I just started Precalculus and need help with this problem: Given f(x)=7-x^2, find each value if f(x)=6+n?

Question

the answer is f(x)=-29+12n+n^2. but I don't understand how they got that. Please include steps.

Answer 1

Given f(x)=7-x^2
f(x)=6+n

So
f(x)=7-x^2 = 6+n

x^2 = 1+n

this is the answer you would get given the question the way you presented it.

The book may have specifically stated x = 6+n
In this case you substitute 6+n for x.

f(x)=7-x^2

f(x)=7- (6+n) ^2
Expand the square
f(x)=7- (36+ 12n + n^2)

f(x)=7- (36+ 12n + n^2)
f(x)=7- 36 - 12n - n^2
f(x)= - 29 - 12n - n^2

sometimes the book can be wrong!

Answer 2

f(x) = 7 - x^2, but x is also a function of n : f(x) = 6 +n

substitute (6+n) to x

f(x) = 7 - (6+n)^2
f(x) = 7 - (36 + 12n + n^2)
f(x) = 7 - 36 - 12n - n^2
f(x) = -29 - 12n - n^2 or -( 29 +12n + n^2)

Answer 3

substitute in 6+n for x so the problem looks like f(n)=7-(6+n)^2 then simplify with FOIL to 7-36+6n+6n+n^2 then simplify againg to the answer 29+12n+n^2