You are working on a graphing project in your math class, where you are supposed to graph several functions. Things seem to be going well, until you realize that there is a bold, vertical line three units to the left of where you placed your "y" axis! As it turns out, you've accidentally shifted your entire graph. You didn't notice that your instructor had placed a bold line where the "y" axis was supposed to be. And now, all of the points for your graph of the cosine function are three points farther to the right than they are supposed to be along the "x" axis.

You might be able to keep all of your work, if you can find a way to rewrite the equation so that it takes into account the change in your graph.

Can you think of a way to rewrite the function so that the graph is correct the way you plotted it?

### Horizontal Translations

Horizontal translations involve placing a constant inside the argument of the trig function being plotted. If we return to the example of the parabola, \begin{align*}y = x^2\end{align*}, what change would you make to the equation to have it move to the right or left? Many students guess that if you move the graph vertically by adding to the \begin{align*}y-\end{align*}value, then we should add to the \begin{align*}x-\end{align*}value in order to translate horizontally. This is correct, but the graph itself behaves in the opposite way than what you may think.

Here is the graph of \begin{align*}y = (x + 2)^2\end{align*}.

Notice that *adding* 2 to the \begin{align*}x-\end{align*}value shifted the graph 2 units *to the left*, or in the *negative* direction.

To compare, the graph \begin{align*}y = (x - 2)^2\end{align*} moves the graph 2 units *to the right* or in the *positive* direction.

We will use the letter \begin{align*}C\end{align*} to represent the horizontal shift value. Therefore, **subtracting** \begin{align*}C\end{align*} from the \begin{align*}x-\end{align*}value will shift the graph to the **right** and **adding** \begin{align*}C\end{align*} will shift the graph \begin{align*}C\end{align*} units to the **left**.

Adding to our previous equations, we now have \begin{align*}y=D \pm \sin (x \pm C)\end{align*} and \begin{align*}y=D \pm \cos (x \pm C)\end{align*} where \begin{align*}D\end{align*} is the vertical translation and \begin{align*}C\end{align*} is the *opposite sign* of the horizontal shift.

#### Sketching Graphs

1. Sketch \begin{align*}y=\sin \left( x - \frac{\pi}{2} \right)\end{align*}

This is a sine wave that has been translated \begin{align*}\frac{\pi}{2}\end{align*} units to the *right*.

Horizontal translations are also referred to as **phase shifts**. Two waves that are identical, but have been moved horizontally are said to be “out of phase” with each other. Remember that cosine and sine are really the same waves with this phase variation.

\begin{align*}y = \sin x\end{align*} can be thought of as a cosine wave shifted horizontally to the right by \begin{align*}\frac{\pi}{2}\end{align*} radians.

Alternatively, we could also think of cosine as a sine wave that has been shifted \begin{align*}\frac{\pi}{2}\end{align*} radians to the left.

2. Draw a sketch of \begin{align*}y = 1 + \cos (x - \pi)\end{align*}

This is a cosine curve that has been translated up 1 unit and \begin{align*}\pi\end{align*} units to the right. It may help you to use the quadrant angles to draw these sketches. Plot the points of \begin{align*}y = \cos x\end{align*} at \begin{align*}0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{align*} (as well as the negatives), and then translate those points before drawing the translated curve. The blue curve below is the final answer.

#### Graphing Functions

Graph \begin{align*}y=-2+\sin \left(x+\frac{3\pi}{2}\right)\end{align*}

This is a sine curve that has been translated 2 units down and moved \begin{align*}\frac{3\pi}{2}\end{align*} radians to the left. Again, start with the quadrant angles on \begin{align*}y = \sin x\end{align*} and translate them down 2 units.

Then, take that result and shift it \begin{align*}\frac{3\pi}{2}\end{align*} to the left. The blue graph is the final answer.

### Examples

#### Example 1

Earlier, you were asked if you can think of a way to rewrite the function so that the graph is correct the way you plotted it.

As you've now seen by reading this Concept, it is possible to shift an entire graph to the left or the right by changing the argument of the graph.

So in this case, you can keep your graph by changing the function to \begin{align*}y = \cos(x - 3)\end{align*}

#### Example 2

Draw a sketch of \begin{align*}y = 3 + \cos (x - \frac{\pi}{2})\end{align*}

As we've seen, the 3 shifts the graph vertically 3 units, while the \begin{align*}-\frac{\pi}{2}\end{align*} shifts the graph to the right by \begin{align*}\frac{\pi}{2}\end{align*} units.

#### Example 3

Draw a sketch of \begin{align*}y = \sin (x + \frac{\pi}{4})\end{align*}

The \begin{align*}\frac{\pi}{4}\end{align*} shifts the graph to the left by \begin{align*}\frac{\pi}{4}\end{align*}.

#### Example 4

Draw a sketch of \begin{align*}y = 2 + \cos (x + 2\pi)\end{align*}

The 2 added to the function shifts the graph up by 2 units, and the \begin{align*}2\pi\end{align*} added in the argument of the function brings the function back to where it started, so the cosine graph isn't shifted horizontally at all.

### Review

Graph each of the following functions.

- \begin{align*}y=\cos(x-\frac{\pi}{2})\end{align*}
- \begin{align*}y=\sin(x+\frac{3\pi}{2})\end{align*}
- \begin{align*}y=\cos(x+\frac{\pi}{4})\end{align*}
- \begin{align*}y=\cos(x-\frac{3\pi}{4})\end{align*}
- \begin{align*}y=-1+\cos(x-\frac{\pi}{4})\end{align*}
- \begin{align*}y=1+\sin(x+\frac{\pi}{2})\end{align*}
- \begin{align*}y=-2+\cos(x+\frac{\pi}{4})\end{align*}
- \begin{align*}y=3+\cos(x-\frac{3\pi}{2})\end{align*}
- \begin{align*}y=-4+\sec(x-\frac{\pi}{4})\end{align*}
- \begin{align*}y=3+\csc(x-\frac{\pi}{2})\end{align*}
- \begin{align*}y=2+\tan(x+\frac{\pi}{4})\end{align*}
- \begin{align*}y=-3+\cot(x-\frac{3\pi}{2})\end{align*}
- \begin{align*}y=1+\cos(x-\frac{3\pi}{4})\end{align*}
- \begin{align*}y=5+\sec(x+\frac{\pi}{2})\end{align*}
- \begin{align*}y=-1+\csc(x+\frac{\pi}{4})\end{align*}
- \begin{align*}y=3+\tan(x-\frac{3\pi}{2})\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.13.