## Comparing Data in Different Formats

This post about comparing data in different formats is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Liam’s Pay

This graph shows how much Liam gets paid as a function of the number of hours he works. Sarah, on the other hand, is paid according to the formula S=(75/4)h, where S represents Sarah’s pay and h represents the number of hours she has worked.

Is Liam’s rate of pay more or less than Sarah’s, and by how much?

## Solution

You’re being asked to compare data that measure similar things but are given to you in different formats.

The trick is to convert one of those expressions into the same format as the other. Or, if you can’t do that, then convert both to some third format.

P=(75/4)h looks . . . almost straightforward. That fraction may be inconvenient. 75/4 = 18.75, so let’s say S = 18.75h represents Sarah’s rate of pay.

Now look at the graph. Can you represent Liam’s pay as something like L=kh, where k is Liam’s rate of pay? Since the graph goes through the origin, its equation will be of the form L=something*and “something” is the slope. To find it from the graph, pick two points on the line that are easy to read, and calculate. The origin — with coordinates (0, 0) — looks easy. And the point (2, 40) doesn’t look bad. The slope is the change in y divided by the change in for those two points:

$\fn_jvn&space;\frac{40-0}{2-0}=\frac{40}{2}=20$

So Liam’s pay is L=20h.

You’re asked for the difference between Liam’s rate of pay and Sarah’s. That is

$\fn_jvn&space;\20.00/hr&space;-&space;\18.75/hr&space;=&space;\1.25/hr$

Answer: Liam gets paid \$1.25 per hour more than Sarah.

For more about relating a line’s equation to its graph, see “Finding a Line’s Equation from the Graph of Its Perpendicular.

This question is similar to question number 5 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## Question

This is the graph of $y=f\left&space;(&space;x&space;\right&space;)$

Sketch the graph of $y=f(x-1)$.

## Solution

The change from $y=f\left&space;(&space;x&space;\right&space;)$ to $y=f(x-1)$ is called a horizontal transformation; it slides the function sideways. The graph of $y=f(x-1)$ lies one unit to the right of the graph of $y=f\left&space;(&space;x&space;\right&space;)$. This may be counterintuitive, so let’s look at some numbers. At each x-value, the graph of $y=f(x-1)$ takes the y-value that the graph of $y=f\left&space;(&space;x&space;\right&space;)$ takes one unit earlier. Some values on the two graphs look like this:

The y-values from the original graph have all chugged along to correspond to higher x-values than they did before.

At any x-value on the graph, the function  operates on the  value 1 unit to the left of that x-value. That has the effect of moving the curve 1 unit to the right.

Similarly, $y=f(x+1)$ graphs one unit to the left of $y=f\left&space;(&space;x&space;\right&space;)$. Note that this is different from $y=f\left&space;(&space;x&space;\right&space;)+1,$ which graphs one unit up from $y=f\left&space;(&space;x&space;\right&space;)$ because the 1 that gets added to the function means 1 gets added to every y-value. Similarly, $y=f\left&space;(&space;x&space;\right&space;)-1$ graphs one unit below $f\left&space;(&space;x&space;\right&space;).$ If you don’t believe me on any of these, try playing around with values of a simple function to see what happens.