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Question 81712: Given f(x) = –2x – 1, find f(a – 3).
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! Given f(x)=-2x-1, find f(a-3)
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f(a-3) tells you to replace x with the quantity (a-3). When you do that replacement,
the original equation for f(x) becomes:
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f(a-3) = -2(a-3) - 1
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Do the distributed multiplication on the right side by multiplying -2 times each of the two
terms in the parentheses to get:
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f(a-3) = -2a + 6 - 1
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Simplify by combining the +6 and -1 to get:
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f(a-3) = -2a + 5
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To graph this equation, replace f(a-3) by y and you have:
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y = -2a + 5
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Call the vertical axis the y-axis and the horizontal axis the a-axis.
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The equation y = -2a + 5 is in the slope intercept form. If you set "a" equal to zero, the
equation reduces to y = +5. This tells you that the point (0, 5) is on the graph.
But this point identifies the value where the graph intersects the y-axis.
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You can also set y equal to zero, and when you do that the equation becomes:
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0 = -2a + 5
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Add +2a to both sides to eliminate the -2a on the right side, and the equation changes to:
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2a = +5
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Divide both sides by the multiplier of the a on the left side to get:
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a = 5/2 = 2.5
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This tells you that when y = 0 then a = 2.5 so the point (2.5, 0) is on the graph, and this
point is on the a-axis at 2.5.
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You now have two points on the graph. If you plot these two points, you can then use a
straight edge to extend a line through them. Note that from the slope intercept form of
the equation [y = -2a + 5] the slope is the multiplier of the "a" term. In this case that
multiplier is -2. So the graph you have created has a slope of -2 and you can check
your graphed line to make sure that for any point on the line, if you move horizontally
1 unit from that point and then move 2 units down you should be back on the graphed line.
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When you get done, you should have a graph that looks like this except the label on the
horizontal axis should be "a" instead of "x":
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Hope this helps you to understand the problem a little better.
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