Finding $f(m-1)$ where $f(x) = x^2 + 2x+2$ [closed]

If $f(x) = x^2 + 2x + 2$, find $f(m - 1)$.

Can someone please solve this for me, with the steps.

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closed as off-topic by Jonas Meyer, RecklessReckoner, hardmath, Ivo Terek, TravisJan 15 at 2:09

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What have you tried so far? Where are you getting stuck? If this is homework, you should use the "homework" tag. –  JavaMan Mar 14 '11 at 17:01
"Can someone solve this for me" is, I think, the worst way to ask for help! –  Mariano Suárez-Alvarez Mar 14 '11 at 19:44
@Mar: If you submit the answer "Yes, I can" to that question, I'll vote for you! I don't have the rep to spare ;) –  The Chaz 2.0 Mar 14 '11 at 23:30
Replace each $x$ in the expression with a $(m-1)$. –  ncmathsadist Jan 15 at 2:00
I think that closure (and often then deletion) of "old" questions because they do meet current wishes is unreasonable. –  André Nicolas Jan 15 at 2:04

Evaluating functions with inputs that aren't just "x" is an important skill.

Let me walk you through some examples, using this function:

$f(x) = x^2 + 2x + 2$ We will just plug in some numbers and letters, using parenthesis to show where we substituted...

$f(1) = (1)^2 + 2*(1) + 2 = 1 + 2 + 2 = 5$
$f(2) = (2)^2 + 2*(2) + 2 = 4 + 4 + 2 = 10$
$f(3a) = (3a)^2 + 2*(3a) + 2 = 9a^2 + 6a + 2$ Notice that our answer isn't just a number; it is a variable expression.

Now $f(m - 1) = (m - 1)^2 + 2*(m - 1) + 2$.

I'll leave it to you to expand/distribute and combine like terms.

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Side question - What is Latex/markup for a simple dot (for multiplication)? –  The Chaz 2.0 Mar 14 '11 at 17:01
@Chaz, "\cdot" is the appropriate symbol. –  JavaMan Mar 14 '11 at 17:03
@DJC: 10-4!.... –  The Chaz 2.0 Mar 14 '11 at 17:12

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

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