1. f(x) = 1/4x − 1. B)The line will be less steep. The slope is lower now
2. C)x-intercept (−1/3 , 0). Replacing f(x) = 0 we get x = -1/3
3. B)horizontal line with a slope of zero. The slope is 0 because there is no x in the equation.
4. D)Line changes from decreasing to increasing. The slope changed from negative to positive.
5. See first graph attached
6. See second graph attached. g(x) = (1/3)*x
7. C)The rate of change is different, but the y-intercepts are the same. The rate of change of Mike is negative because he withdraws $25 each week. The y-intercepts of Mike is $200 because that was his initial deposit.
8. B)shifts the line 7 units down. f(x-7) translates f(x) 7 units to the right, given the slope is positive ( = 1) then f(x) is translated 7 units down.
9. No graph is shown
10. C)g(x) = x - 3. g(x) = f(x) - 3, so g(x) is f(x) translated 3 units down.
1. f(x) = 1/4x − 1. B)The line will be less steep. The slope is lower now
2. C)x-intercept (−1/3 , 0). Replacing f(x) = 0 we get x = -1/3
3. B)horizontal line with a slope of zero. The slope is 0 because there is no x in the equation.
4. D)Line changes from decreasing to increasing. The slope changed from negative to positive.
5. See first graph attached
6. See second graph attached. g(x) = (1/3)*x
7. C)The rate of change is different, but the y-intercepts are the same. The rate of change of Mike is negative because he withdraws $25 each week. The y-intercepts of Mike is $200 because that was his initial deposit.
8. B)shifts the line 7 units down. f(x-7) translates f(x) 7 units to the right, given the slope is positive ( = 1) then f(x) is translated 7 units down.
9. No graph is shown
10. C)g(x) = x - 3. g(x) = f(x) - 3, so g(x) is f(x) translated 3 units down.
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