Copyright © Cengage Learning. All rights reserved. 3 Functions and Graphs 3.2 Functions and Graphs Copyright © Cengage Learning. All rights reserved.

Rectangular Coordinate Systems We shall now show how to assign an ordered pair (a, b) of real numbers to each point in a plane. Although we have also used the notation (a, b) to denote an open interval, there is little chance for confusion, since it should always be clear from our discussion whether (a, b) represents a point or an interval.

Rectangular Coordinate Systems We introduce a rectangular, or Cartesian, coordinate system in a plane by means of two perpendicular coordinate lines, called coordinate axes, that intersect at the origin O, as shown in Figure 1. Figure 1

Rectangular Coordinate Systems We often refer to the horizontal line as the x-axis and the vertical line as the y-axis and label them x and y, respectively. The plane is then a coordinate plane, or an xy-plane. The coordinate axes divide the plane into four parts called the first, second, third, and fourth quadrants, labeled I, II, III, and IV, respectively (see Figure 1). Points on the axes do not belong to any quadrant. Each point P in an xy-plane may be assigned an ordered pair (a, b), as shown in Figure 1.

Rectangular Coordinate Systems We call a the x-coordinate (or abscissa) of P, and b the y-coordinate (or ordinate). We say that P has coordinates (a, b) and refer to the point (a, b) or the point P (a, b). Conversely, every ordered pair (a, b) determines a point P with coordinates a and b. We plot a point by using a dot, as illustrated in Figure 2. Figure 2

Rectangular Coordinate Systems We may use the following formula to find the distance between two points in a coordinate plane.

Example 1 – Finding the distance between points Plot the points A(–3, 6) and B(5, 1), and find the distance d (A, B). Solution: The points are plotted in Figure 4. Figure 4

Example 1 – Solution By the distance formula, d (A, B) = =  9.43. cont’d By the distance formula, d (A, B) = =  9.43.

Example 4 – Finding a formula that describes a perpendicular bisector Given A(1, 7) and B(–3, 2), find a formula that expresses the fact that an arbitrary point P (x, y) is on the perpendicular bisector l of segment AB. Solution: By condition 2 of Example 3, P (x, y) is on l if and only if d (A, P) = d (B, P); that is,

Example 4 – Solution cont’d To obtain a simpler formula, let us square both sides and simplify terms of the resulting equation, as follows: (x – 1)2 + (y – 7)2 = [x – (– 3)]2 + (y – 2)2 x2 – 2x + 1 + y2 – 14y + 49 = x2 + 6x + 9 + y2 – 4y + 4 –2x + 1 – 14y + 49 = 6x + 9 – 4y + 4 –8x – 10y = –37 8x + 10y = 37

Example 4 – Solution cont’d Note that, in particular, the last formula is true for the coordinates of the point C(4, ) in Example 3, since if x = 4 and y = , substitution in 8x + 10y gives us 8  4 + 10  = 37.

Rectangular Coordinate Systems We can find the midpoint of a line segment by using the following formula.

Rectangular Coordinate Systems To apply the midpoint formula, it may suffice to remember that the x-coordinate of the midpoint = the average of the x-coordinates, and that the y-coordinate of the midpoint = the average of the y-coordinates.

Example 5 – Finding a midpoint Find the midpoint M of the line segment from P1(–2, 3) to P2(4, –2), and verify that d(P1, M) = d(P2, M). Solution: By the midpoint formula, the coordinates of M are

Example 5 – Solution cont’d The three points P1, P2, and M are plotted in Figure 8. Figure 8

Example 5 – Solution By the distance formula, d(P1, M) = d(P2, M) = cont’d By the distance formula, d(P1, M) = d(P2, M) = Hence, d(P1, M) = d(P2, M).

Rectangular Coordinate Systems The term graphing utility refers to either a graphing calculator or a computer equipped with appropriate software packages. The viewing rectangle of a graphing utility is the portion of the xy-plane shown on the screen. The boundaries (sides) of the viewing rectangle can be manually set by assigning a minimum x value (Xmin), a maximum x value (Xmax), the difference between the tick marks on the x-axis (Xscl), a minimum y value (Ymin), a maximum y value (Ymax), and the difference between the tick marks on the y-axis (Yscl).

Rectangular Coordinate Systems In examples, we often use the standard (or default) values for the viewing rectangle. These values depend on the dimensions (measured in pixels) of the graphing utility screen. If we want a different view of the graph, we use the phrase “using [Xmin, Xmax, Xscl] by [Ymin, Ymax, Yscl]” to indicate the change in the viewing rectangle. If Xscl and/or Yscl are omitted, the default value is 1.

Graphs of Equations Two quantities are sometimes related by means of an equation or formula that involves two variables. In this section we discuss how to represent such an equation geometrically, by a graph in a coordinate plane. The graph may then be used to discover properties of the quantities that are not evident from the equation alone.

Graphs of Equations The following chart introduces the basic concept of the graph of an equation in two variables x and y. Of course, other letters can also be used for the variables. For each solution (a, b) of an equation in x and y there is a point P (a, b) in a coordinate plane. The set of all such points is called the graph of the equation.

Graphs of Equations To sketch the graph of an equation, we illustrate the significant features of the graph in a coordinate plane. In simple cases, a graph can be sketched by plotting few, if any, points. For a complicated equation, plotting points may give very little information about the graph. In such cases, methods of calculus or computer graphics are often employed. Let us begin with a simple example.

Example 1 – Sketching a simple graph by plotting points Sketch the graph of the equation y = 2x – 1. Solution: We wish to find the points (x, y) in a coordinate plane that correspond to the solutions of the equation. It is convenient to list coordinates of several such points in a table, where for each x we obtain the value for y from y = 2x – 1:

Example 1 – Solution cont’d The points with these coordinates appear to lie on a line, and we can sketch the graph in Figure 1. Figure 1

Example 1 – Solution cont’d Ordinarily, the few points we have plotted would not be enough to illustrate the graph of an equation; however, in this elementary case we can be reasonably sure that the graph is a line.

Graphs of Equations It is impossible to sketch the entire graph in Example 1, because we can assign values to x that are numerically as large as desired. Nevertheless, we call the drawing in Figure 1 the graph of the equation or a sketch of the graph. Figure 1

Graphs of Equations In general, the sketch of a graph should illustrate its essential features so that the remaining (unsketched) parts are self-evident. For instance, in Figure 1, the end behavior—the pattern of the graph as x assumes large positive and negative values (that is, the shape of the right and left ends)—is apparent to the reader.

Graphs of Equations For written work, we use arrow notation from the following chart when describing functions and their end behavior.

Graphs of Equations The symbols (read “infinity”) and (read “minus infinity”) do not represent real numbers; they simply specify certain types of behavior of functions and variables. If a graph terminates at some point (as would be the case for a half-line or line segment), we place a dot at the appropriate endpoint of the graph.

Graphs of Equations As a final general remark, if ticks on the coordinate axes are not labeled (as in Figure 1), then each tick represents one unit. Figure 1

Graphs of Equations We shall label ticks only when different units are used on the axes. For arbitrary graphs, where units of measurement are irrelevant, we omit ticks completely.

Graphs of Equations The graph in Figure 2 is a parabola, and the y-axis is the axis of the parabola. Figure 2

Graphs of Equations The lowest point (0, –3) is the vertex of the parabola, and we say that the parabola opens upward. If we invert the graph, then the parabola opens downward and the vertex is the highest point on the graph. In general, the graph of any equation of the form y = ax2 + c with a ≠ 0 is a parabola with vertex (0, c), opening upward if a > 0 or downward if a < 0. If c = 0, the equation reduces to y = ax2 and the vertex is at the origin (0, 0). Parabolas may also open to the right or to the left or in other directions.

Graphs of Equations We shall use the following terminology to describe where the graph of an equation in x and y intersects the x-axis or the y-axis. Intercepts of the Graph of an Equation in x and y

Graphs of Equations An x-intercept is sometimes referred to as a zero of the graph of an equation or as a root of an equation. When using a graphing utility to find an x-intercept, we will say that we are using a root feature.

Example 3 – Finding x-intercepts and y-intercepts Find the x- and y-intercepts of the graph of y = x2 – 3. Solution: Substituting values for x and finding the corresponding values of y using y = x2 – 3, we obtain a table of coordinates for several points on the graph:

Example 3 – Solution cont’d Larger values of |x| produce larger values of y. For example, the points (4, 13), (5, 22), and (6, 33) are on the graph, as are (–4, 13), (–5, 22), and (–6, 33). Plotting the points given by the table and drawing a smooth curve through these points (in the order of increasing values of x) gives us the sketch in Figure 2. We find the intercepts as stated in the preceding chart. Figure 2

Example 3 – Solution (1) x-intercepts: y = x2 – 3 0 = x2 – 3 x2 = 3 cont’d (1) x-intercepts: y = x2 – 3 0 = x2 – 3 x2 = 3 x =   1.73 Thus, the x-intercepts are and . The points at which the graph crosses the x-axis are and . given let y = 0 equivalent equation take the square root

Example 3 – Solution (2) y-intercepts: y = x2 – 3 y = 0 – 3 = –3 cont’d (2) y-intercepts: y = x2 – 3 y = 0 – 3 = –3 Thus, the y-intercept is –3, and the point at which the graph crosses the y-axis is (0, –3). given let x = 0

Graphs of Equations If the coordinate plane in Figure 2 is folded along the y-axis, the graph that lies in the left half of the plane coincides with that in the right half, and we say that the graph is symmetric with respect to the y-axis. Figure 2

Graphs of Equations A graph is symmetric with respect to the y-axis provided that the point (–x, y) is on the graph whenever (x, y) is on the graph. The graph of y = x2 – 3 has this property, since substitution of –x for x yields the same equation: y = (–x)2 – 3 = x2 – 3 This substitution is an application of symmetry test 1 in the next chart. Two other types of symmetry and the appropriate tests are also listed.

Graphs of Equations Symmetries of Graphs of Equations in x and y

Graphs of Equations Symmetries of Graphs of Equations in x and y cont’d Symmetries of Graphs of Equations in x and y

Graphs of Equations If a graph is symmetric with respect to an axis, it is sufficient to determine the graph in half of the coordinate plane, since we can sketch the remainder of the graph by taking a mirror image, or reflection, through the appropriate axis.

Example 5 – A graph that is symmetric with respect to the x-axis Sketch the graph of the equation y2 = x. Solution: Since substitution of –y for y does not change the equation, the graph is symmetric with respect to the x-axis (see symmetry test 2). Hence, if the point (x, y) is on the graph, then the point (x, –y) is on the graph. Thus, it is sufficient to find points with nonnegative y-coordinates and then reflect through the x-axis.

Example 5 – Solution The equation y2 = x is equivalent to y = . cont’d The equation y2 = x is equivalent to y = . The y-coordinates of points above the x-axis (y is positive) are given by y = , whereas the y-coordinates of points below the x-axis (y is negative) are given by y = . Coordinates of some points on the graph are listed below.

Example 5 – Solution The graph is sketched in Figure 3. cont’d The graph is sketched in Figure 3. The graph is a parabola that opens to the right, with its vertex at the origin. In this case, the x-axis is the axis of the parabola. Figure 3

Graphs of Equations If C (h, k) is a point in a coordinate plane, then a circle with center C and radius r > 0 consists of all points in the plane that are r units from C. As shown in Figure 5, a point P(x, y) is on the circle provided d(C, P) = r or, by the distance formula, The above equation is equivalent to the next equation, which we will refer to as the standard equation of a circle. Figure 3

Graphs of Equations If h = 0 and k = 0, this equation reduces to x2 + y2 = r2, which is an equation of a circle of radius r with center at the origin (see Figure 6). If r = 1, we call the graph a unit circle. Figure 6

Example 7 – Finding an equation of a circle Find an equation of the circle that has center C(–2, 3) and contains the point D(4, 5). Solution: The circle is shown in Figure 7. Since D is on the circle, the radius r is d (C, D). By the distance formula, Figure 7

Example 7 – Solution cont’d Using the standard equation of a circle with h = –2, k = 3, and r = , we obtain (x + 2)2 + (y – 3)2 = 40. By squaring terms and simplifying the last equation, we may write it as x2 + y2 + 4x – 6y – 27 = 0.

Graphs of Equations As in the solution to Example 7, squaring terms of an equation of the form (x – h)2 + (y – k)2 = r2 and simplifying leads to an equation of the form x2 + y2 + ax + by + c = 0, where a, b, and c are real numbers. Conversely, if we begin with this equation, it is always possible, by completing squares, to obtain an equation of the form (x – h)2 + (y – k)2 = d.

Graphs of Equations This method will be illustrated in Example 8. If d > 0, the graph is a circle with center (h, k) and radius r = . If d = 0, the graph consists of only the point (h, k). Finally, if d < 0, the equation has no real solutions, and hence there is no graph.

Example 8 – Finding an equation of a circle Find the center and radius of the circle with equation 3x2 + 3y2 – 12x + 18y = 9. Solution: Since it is easier to complete the square if the coefficients of x2 and y2 are 1, we begin by dividing the given equation by 3, obtaining x2 + y2 – 4x + 6y = 3.

Example 8 – Solution cont’d Next, we rewrite the equation as follows, where the underscored spaces represent numbers to be determined: (x2 – 4x + _) + (y2 + 6y + _) = 3 + _ + _ We then complete the squares for the expressions within parentheses, taking care to add the appropriate numbers to both sides of the equation. To complete the square for an expression of the form x2 + ax, we add the square of half the coefficient of x (that is, (a/2)2) to both sides of the equation.

Example 8 – Solution cont’d Similarly, for y2 + by, we add (b/2)2 to both sides. In this example, a = –4, b = 6, (a/2)2 = (–2)2 = 4, and (b/2)2 = 32 = 9. These additions lead to (x2 – 4x + 4) + (y2 + 6y + 9) = 3 + 4 + 9 (x – 2)2 + (y + 3)2 = 16. Comparing the last equation with the standard equation of a circle, we see that h = 2 and k = –3 and conclude that the circle has center (2, –3) and radius . completing the squares equivalent equation

Example 8 – Solution A sketch of this circle is shown in Figure 8. cont’d A sketch of this circle is shown in Figure 8. Figure 8

Graphs of Equations In some applications it is necessary to work with only one-half of a circle—that is, a semicircle. The next example indicates how to find equations of semicircles for circles with centers at the origin.

Example 9 – Finding equations of semicircles Find equations for the upper half, lower half, right half, and left half of the circle x2 + y2 = 81 Solution: The graph of x2 + y2 = 81 is a circle of radius 9 with center at the origin (see Figure 9). Figure 9

Example 9 – Solution cont’d To find equations for the upper and lower halves, we solve for y in terms of x: x2 + y2 = 81 y2 = 81 – x2 y =  given subtract x2 take the square root

Example 9 – Solution cont’d Since  0, it follows that the upper half of the circle has the equation y = (y is positive) and the lower half is given by y = (y is negative), as illustrated in Figure 10(a) and (b). (a) (b) Figure 10

Example 9 – Solution cont’d Similarly, to find equations for the right and left halves, we solve x2 + y2 = 81 for x in terms of y, obtaining

Example 9 – Solution cont’d Since , it follows that the right half of the circle has the equation (x is positive) and the left half is given by the equation (x is negative), as illustrated in Figure 10(c) and (d). (c) (d) Figure 10