Question 447691: I have the following points (25, 16) and (40,26) and thanks to the very generous help of scott8148 I was able to figure out the first two part of this problem. I figured out the slope as m=2/3, and with that I figured out the y-intercept as
- 2/3. Using this equation that he graciously provided. This points represented the height a ball was dropped from as the x-axis and the bounce heights as the y axis.
So given the information above how would I set up an equation to determine the value of x if y=100. I also know that on this same graph is the point (50,33) So would I be correct in assuming then that by increasing the the value of x to something like 151 I would thus attain a y value of approximately 100?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Just so that no one who reads this later gets lost, let's take it from the top:
You are looking for a linear function, ) such that if  then  -- that gives you the point (26,16). Also if  then  -- the point (40, 26).
You did the slope calculation correctly: 
And then, using the point-slope form of the equation of a line:
 )
where ) are the coordinates of a point on the line and  is the slope:
)
from which you derived:
And correctly discovered that the  -intercept is 
But that IS your function:
\ =\ \frac{2}{3}x\ -\ \frac{2}{3})
Pick any value you like for  , insert that value in place of , and do the arithmetic.
Let's check it out:
\ -\ \frac{2}{3}\ =\ \frac{48}{3}\ =\ 16)
That one checks, so:
\ -\ \frac{2}{3}\ =\ \frac{78}{3}\ =\ 26)
and this checks, so:
\ -\ \frac{2}{3}\ =\ \frac{98}{3}\ =\ 32\frac{2}{3}\ \neq\ 33)
Well, almost, so
\ -\ \frac{2}{3}\ =\ \frac{??}{3}\ =\ ??)
I'll let you do the arithmetic for the last one.
John
My calculator said it, I believe it, that settles it
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