# How do you find h(-2n) given h(n)=n^2+n?

Mar 2, 2017

See the entire solution process below:

#### Explanation:

For each occurrence of $\textcolor{red}{n}$ in the function $h \left(n\right)$ replace it with $\textcolor{red}{- 2 n}$:

$h \left(\textcolor{red}{n}\right) = {\textcolor{red}{n}}^{2} + \textcolor{red}{n}$ becomes:

$h \left(\textcolor{red}{- 2 n}\right) = {\left(\textcolor{red}{- 2 n}\right)}^{2} + \textcolor{red}{- 2 n}$

$h \left(\textcolor{red}{- 2 n}\right) = 4 {n}^{2} - 2 n$

Or

$h \left(\textcolor{red}{- 2 n}\right) = 2 \left(2 {n}^{2} - n\right)$