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Suppose f(3)=4. Name a point that must be on the graph of y=2f(x/3)+17

 May 27, 2024
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Alright! I'll tackle this problem!

 

First off, we notice there is an f(x) function inside the second equation. 

 

We know f(3)=4, we let's let x/3 in the second equation be 3 so that we have an integer. 

 

Setting x=9 would get us this desired equation. 

 

We have \(y=2f(3) + 17\), which is essentially \(2(4)+17\)

 

This simplifies to 25, meaning when x equals 9, we have y equals 25. 

 

This means that (9, 25) HAS to be on the graph for y=2f(x/3)+17!

 

Thanks! :)

 May 27, 2024

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