Worked Examples of Graphing Translated Exponential Functions

This is the standard exponential, except that the "+ 4" pushes the graph up so it is four units higher than usual.

First I compute some points:

Then I plot those points:

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Students, when they're just starting out with exponential graphs, very often only compute y-values for a few x-values that are close to zero, and sometimes they'll only compute values for x-values that are zero or positive. Then either they have no idea where the graph goes on the left-hand side, and just leave it hanging there:

This graph is WRONG!

...or else they take the graph down to the x-axis, as is usual for the standard exponential graph:

This graph is also WRONG!

But y = 2^x + 4 isn't the standard exponential graph; it is the standard exponential graph shifted upward by four units. When x is negative, y = 2^x + 4 won't be very close to zero; instead, it will be very close to 4, because the values will be "a teensy-tiny little number, plus four", which works out to be a teensy-tiny bit more than four.

To help me with my graph, and to indicate that I know that y = 2^x + 4 never goes below (or even touches, for that matter) the line y = 4, I will drawn a dashed line at y = 4:

Drawing my dashed line:

This dashed-in line, indicating where the graph goes as x heads off to the left, is called a "horizontal asymptote", or just an "asymptote". It is not required that you draw it in, but it can be helpful for graphing, and can point out to your teacher on the test that you do know what you're doing.

Now that I have my plotted points and my asymptote, I'll draw my exponential curve:

Graph of y = 2^x + 4

I need to remember that the "minus" exponent reverses the location (along the x-axis) in which the power on 5 is negative. When the x-values are negative (that is, when I'm on the left-hand side of the graph), the value of −x will be positive, so the graph will grow quickly on the left-hand side.

On the other hand, when the x-values are positive (that is, on the right-hand side of the graph), then the corresponding values of −x will be negative. This means that the graph will stay very close to the x-axis as the graph goes off to the right.

In other words, the standard values are reversed:

Then y = 5^−x graphs as:

This equation is not the same as y = 2^x + 3. In 2^x + 3, the standard exponential is shifted up three units. In this case, the shift is inside the exponent. Instead of the + 3 shifting the 2^x up by three, the + 3 shifts the 2^x over sideways by three. The only question is: shifts sideways which way, left or right?

The way I keep it straight is to consider one of the basic points on any exponential. When the power is zero, the exponential is 1. For 2^(x + 3), when is the power zero? It will be zero when x + 3 = 0, so x = −3. That is, the basic plot point (0, 1) has been shifted to the point (−3, 1), so the graph has been shifted three points to the left:

There is no upward or downward shift for this graph, so the left-hand side will get and stay close to the x-axis as the graph moves off to the left. On the right-hand side, the graph will shoot up through the top.

So my graph looks like this: