1m22 the equation 3 x 2 y 4 of the other line may be

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1.M.22: The equation 3 x 2 y = 4 of the other line may be written in the form y = 3 2 x 2, revealing that it and L have slope 3 2 . Hence an equation of L is y ( 3) = 3 2 ( x 2); that is, y = 3 2 x 6 . 1.M.23: The equation y 2 x = 10 may be written in the form y = 2 x + 10, showing that it has slope 2. Hence the perpendicular line L has slope 1 2 . Therefore an equation of L is y 7 = 1 2 ( x ( 3)); that is, x + 2 y = 11 . 1.M.24: The segment S joining (1 , 5) and (3 , 1) has slope ( 1 ( 5)) / (3 1) = 2 and midpoint (2 , 3), and hence L has slope 1 2 and passes through (2 , 3). So an equation of L is y ( 3) = 1 2 ( x 2); that is, x + 2 y = 4 . 1.M.25: The graph of y = f ( x ) = 2 2 x x 2 is a parabola opening downward. The only such graph is shown in Fig. 1.MP.6. 1.M.26: Given: f ( x ) = x 3 4 x 2 + 5. Because f ( 1) = 0, f (1) = 2 > 0 > 3 = f (2), and f (3) = 4 < 0 < 5 = f (4), the graph of f crosses the x -axis at x = 1, between x = 1 and x = 2, and between x = 3 and x = 4. Hence the graph of f is the one shown in Fig. 1.MP.9. 1.M.27: Given: f ( x ) = x 4 4 x 3 + 5. Because the graph of f has no vertical asymptotes and because f ( x ) approaches + as x approaches either + or −∞ , the graph of f must be the one shown in Fig. 1.MP.4. 41
1.M.28: Given: f ( x ) = 5 x 2 x 6 = 5 ( x 3)( x + 2) . The denominator in f ( x ) is zero when x = 3 and when x = 2 (and the numerator is not zero), so the graph of y = f ( x ) has vertical asymptotes at x = 2 and at x = 3. Also f ( x ) approaches zero as x approaches either + or −∞ . Therefore the graph of y = f ( x ) must be the one shown in Fig. 1.MP.11. 1.M.29: Given: f ( x ) = 5 x 2 x + 6 = 20 4 x 2 4 x + 1 + 23 = 20 (2 x 1) 2 + 23 . The algebra displayed here shows that the denominator in f ( x ) is never zero, so there are no vertical asymptotes. It also shows that the maximum value of f ( x ) occurs when the denominator is minimal; that is, when x = 1 2 . Finally, f ( x ) approaches zero as x approaches either + or −∞ . So the graph of y = f ( x ) must be the one shown in Fig. 1.MP.3. 1.M.30: If y = f ( x ) = 8 + 2 x x 2 , then y 2 = 8 + 2 x x 2 ; x 2 2 x + 1 + y 2 = 9; ( x 1) 2 + ( y 0) 2 = 3 2 . The last is the equation of a circle with center (1 , 0) and radius 3. But y 0, so the graph of f is the upper half of that circle, and it is shown in Fig. 1.MP.10. 1.M.31: Given: f ( x ) = 2 x 1. The graph of y = 2 x is an increasing exponential function, so the graph of y = 2 x is a decreasing exponential function, approaching 0 as x approaches + . So the graph of f approaches 1 as x approaches + . Moreover, f (0) = 0. Therefore the graph of f is the one shown in Fig. 1.MP.7. 1.M.32: The graph of f ( x ) = log 10 ( x + 1) is obtained from the graph of g ( x ) = log 10 x by translation one unit to the left; note also that f (0) = 0. Therefore the graph of f is the one shown in Fig. 1.MP.2. 1.M.33: The graph of y = 3 sin x oscillates between its minimum value 3 and its maximum value 3, so the graph of f ( x ) = 1 + 3 sin x oscillates between 2 and 4. This graph is shown in Fig. 1.MP.8.
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