# How do you graph y=3^x and y=3^(x+1) and how do the graphs compare?

Dec 25, 2016

See explanation.

#### Explanation:

The easiest way to graph exponential functions is to simply find a few $\left(x , y\right)$ pairs that are easy to plot.

Let's start with $y = {3}^{x}$.
When $x = 0$ (i.e. when we're on the $y$-axis), $y = {3}^{0} = 1$. So the point $\left(0 , 1\right)$ is on our graph. From here, every time $x$ goes up by $1$, $y$ will triple. That is, for every one step right, our step up is 3 times larger than it was before.

$x = 1 \text{ "=>" } y = 3 \cdot 1 = \textcolor{b l u e}{3}$
$x = 2 \text{ "=>" } y = 3 \cdot \textcolor{b l u e}{3} = \textcolor{g r e e n}{9}$
$x = 3 \text{ "=>" } y = 3 \cdot \textcolor{g r e e n}{9} = 27$
and so on.

graph{3^x [-17.15, 8.16, -1.75, 10.9]}

For the graph of $y = {3}^{x + 1}$, we can actually recognize this as a translation of the first graph. The $x$-value is being increased by $1$; what this means is, at every $x$-value, $y = {3}^{x + 1}$ looks to the right of $y = {3}^{x}$ by one unit, finds the $y$-value there, and makes that its own $y$-value at the same $x$. This results in the original graph of $y = {3}^{x}$ getting shifted to the left by one unit.

Think of it this way: when you look to your left, what was in your field of vision "moves right". Similarly, when an $x$-input decreases by $h$ (a.k.a. $\left(x - h\right)$), the parent graph shifts right by $h$. That's why all these math formulas like $y = a \sin \left[b \left(x - h\right)\right] + k$ and $y = a {\left(x - h\right)}^{2} + k$ have "$- h$" instead of "$+ h$": because shifting a graph to the right requires a decrease in $x$ input, not an increase.

In short: $y = {3}^{x + 1}$ will look identical to the graph of $y = {3}^{x}$, just shifted one unit to the left.

graph{3^(x+1) [-17.15, 8.16, -1.75, 10.9]}