How do you find $f (n - 1)$ $f(n-1)$ if $f (x) = 2 x^{2} - x + 9$ $f(x)=2x^2-x+9$ ?

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Answer 1 · 2017-10-20T05:40:30

We just subsitute $x$ $x$ with $n - 1$ $n-1$ and carrying out all algebraic operations we get $f (n - 1) = 2 n^{2} - 5 n + 12$ $f(n-1)=2n^2-5n+12$

We just subsitute $x$ $x$ with $n - 1$ $n-1$ and carry out all algebraic operations.

As $f (x) = 2 x^{2} - x + 9$ $f(x)=2x^2-x+9$

$f (n - 1) = 2 {(n - 1)}^{2} - (n - 1) + 9$ $f(n-1)=2(n-1)^2-(n-1)+9$

= $2 (n^{2} - 2 n + 1) - n + 1 + 9$ $2(n^2-2n+1)-n+1+9$

= $2 n^{2} - 4 n + 2 - n + 1 + 9$ $2n^2-4n+2-n+1+9$

= $2 n^{2} - 5 n + 12$ $2n^2-5n+12$

How do you find f(n−1)f(n-1) if f(x)=2x2−x+9f(x)=2x^2-x+9?