Function problem help

Zeal

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Mar 11, 2019
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3
If y=f(x)
How do i get the function use to derive these.

100=f(100)

50=f(70)
 

HallsofIvy

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Jan 27, 2012
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Given any finite number of values, there exist an infinite number of functions that take on those values. Here, you have two values and you might remember that "there is a unique straight line passing through two given points". So there exist a unique linear function taking on those two values. Such a function can be written as "f(x)= ax+ b". f(100)= 100a+ b= 100 and f(70)= 70a+ b= 50.

Subtracting the second equation from the first eliminates b and gives 30a= 50 so a= 5/3. Then 100a+ b= 500/3+ b= 100 so b= 100- 500/3= -200/3. f(x)= (5/3)x- 200/3= (5x- 200)/3.

Checking, f(100)= (500- 200)/3= 300/3= 100 and f(70)= (350- 200)/3= 150/3= 50.
 

Zeal

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Mar 11, 2019
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Hi hallsoflvy, could you elaborate a bit im lost.
Thank
 

MarkFL

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Hi hallsoflvy, could you elaborate a bit im lost.
Thank
It would be better if you state which part(s) of the explanation you don't understand. You are given two points on the graph of the function:

(70,50) and (100,100)

As HallsofIvy pointed out, there are an infinite number of functions we could find passing through those two points, but the simplest of these is going to be a linear function (\(n\) points can be uniquely described by a polynomial of degree \(n-1\)). Using the definition of slope, and the point-slope equation for a line, we then find:

\(\displaystyle f(x)=\frac{100-50}{100-70}(x-70)+50=\frac{5}{3}x-70\cdot\frac{5}{3}+50=\frac{5}{3}x-\frac{200}{3}=\frac{5}{3}(x-40)\)
 

mmm4444bot

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… could you elaborate …
We could, if you could explain where you're stuck.

If you're not familiar with linear equations (eg: lines, slopes, equation forms), let us know. We can help you find online videos and lesson links.

Please read the forum's submission guidelines. You can start with this summary. Thank you.

😎
 

Zeal

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Mar 11, 2019
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It would be better if you state which part(s) of the explanation you don't understand. You are given two points on the graph of the function:

(70,50) and (100,100)

As HallsofIvy pointed out, there are an infinite number of functions we could find passing through those two points, but the simplest of these is going to be a linear function (\(n\) points can be uniquely described by a polynomial of degree \(n-1\)). Using the definition of slope, and the point-slope equation for a line, we then find:

\(\displaystyle f(x)=\frac{100-50}{100-70}(x-70)+50=\frac{5}{3}x-70\cdot\frac{5}{3}+50=\frac{5}{3}x-\frac{200}{3}=\frac{5}{3}(x-40)\)

Thanks, this cleared my confusion
 
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