There is no other function. Those are the only two things given to me.
The question asked is:
Describe the transformation that must be applied to the graph of the base function f(x) to obtain the transforming function. Write the transformed equation in simplified form.
f(x) = x^2, y = 7f(-1/6(x-1))+1
There is no other function, equation, or anything. Nothing to plug it into.
And on that note, wouldn't "Plugging -1/6(x-1) into the function f, then multiplying by 7 and adding 1,", if that means "Replace f(x) = x^2 with f(x) = (-1/6(x-1)^2, then multiply by 7 and add 1", basically result in an equation that looks like y = 7f(-1/6(x-1))^2+1? Or possibly y = 7f((-1/6(x-1))^2)+1. *
And finally, as I said, it's transforming the graphed f(x) = x^2. Nothing algebraic there, as far as I can tell.
I'm not saying you're wrong. These are my assumptions, they're fully capable of being wrong. But they're the assumptions I got from the rest of the book, so if they're wrong I'm either a dumbass or the book is terrible at teaching. Or both :v
*I ran with it and simplified y = 7f((-1/6(x-1))^2)+1, and the result is exactly the same as simplifying y = 7f(-1/6(x-1))^2+1 (the equation I assumed the book -meant- to write and forgot the exponent to), that is to say, y = (-7(x-1/6)^2)+1