|
Question 81747: would someone help me to figure out how to find the second part of this problem and graph it please. Thanks
Given f(x)=-x-4, find f(a-3)
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Given f(x)=-x-4, find f(a-3)
.
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:
.
f(a-3) = -(a-3) - 4
.
When you remove the parentheses on the right side, the minus sign in front of the parentheses
require you to change the signs of all the terms in the parentheses. Applying this rule
changes the equation to:
.
f(a-3) = -a + 3 -4
.
Finally, simplify by combining the +3 and -4 to get:
.
f(a-3) = -a - 1
.
To graph this equation, replace f(a-3) by y and you have:
.
y = -a - 1
.
Call the vertical axis the y-axis and the horizontal axis the a-axis.
.
The equation y = -a - 1 is in the slope intercept form. If you set a equal to zero, the
equation reduces to y = -1. This tells you that the point (0, -1) is on the graph.
But this point identifies the value where the graph intersects the y-axis.
.
You can also set y equal to zero, and when you do that the equation becomes:
.
0 = -a - 1
.
Add +a to both sides to eliminate the -a on the right side, and the equation changes to:
.
a = -1
.
This tells you that when y = 0 then a = -1 so the point (-1, 0) is on the graph, and this
point is on the a-axis at -1.
.
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 = -a -1] the slope is the multiplier of the "a" term. In this case that
multiplier is -1. So the graph you have created has a slope of -1 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 1 unit down you should be back on the graphed line.
.
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":
.
.
Hope this helps you to understand the problem a little better.
.
|
|
|
| |
