## 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 post about how to find a quadratic function’s equation from its graph is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

This graph represents a polynomial function $\inline \fn_jvn y=g(x)$. Which of the following could be g‘s equation?

1. $\inline \fn_jvn g(x)=x^2-2x-3$
2. $\inline \fn_jvn g(x)=-x^2+2x+3$
3. $\inline \fn_jvn g(x)=(x-1)(x+3)$
4. $\inline \fn_jvn g(x)=-(x-1)^2-4$

## Solution

What can you tell from the graph?

You can read the coordinates of the x-intercepts, the y-intercept, and the turning point.

 x-intercepts (-1, 0), (3, 0) y-intercept (0, -3) turning point (1, -4)

The x-intercepts are clues to g‘s factors: The x-intercept (-1, 0) tells you that $\fn_phv x+1$ is a zero, and the x-intercept (3, 0) tells you that $\fn_phv x-3$ is a factor.

Sidebar: The factor theorem says that a polynomial that has a polynomial has a factor $\fn_phv x-c$ has a root at c, and a polynomial that has a root at c has a factor $\fn_phv x-c$. A root is the x-value at the x-intercept: for this example, -1 and 3. So the factors are $\fn_phv x+1$ and $\fn_phv x-3$.

Multiply those factors together and see what you get:

$\fn_jvn {\color{DarkBlue} (x+1)(x-3)=x^2-2x-3}$

The graph looks like a parabola, a second-degree curve. And $\fn_phv x^2-2x-3$ is second degree. So you may be done. How can you check? Use one of the two other points you can read from the graph. The y-intercept is (0, -3). Plug in 0 for x and see if the equation gives you -3, the y-intercept.

If

$\fn_jvn {\color{DarkBlue} g(x)=x^2-2x-3}$

then

$\fn_jvn {\color{DarkBlue} g(0)=0^2-2 \cdot 0-3}$

And as we saw from the graph, the y-intercept is (0, -3). Check. So answer choice #1 is the correct one.

That is one way to find a quadratic function’s equation from its graph.

Alternatively, since this question is multiple choice, you could try each answer choice. The easiest way to do that, I think, is to substitute 0 for and see if you get that $\inline&space;\fn_phv&space;g\left&space;(&space;0&space;\right&space;)=-3$, the y-intercept:

1. $\inline \fn_jvn g(x)=x^2-2x-3$.  Then $\inline \fn_jvn g(0)=-3$. This is the answer we found. It works.
2. $\inline \fn_jvn g(x)=-x^2+2x+3$$\inline \fn_jvn g(0)=3\neq -3$. Not this one.
3. $\inline \fn_jvn g(x)=(x-1)(x+3)$$\inline \fn_jvn g(0)=-3$.  Looks like this works.
4. $\inline \fn_jvn g(x)=-(x-1)^2-4$$\inline \fn_jvn g(0)=-5$. Does not work.

So we have two contenders, answer choice #1 and answer choice #3. That means we have to try at least one more point besides the y-intercept. We have already done that for answer choice #1; we know the x-intercepts work for it. Let’s see if $\inline \fn_jvn y=0$ at $\inline \fn_jvn x=-1$ for answer choice #3, $\inline \fn_jvn g(x)=(x-1)(x+3)$.

$\fn_jvn {\color{DarkBlue} g(-1)=(-1-1)(-1+3)=-2 \cdot 2 = -4 \neq 0}$

So answer choice #3 does not work after all, and answer choice #1 is the correct one.

For more about the end behavior of polynomial functions, see “End Behavior, Degree, and Leading Coefficient.”

The Accuplacer sample problem on which this one is based is #10 in the sample questions for the Accuplacer Advanced Algebra and Functions test.