Please help me solve this?

Question

find f(n-1) if f(x)=(2x^2)-x+9

Answer 1

f(n-1)=[2*(n-1)²]-(n-1)+9

=2*(n²-2n+1)-n+1+9

=2n²-4n+2-n+10

=2n²-5n+12

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Answer 2

f(x) = 2x^2 - x + 9

Solve for f(n-1).

Simply substitute n-1 for x in the function to find f(n-1).

2(n-1)^2 - (n-1) + 9

Follow PEMDAS (the order of operations).

2(n-1)^2 - n + 1 + 9 [distribute the negative sign to (n-1)]

2(n^2-2n+1) - n + 10 [square the binomial; see below*]

2n^2 - 4n + 2 -n + 10 [distribute]

2n^2 - 5n + 12

.: f(n-1) = 2n^2 - 5n + 12

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* Square this binomial by multiplying by itself:

(a+b)(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2

(n-1)(n-1) = (n-1)^2 = n^2 + 2(n)(-1) + (-1)^2 = n^2 - 2n + 1