# How do you graph y=log_2(x+1)+3?

Jan 19, 2017

#### Explanation:

As the equation is $y = {\log}_{2} \left(x + 1\right) + 3$,

the domain of $y$ is the set of all $x$ values such that $x + 1 > 0$ or $x > - 1$

and the range of $y$ is the interval $\left(- \infty , + \infty\right)$.

Hence as $x \to - 1$, $y \to - \infty$ i.e. $x + 1 = 0$ is the asymptote.

Now let us select a few values of $x$, choosing so that calculating ${\log}_{2} \left(x + 1\right)$ is easy. Hence, let $x$ take values $\left\{0 , 1 , 3 , 7 , 15\right\}$ and for these values, $y$ takes values $\left\{3 , 4 , 5 , 6 , 7\right\}$ and hence points on the graph are

$\left(0 , 3\right)$; $\left(1 , 4\right)$; $\left(3 , 5\right)$; $\left(7 , 6\right)$ and $\left(15 , 7\right)$

and graph appears as follows:
graph{2^(y-3)=x+1 [-4.87, 15.13, -1.92, 8.08]}