261 2.5 Equations of Lines and Linear Models Graphical Solution of Linear Equations in One Variable Suppose that y1 and y2 are linear expressions in x. We can solve the equation y1 = y2 graphically as follows (assuming it has a unique solution). 1. Rewrite the equation as y1 - y2 = 0. 2. Graph the linear function y3 = y1 - y2. 3. Find the x-intercept of the graph of the function y3. This x-value is the solution of y1 = y2. Some calculators use the term zero to identify the x-value of an x-intercept. In general, if ƒ1a2 = 0, then a is a zero of ƒ. 7 EXAMPLE 8 Solving an Equation with a Graphing Calculator Use a graphing calculator to solve -2x - 412 - x2 = 3x + 4. SOLUTION We write an equivalent equation with 0 on one side. -2x - 412 - x2 - 3x - 4 = 0 Subtract 3x and 4. Then we graph y = -2x - 412 - x2 - 3x - 4 to find the x-intercept. The standard viewing window cannot be used because the x-intercept does not lie in the interval 3-10, 104. As seen in Figure 53, the solution of the equation is -12, and the solution set is 5-126. S Now Try Exercise 69. y1=−2x−4(2−x)−3x−4 −20 −15 10 5 Figure 53 2.5 Exercises CONCEPT PREVIEW Fill in the blank(s) to correctly complete each sentence. 1. The graph of the line y - 3 = 41x - 82 has slope and passes through the point 18, __2. 2. The graph of the line y = -2x + 7 has slope and y-intercept . 3. The vertical line through the point 1-4, 82 has equation = -4. 4. The horizontal line through the point 1-4, 82 has equation _______ = 8. 5. Any line parallel to the graph of 6x + 7y = 9 must have slope . 6. Any line perpendicular to the graph of 6x + 7y = 9 must have slope . CONCEPT PREVIEW Match each equation with its graph in A–D. 7. y = 1 4 x + 2 8. 4x + 3y = 12 9. y - 1-12 = 3 2 1x - 12 10. y = 4 A. (0, 4) x y 0 B. (0, 4) (3, 0) x y 0 C. (1, –1) (–1, –4) x y 0 3 2 D. x y 0 (0, 2) 1 4
RkJQdWJsaXNoZXIy NjM5ODQ=