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Section 2.1 Linear Equations in Two Variables 135 Notice that the x- and y-intercepts are both located at the origin (0, 0). In this case, the intercepts do not yield two distinct points. Therefore, another point is necessary to draw the line.We may pick any value for either x or y.However, for this equation, it would be particularly convenient to pick a value for x that is a multiple of 4 such as 5 4 3 2 1 (4, 1) 543 1 2 3 4 5 21 1 2 3 4 5 1 4 x 1 4 142 y x 4. y x 4. y 1 The point (4, 1) is a solution to the equation (Figure 2-9). Figure 2-9 Skill Practice 8. Given y 5x, find the x- and y-intercepts.Then graph the equation. x y (0, 0) y x 1 4 y 14 x Answer 8. 5 4 3 2 1 543 21 1 2 3 4 5 1 2 3 4 5 x y y 5x (0, 0) Substitute Interpreting the x- and y-Intercepts of a Line Example 7 Companies and corporations are permitted to depreciate assets that have a known useful life span. This accounting practice is called straight-line depreciation. In this procedure the useful life span of the asset is determined, and then the asset is depreciated by an equal amount each year until the taxable value of the asset is equal to zero. The J. M. Gus trucking company purchases a new truck for $65,000. The truck will be depreciated at $13,000 per year. The equation that describes the depreciation line is y 65,000 13,000x where y represents the value of the truck in dollars and x is the age of the truck in years. a. Find the x- and y-intercepts. Plot the intercepts on a rectangular coordinate system, and draw the line that represents the straight-line depreciation. b. What does the x-intercept represent in the context of this problem? c. What does the y-intercept represent in the context of this problem? Solution: a. To find the x-intercept, substitute To find the y-intercept, substitute y 0. x 0. 0 65,000 13,000x y 65,000 13,000102 13,000x 65,000 y 65,000 x 5 The x-intercept is (5, 0). The y-intercept is (0, 65,000). Avoiding Mistakes You can always find a third point on a line to check the accuracy of your graph. For example, the point 14, 12 satisfies the equation and lines up with the other points from Example 6.


miller_intermediate_algebra_4e_ch1_3
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