+171 Suppose f(3)=4. Name a point that must be on the graph of y=2f(x/3)+17
+907 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! :)
