1 Using a Graphing Calculator A graphing calculator or computer displays a rectangular portion of the graph of an equation in a display window or viewing.

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Presentation transcript:

1 Using a Graphing Calculator A graphing calculator or computer displays a rectangular portion of the graph of an equation in a display window or viewing screen, which we call a viewing rectangle. The default screen often gives an incomplete or misleading picture, so it is important to choose the viewing rectangle with care.

2 Using a Graphing Calculator We refer to this as the [a, b] by [c, d ] viewing rectangle. It plots points of the form (x, y) for a certain number of values of x, equally spaced between a and b. If the equation is not defined for an x-value or if the corresponding y-value lies outside the viewing rectangle, the device ignores this value and moves on to the next x-value.

3 Example 1 – Choosing an Appropriate Viewing Rectangle Graph the equation y = x in an appropriate viewing rectangle. Solution: Let’s experiment with different viewing rectangles. We start with the viewing rectangle [–2, 2] by [–2, 2], so we set Xmin = –2 Ymin = –2 Xmax = 2 Ymax = 2

4 Example 1 – Solution The resulting graph in Figure 2(a) is blank! This is because x 2  0, so x  3 for all x. Thus, the graph lies entirely above the viewing rectangle, so this viewing rectangle is not appropriate. cont’d Figure 2(a) Graphs of y = x 2 + 3

5 Example 1 – Solution If we enlarge the viewing rectangle to [–4, 4] by [–4, 4], as in Figure 2(b), we begin to see a portion of the graph. cont’d Figure 2(b) Graphs of y = x 2 + 3

6 Example 1 – Solution Now let’s try the viewing rectangle [–10, 10] by [–5, 30]. The graph in Figure 2(c) seems to give a more complete view of the graph. cont’d Figure 2(c) Graphs of y = x 2 + 3

7 Example 1 – Solution If we enlarge the viewing rectangle even further, as in Figure 2(d), the graph doesn’t show clearly that the y-intercept is 3. So the viewing rectangle [–10, 10] by [–5, 30] gives an appropriate representation of the graph. cont’d Figure 2(d) Graphs of y = x 2 + 3

8 Solving Equations Graphically Note that from the graph we obtain an approximate solution. We summarize these methods in the box below.

9 Example 4 – Solving a Quadratic Equation Algebraically and Graphically Solve the quadratic equations algebraically and graphically. (a) x 2 – 4x + 2 = 0 (b) x 2 – 4x + 4 = 0 (c) x 2 – 4x + 6 = 0 Solution 1: Algebraic We use the Quadratic Formula to solve each equation. (a)

10 Example 4 – Solution 1 There are two solutions, x = 2 + and x = 2 –. (b) (c) There is no real solution. cont’d

11 Example 4 – Solution 2 Solution 2: Graphical We graph the equations y = x 2 – 4x + 2, y = x 2 – 4x + 4, and y = x 2 – 4x + 6 in Figure 6. Figure 6(a) y = x 2 – 4x + 2 Figure 6(b) y = x 2 – 4x + 4 Figure 6(c) y = x 2 – 4x + 6

12 Example 4 – Solution 2 By determining the x-intercepts of the graphs, we find the following solutions. (a) x  0.6 and x  3.4 (b) x = 2 (c) There is no x-intercept, so the equation has no solution. cont’d

13 Solving Inequalities Graphically Inequalities can be solved graphically. To describe the method, we solve x 2 – 5x + 6  0 To solve the inequality graphically, we draw the graph of y = x 2 – 5x + 6 Our goal is to find those values of x for which y  0.

14 Solving Inequalities Graphically These are simply the x-values for which the graph lies below the x-axis. From Figure 11 we see that the solution of the inequality is the interval [2, 3]. Figure 11 x 2 – 5x + 6  0

15 Example 8 – Solving an Inequality Graphically Solve the inequality 3.7x x – 1.9  2.0 – 1.4x. Solution: We graph the equations y 1 = 3.7x x – 1.9 and y 2 = 2.0 – 1.4x in the same viewing rectangle in Figure 12. Figure 12 y 1 = 3.7x x – 1.9 y 2 = 2.0 – 1.4x

16 Example 8 – Solution We are interested in those values of x for which y 1  y 2 ; these are points for which the graph of y 2 lies on or above the graph of y 1. To determine the appropriate interval, we look for the x-coordinates of points where the graphs intersect. We conclude that the solution is (approximately) the interval [–1.45, 0.72]. cont’d