# Thread: Find f(3m - 4) + 3

1. ## Find f(3m - 4) + 3

If f(x) = x^2 - 2x + 4, find f(3m - 4) + 3.

2. Originally Posted by magentarita
If f(x) = x^2 - 2x + 4, find f(3m - 4) + 3.
sub 3m-4 in for x: $(3m-4)^2 - 2(3m-4) + 4$

$9m^2 - 24m + 16 - 6m + 8 + 4 +3 = 9m^2 - 30m + 31$

3. Originally Posted by magentarita
If f(x) = x^2 - 2x + 4, find f(3m - 4) + 3.

$f(x) = x^2 - 2x + 4$

$f(3m-4)=(3m-4)^2-2(3m-4)+4$

$=9m^2-24m+16-6m+8+4$

$=9m^2-30m+28$

$
f(3m - 4) + 3=9m^2-30m+28+3=9m^2-30m+31
$

4. ## ok

Great but where does the 3 come in?

f(3m - 4) + 3

5. Simply add it to whatever you get for ${\color{blue}f(3m-4)}$ .

So: ${\color{blue}f(3m-4)} + {\color{red}3} = {\color{blue}9m^2-30m+} \underbrace{{\color{blue}28} + {\color{red}3}}_{\text{Add}} = 9m^2 - 30m + 31$

as what the other posters got.

6. ## ok

Originally Posted by o_O
Simply add it to whatever you get for ${\color{blue}f(3m-4)}$ .

So: ${\color{blue}f(3m-4)} + {\color{red}3} = {\color{blue}9m^2-30m+} \underbrace{{\color{blue}28} + {\color{red}3}}_{\text{Add}} = 9m^2 - 30m + 31$

as what the other posters got.
It all makes perfect sense to me now. Thank you.