Krish Schmitt

## Answered question

2022-10-02

Given is
$x=3\phantom{\rule{0ex}{0ex}}h=2\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{2}+2x+10$
what is the solution of $f\left(x+h\right)$
Is that correct?
$f\left(x\right)={x}^{2}+2x+10=\left(9+4\right)+2\left(3+2\right)+10=13+10+10=33$

### Answer & Explanation

pereishen9g

Beginner2022-10-03Added 7 answers

In the formula
$f\left(x\right)={x}^{2}+2x+10$
think of the "x" as a box you get to fill in, that is,
$f\left(◻\right)=\left(◻{\right)}^{2}+2\left(◻\right)+10$
So you can put an x, or x+h, or anything you want, into the box. In particular, putting x+h into the box gives you
$f\left(x+h\right)=\left(x+h{\right)}^{2}+2\left(x+h\right)+10$
Now since x=3 and h=2, you know $\overline{)x+h}=3+2=\overline{)5}$

Riya Andrews

Beginner2022-10-04Added 4 answers

Your basic idea is correct, but you are asked for $f\left(x+h\right)$ so that is what you should have on the left side, not $f\left(x\right)$. Then it might help to write out the function in terms of the variables first. The first term in the expression is not correct.

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