## Is This Relation a Function?

This post about whether a relation is a function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

One or more of the graphs below represent 𝑦 as a function of 𝑥. Which one or ones?

 A B C D

## Solution

Function/not function?

A function is a relation – that is, a correspondence between x and  y – for which every x-value relates to only one y-value. No vertical line will pass through more than one point on a function’s graph, but a vertical line may pass through more than one point on the graph of a relation that is not a function. Thus you can use the vertical line test on a relation to see whether it is a function.

Look at relation A. Imagine sliding a vertical line across it. At no point will the line intersect the relation more than once. This relation is a function.

Now look at relation B. The y-axis, which is a vertical line, intersects the relation in three places. Thus relation B fails the vertical line test and relation B is not a function.

Relation C: Again, there are places where you could draw a vertical line and cross the relation more than once. For example, the y-axis crosses the relation twice. This relation is not a function.

Relation D: There is only one x-value where it looks like a vertical line could cross this relation twice: the x-value at the right-hand end of the lower piece, which is the x-value at left-hand end of the upper piece. Are there two y-values at this x-value? The lower piece, in which the circle is filled in, does take this x-value. But the upper piece, in which the circle is not filled in, does not take the value. Thus there is no x-value that both pieces take, and there is no x-value for which there is more than one value of  This relation is a function.

𝑨𝒏𝒔𝒘𝒆𝒓: A and D

## Finding a Function’s Range

This post about finding a function’s range from its equation is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Find the range of this function:

$\fn_jvn {\color{DarkBlue} y=-3x^4+5}$

## Solution

A function’s range is the set of values y can take in that function.

You are asked to find a function’s range. What are the values y can take in this function?

Consider the first term, $\fn_jvn -3x^4$. Any real number (unless otherwise stated, it’s safe to assume x is real) to an even power, like 4, has a positive value or zero. It just can’t be negative. The lowest value of $\inline \fn_jvn 3x^4$ is zero — that happens where $\inline \fn_jvn x=0$ — and there is no limit to how high $\inline \fn_jvn 3x^4$ can go, so $\inline \fn_jvn 3x^4$ ranges from 0 to infinity. Then its negative, $\fn_jvn -3x^4$ ranges from negative infinity to 0.

 $\fn_jvn y=x^4$$\fn_jvn y=x^4$ $\fn_jvn y=-x^4$$\fn_jvn y=-x^4$ $\fn_jvn y=-x^4+5$$\fn_jvn y=-x^4+5$

The second term, 5. increases each y-value by 5. At the low end of the range, negative infinity plus 5 is still negative infinity, so the bottom of the range is still negative infinity. At the top end, $\fn_jvn 0+5=5$, so the top of the range is 5.

So the range of $\inline \fn_jvn y=-3x^4+5$ is from negative infinity to 5. In other words, $\inline \fn_jvn y\leq 5$. In interval notation, that’s $\inline \fn_jvn \left ( -\infty , 5 \right )$.

This question is similar to question number 11 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## How Do You Evaluate a Function? II

$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$This post about how to evaluate a function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

For the function

$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$

what is the value of $\inline&space;\fn_jvn&space;f(x-1)$?

A.  $\inline&space;\fn_jvn&space;x^2-2x+3$

B. $\inline&space;\fn_jvn&space;(x-1)^2-2(x-1)+4$$\inline&space;\fn_jvn&space;(x-1)(x^2-2x+3)$

C. $\inline&space;\fn_jvn&space;(x-1)(x^2-2x+3)$

D.  $\fn_jvn&space;x^2+3x+4$

## Solution

f(x-1) is what you get when you replace x with x-1 in $\inline \fn_jvn x^2-2x+4$. That is called “evaluating the function for x-1.

### Sidebar: Function Notation

In algebra, usually when you write two things side by side with one of them in parentheses, that means you multiply those two things together:

$\fn_jvn&space;{\color{Blue}&space;a(b)=a&space;\cdot&space;b}$

So no one can blame you for thinking f(x) means f multiplied by x. But math notation, like other kinds of language, has exceptions. f(x) means function f, carried out on x. And f(something else) means function f carried out on something else. For example, say function g is defined as $\inline \fn_jvn g(x)=x^2$. That means $\inline \fn_jvn g(1)=1^2$; $\inline \fn_jvn g(-2)=\left ( -2 \right )^2$; and $\fn_jvn g(*)$ means $\inline \fn_jvn *^2$ (whatever * means).

It gets interesting when the parentheses in the function definition contain a function. For example, if  $\inline \fn_jvn g(x)=x^2$, what is $\inline \fn_jvn g(x+2)$? Well, we said that if you want to find $\fn_jvn g(*)$, you insert * into the function definition. To find $\inline \fn_jvn g(x+2)$, insert $\inline \fn_jvn x+2$ into the function definition: if $\inline \fn_jvn g(x)=x^2$, then $\inline \fn_jvn g(x+2)=(x+2)^2$.

So to find f(x-1) in the expression

$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$

insert x-1 wherever you see x. You should get:

$\fn_jvn&space;{\color{Blue}&space;(x-1)^2-2(x-1)+4}$

Usually an answer like this would be expressed as $\inline \fn_jvn x^2-4x+7$, but this question is multiple choice so you go with what you’ve got.

Another example of how to evaluate a function is here.

This question is similar to question number 13 in the sample questions for the Accuplacer Advanced Algebra and Functions test.

## How Do You Evaluate a Function? I

This post about how you evaluate a function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

To evaluate a function means to find the function’s y-value if you are given an x-value.

## Question

Let’s say you’re given this function:

$\fn_jvn {\color{Blue} f(x)=4\left ( x-2 \right )}$

and you’re asked to find f(10) — in other words, the value of f where x=10.

## Solution

To do that, substitute 10 into the equation for f where you see x:

$\fn_jvn {\color{Blue} f(10)=4\left ( 10-2 \right )}$

Now clean it up: Substitute 8 for 10-2:

$\fn_jvn {\color{Blue} f(10)=4 \cdot 8}$

And multiply 4 by 8:

$\fn_jvn&space;{\color{Blue}&space;f(10)=32}$

So f(10)=32..

And that’s all there is to evaluating this function.

The example in this post is a linear function.

See How Do You Evaluate a Function? II for an example of how to evaluate a quadratic function.

This question is similar to question number 1 in the sample questions for the Accuplacer Advanced Algebra and Functions test.