints A(4, 7) and B( 10, 7).16

Solution.Plugging into the midpoint formula we find 2 y1 y2, 2 Midpoint x1 x24 ( 10) 7 7 , 22 ( 3, 7)Graph of an EquationGiven an equation involving the two variables x and y. The graph of anequation is the set of ordered pairs (x, y) that satisfy the equation.A typical procedure for graphing an equation is to plot points and thenconnect them with a continuous curve as shown in the next examples.Example 2.5Graph the equation by plotting points: 2x y 1.Solution.Writing y in terms of x we find y 1 2x. The table below shows somepoints on the graph of the equation.x -2 -1 0 1 2y 3 1 -1 -3 -5Next, plot the points and draw a curve through them. See Figure 2.4.Figure 2.4Example 2.6Graph the equation by plotting points: y x 3 2.17

Solution.The table below shows some points on the graph of the equation.x -6y 1-5 -4 -3 -2 -1 00 -1 -2 -1 0 1Next, plot the points and draw a curve through them. See Figure 2.5.Figure 2.5Example 2.7Graph the equation y x2 2x 8.Solution.The table below shows some points on the graph of the equation.x -3 - 2 -1 0 1 2 3 4 5y 7 0-5 -8 -9 -8 -5 0 7Next, plot the points and draw a curve through them. See Figure 2.6.18

Figure 2.6InterceptsA point (x, 0) on the graph of an equation is called the x-intercept. Geometrically, the x-intercept is the point where the graph crosses the x-axis.Similarly, a point of the form (0, y) is called the y-intercept. This is thepoint where the graph crosses the y-axis.Example 2.8Find the x- and y-intercepts of the graph of x2 y 2 4.Solution.Letting y 0 in the given equation we find x2 4. Solving for x to obtainx 2. Thus, the x-intercepts are the points ( 2, 0) and (2, 0). Similarly,setting x 0 to obtain y 2 4. Solving for y we obtain y 2. So the points(0, 2) and (0, 2) are the y-intercepts.The Equation of a CircleBy a circle we mean the collection of all points in the plane that are at anequal distance to a fixed point called the center of the circle. The distanceof a point on a circle to its center is called the radius. The diameter of a19

circle is the length of a line segment crossing the center and with endpointson the circle. Thus, the center is the midpoint and as a result a diameter istwice the radius.Next, we want to find the equation of a circle with center C(a, b) and radiusr. For this, let M (x, y) be an arbitrary point on the circle. Then d(C, M ) r.By the distance formula, we have(x a)2 (y b)2 r2 .This equation is called the standard form of the equation of a circle.Example 2.9Determine the center and the radius of the circle with equation: (x 2)2 (y 4)2 25.Solution. The center is the point (2, 4) and the radius is r 25 5.Example 2.10Find the equation of the circle with center C(5, 3) and radius r 4. Writethe answer in standard form.Solution.The equation of the circle is given by(x 5)2 (y 3)2 16.Example 2.11Find the equation of the circle with center C( 2, 5) and passing through thepoint M (1, 7).Solution.p The radius of the circle is r d(C, M ) (7 5)2 (1 ( 2))2 13.Thus, the equation of the circle is(x 2)2 (y 5)2 13.Another form of the equation of a circle is known as the general form andis given by the equationx2 y 2 Ax By C 0.To find the standard form from the general form we use the process of completing the square as shown in the following example.20

Example 2.12Find the center and the radius of the circle: x2 y 2 6x 4y 12 0.Solution.We use the method of completing the square:(x2 6x) (y 2 4y) 12(x 6x 9) (y 2 4y 4) 12 9 4(x 3)2 (y 2)2 1.2Thus, the center is (3, 2) and the radius is r 1.Example 2.13Find the equation of a circle that has diameter with endpoints (7, 2) and( 3, 5). Write your answer in standard form.Solution.The center of the circle is the midpoint of the given diameter. By the mid, 2 5) (2, 32 ). Thepoint formula, the coordinates of the center are ( 7 322radius of the circle is the distance between the center and one of the endpoints. This can be found by using the distance formular 3149d (2 7)2 ( 2)2 .22The equation of the circle is3149(x 2)2 (y )2 .24Example 2.14Find an equation of a circle that has its center at ( 2, 3) and is tangent tothe y-axis. Write your answer in standard form.Solution.The radius of the circle is the distance from the center to the y-axis whichis the absolute value of the x-coordinate of the the center, i.e. r 2. Hence,the equation of the circle is given by(x 2)2 (y 3)2 4.21

Review ProblemsExercise 2.1Plot the points whose coordinates are given on a Cartesian coordinate system.(a) (2, 4), (0, 3), ( 2, 1), ( 5, 3).(b) ( 3, 5), ( 4, 3), (0, 2), ( 2, 0).Exercise 2.2Find the distance between the points whose coordinates are given.(a) (6, 4), ( 8, 11).(b) (5,0). (0, 8),(c) ( 3, 8), ( 12, 27).(d) (x, 4x), ( 2x, 3x), x 0.Exercise 2.3Find the midpoint of the line segment with the following endpoints.(a) (1, 1), (5, 5).(b) (6, 3), (6, 11).(c) (1.75, 2.25), ( 3.5, 5.57).Exercise 2.4Graph each equation by plotting points that satisfy the equation.(a) x y 4.(b) y 2 x 3 .(c) y 21 (x 1)2 .(d) y x2 2x 8.Exercise 2.5Find the x- and y-intercepts of each equation.(a) 2x 5y 12.(b) x y 4.(c) x y 4.(d) x 4y 8.22

Exercise 2.6Determine the center and the radius of the circle with the given equation.(a) x2 y 2 36.(b) (x 2)2 (y 5)2 25.(c) (x 8)2 y 2 14 .Exercise 2.7Find the equation of the circle with center C(4, 1) and radius r 2. Writethe answer in standard form.Exercise 2.8Find the equation of the circle with center C(0, 0) and passing through thepoint M ( 3, 4).Exercise 2.9Find the equation of the circle with center C(1, 3) and passing through thepoint M (4, 1).Exercise 2.10Find the center and the radius of each of the following circles.(a) x2 y 2 6x 5 0.(b) 4x2 4y 2 4x 63 0. 0.(c) x2 y 2 x 3y 154Exercise 2.11Find the equation of a circle that has diameter with endpoints (2, 3) and( 4, 11). Write your answer in standard form.Exercise 2.12Find an equation of a circle that has its center at (7, 11) and is tangent tothe x-axis. Write your answer in standard form.Exercise 2.13Given the midpoint M (9, 3) of a line segement with endpoints A(x, y) andB(5, 1). Find the coordinates of A.23

Exercise 2.14Find a formula for the set of all points (x, y) for which the distance from(x, y) to (3, 4) is 5.Exercise 2.15Find an equation of a circle that is tangent to both axes, has its center inthe second quadrant, and has a radius 3.24

3Functions and Function NotationFunctions play a crucial role in mathematics. A function describes how onequantity depends on others. More precisely, when we say that a quantity yis a function of a quantity x we mean a rule that assigns to every possiblevalue of x exactly one value of y. We call x the input and y the output. Infunction notation we writey f (x).Since y depends on x it makes sense to call x the independent variableand y the dependent variable.In applications of mathematics, functions are often representations of realworld phenomena. Thus, the functions in this case are referred to as mathematical models. If the set of input values is a finite set then the modelsare known as discrete models. Otherwise, the models are known as continuous models. For example, if H represents the temperature after t hoursfor a specific day, then H is a discrete model. If A is the area of a circle ofradius r then A is a continuous model.There are four common ways in which functions are presented and used: Byverbal descriptions, by tables, by graphs, and by formulas.Example 3.1The sales tax on an item is 6%. So if p denotes the price of the item and Cthe total cost of buying the item then if the item is sold at 1 then the costis 1 (0.06)(1) 1.06 or C(1) 1.06. If the item is sold at 2 then thecost of buying the item is 2 (0.06)(2) 2.12, or C(2) 2.12, and so on.Thus, we have a relationship between the quantities C and p such that eachvalue of p determines exactly one value of C. In this case, we say that C isa function of p. Describes this function using words, a table, a graph, and aformula.Solution. Words: To find the total cost, multiply the price of the item by 0.06 andadd the result to the price. Table: The chart below gives the total cost of buying an item at price p asa function of p for 1 p 6.pC1234561.06 2.12 3.18 4.24 5.30 6.3625

Graph: The graph of the function C is obtained by plotting the data inthe above table. See Figure 3.1. Formula: The formula that describes the relationship between C and p isgiven byC(p) 1.06p.Figure 3.1Recognizing a Function from a TableA table can be viewed as a collection of ordered pairs (x, y). Thus, for a collection of data to define a function we need to show that every first componentx corresponds to exactly one component y. Thus, if there are ordered pairswith the same x value but different y values then the collection of orderedpairs in not a function.Example 3.2Identify the set of ordered pairs (x, y) that define y as a function of x.(a) {(5, 10), (3, 2), (4, 7), (5, 8)}.(b) {(2, 2), (3, 3), (7, 2)}.Solution.(a) The first set does not define a function since the ordered pairs (5, 10) and(5, 8) have the same first component with different second components.(b) This set defines a function since all the first components are different.26

Recognizing a Function from an EquationSuppose that an equation in the variables x and y is given. If for a givenvalue of x, you solve the equation for y and you get exactly one value thenthe equation defines a function.Example 3.3Identify the equations that define y as a function of x.(a) x2 2y 2.(b) x2 y 2 1.Solution.2(a) Solving the equation for y we find y x2 1. Thus, each value of x yieldsexactly one value of y. This shows that y is a function of x.(b) Solving for y to obtain y 1 x2 . Thus, if we let x 0 then y 1.Hence, y is not a function of x.Recognizing a Function from a GraphNext, suppose that the graph of a relationship between two quantities x andy is given. To say that y is a function of x means that for each value ofx there is exactly one value of y. Graphically, this means that each verticalline must intersect the graph at most once. Hence, to determine if a graphrepresents a function one uses the following test:Vertical Line Test: A graph is a function if and only if every verticalline crosses the graph at most once.According to the vertical line test and the definition of a function, if a vertical line cuts the graph more than once, the graph could not be the graphof a function since we have multiple y values for the same x-value and thisviolates the definition of a function.27

Example 3.4Which of the graphs (a), (b), (c) in Figure 3.2 represent y as a function ofx?Figure 3.2Solution.By the vertical line test, (b) represents a function whereas (a) and (c) fail torepresent functions since one can find a vertical line that intersects the graphmore than once.Evaluating a FunctionBy evaluating a function, we mean figuring out the output value corresponding to a given input value. Thus, notation like f (10) 4 means that thefunction’s output, corresponding to the input 10, is equal to 4.If the function is given by a formula, say of the form y f (x), then to findthe output value corresponding to an input value a we replace the letter xin the formula of f by the input a and then perform the necessary algebraicoperations to find the output value.Example 3.52 1Let g(x) x5 x. Evaluate the following expressions:(a) g(2) (b) g(a) (c) g(a) 2 (d) g(a) g(2).28

Solution.2 1 57(a) g(2) 25 22 1(b) g(a) a5 a2 1(c) g(a) 2 a5 a 2 5 a 5 a(d) g(a) g(2) a2 15 a 57 a2 2a 95 a7(a2 1) 75 5 a7(5 a)5 a 7a2 5a 18.7a 35Domain and Range of a FunctionIf we try to find the possible input values that can be used in the functiony x 2 we see that we must restrict x to the interval [2, ), that isx 2. Similarly, the function y x12 takes only certain values for the output, namely, y 0. Thus, a function is often defined for certain values of xand the dependent variable often takes certain values.The above discussion leads to the following definitions: By the domain ofa function we mean all possible input values that yield one output value.Graphically, the domain is part of the horizontal axis. The range of a function is the collection of all possible output values. The range is part of thevertical axis.When finding the domain of a function, ask yourself what values can’t beused. Your domain is everything else. There are simple basic rules to consider: The domain of all polynomial functions, i.e. functions of the form f (x) an xn an 1 xn 1 · · · a1 x a0 , where n is nonnegative integer, is the Realnumbers R. Square root functions can not contain a negative underneath the radical.Set the expression under the radical greater than or equal to zero and solvefor the variable. This will be your domain. Fractional functions, i.e. ratios of two functions, determine for which inputvalues the numerator and denominator are not defined and the domain iseverything else. For example, make sure not to divide by zero!Example 3.6Find, algebraically, the domain and the range of each of the following functions. Write your answers in interval notation:1(c) y 2 x1 .(a) y πx2 (b) y x 4Solution.(a) Since the function is a polynomial then its domain is the interval ( , ).29

To findrange, solve the given equation for x in terms of y obtainingptheyx π . Thus, x exists for y 0. So the range is the interval [0, ).1(b) The domain of y x 4consists of all numbers x such that x 4 0 orx 4. That is, the interval (4, ). To find the range, we solve for x in termsof y 0 obtaining x 4 y12 . x exists for all y 0. Thus, the range is theinterval (0, ).(c) The domain of y 2 x1 is the interval ( , 0) (0, ). To find the range,1write x in terms of y to obtain x y 2. The values of y for which this laterformula is defined is the range of the given function, that is, ( , 2) (2, ).Piecewise Defined FunctionsPiecewise-defined functions are functions defined by different formulasfor different intervals of the independent variable.Example 3.7 (The Absolute Value Function)(a) Show that the function f (x) x is a piecewise defined function.(b) Graph f (x).Solution.(a) The absolute value function x is a piecewise defined function since x for x 0, x x for x 0.(b) The graph is given in Figure 3.3.Figure 3.330

Example 3.8 (The Ceiling Function)The Ceiling function f (x) dxe is the piecewise defined function given bydxe smallest integer greater than or equal to x.Sketch the graph of f (x) on the interval [ 3, 3].Solution.The graph is given in Figure 3.4. An open circle represents a point which isnot included.Figure 3.4Example 3.9 (The Floor Function)The Floor function f (x) bxc is the piecewise defined function given bybxc greatest integer less than or equal to x.Sketch the graph of f (x) on the interval [ 3, 3].Solution.The graph is given in Figure 3.5.31

Figure 3.5Example 3.10Sketch the graph of the piecewise defined function given by for x 2, x 42for 2 x 2,f (x) 4 xfor x 2.Solution.The following table gives values of f (x).x-3 -2f(x) 1 2-1 0 1 2 32 2 2 2 1The graph of the function is given in Figure 3.6.Figure 3.632

We next give a real-world situation where piecewise functions can be used.Example 3.11The charge for a taxi ride is 1.50 for the first 15 of a mile, and 0.25 for eachadditional 15 of a mile (rounded up to the nearest 15 mile).(a) Sketch a graph of the cost function C as a function of the distance traveled x, assuming that 0 x 1.(b) Find a formula for C in terms of x on the interval [0, 1].(c) What is the cost for a 54 mile ride?Solution.(a) The graph is given in Figure 3.7.Figure 3.7(b) A formula of C(x) is 0 1.50 1.75C(x) 2.00 2.25 2.50(c) The cost for a45if x 0if 0 x 51 ,if 51 x 25 ,if 25 x 35 ,if 35 x 45 ,if 45 x 1.ride is C( 45 ) 2.25.33

Increasing and Decreasing FunctionsWe say that a function is increasing if its graph climbs as x moves from leftto right. That is, the function values increase as x increases. It is said to bedecreasing if its graph falls as x moves from left to right. This means thatthe function values decrease as x increases.Example 3.12Determine the intervals where the function, given in Figure 3.8, is increasingand decreasing.Figure 3.8Solution.The function is increas

Trigonometry in modern time is an indispensable tool in Physics, engineer-ing, computer science, biology, and in practically all the sciences. This book consists of my lectures of a freshmen-level mathematics class of-fered at Arkansas Tech University. This book