# Algebraic Representation of a Function

🏆Practice representations of functions

A function is a connection between an independent variable $(X)$ and a dependent variable $(Y)$. The relationship between the variables is called a "correspondence rule".

An algebraic representation of a function is actually a description of the relationship between the dependent variable $(Y)$ and the independent variable $(X)$ by means of an equation.

The following is the typical structure of a graphical representation:

• $Y=X+3$, $Y=2X-5$

For example, if the data is that every month, Daniel earns$20.000$ dollars.

The algebraic representation will be $X$ for the number of months $Y$ $f (X)$ for the amount earned. $f (x) = 20000X$

## Test yourself on representations of functions!

Is the given graph a function?

## Exercises on algebraic representation of a function

### Exercise 1

On the graph of the linear function that passes through the points $A(2,10)$ and $B(-5,-4)$

Find the slope of the graph.

Solution

$m=\frac{y_2-y_1}{x_2-x_1}$

Replace accordingly

$x_1=-5,y_1=-4$

$x_2=2,y_2=10$

$\frac{10-\left(-4\right)}{2-\left(-5\right)}=$

$\frac{14}{7}=2$

$2$

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### Exercise 2

From the following function, find the slope of the line:

$y=-x+1$

Solution

In order to find the slope of the line we must remember how is the algebraic representation of a line:

$y=mx+b$

Where $m$ is the slope, that is, the coefficient of the independent variable is the slope or the slope of the function, therefore in the equation

$y=-x+1$

$m=-1$

And from here we deduce that the slope is $-1$

$m=-1$

### Exercise 3

Assignment

Find the slope of the straight line through the following points $(0,4),(-5,6)$

Solution

$m=\frac{y_2-y_1}{x_2-x_1}$

Replace accordingly according to the data

$x_1=0,y_1=4$

$x_2=-5,y_2=6$

$\frac{6-4}{-5-0}=$

$\frac{2}{-5}=-\frac{2}{5}$

$-\frac{2}{5}$

Do you know what the answer is?

### Exercise 4

Given the linear function whose slope of the graph is $-3$ and passes through the point $(-6,-3)$.

Find the algebraic representation of the function

Solution

$y=m\cdot x+b$

$m=-3$

Replace accordingly

$\left(-6,-3\right)$

$-3=\left(-3\right)\cdot\left(-6\right)+b$

$-3=18+b$

$-21=b$

$y=-3\cdot x-21$

$y=-3\cdot x-21$

### Exercise 5

On the graph of the linear function passing through the points $A(0,7)$ and $B(8,-3)$

Find the slope of the graph

Solution

$m=\frac{y_2-y_1}{x_2-x_1}$

Replace accordingly using the data

$x_1=0,y_1=7$

$x_2=8,y_2=-3$

$\frac{-3-7}{8-0}=$

$\frac{-10}{8}=-\frac{5}{4}$

$-\frac{5}{4}$

## Review questions

How is an algebraic function represented?

In some related articles, it had been mentioned that a function can be represented in different ways, verbally, algebraically, with tables and graphically. In the case of how to represent a function algebraically, in a few words we can say that it will be represented in an equation, which will indicate the rule of correspondence between the dependent variable $Y$ and the independent variable. $X$

What is the algebraic representation of a linear function?

The representation of a linear function is the one where we are going to visualize an equation where it represents a straight line, that is to say the independent variable $X$ this with an exponent one, that is to say, of first degree.

Examples

Some algebraic representations of a linear function are the following:

$y=x+4$

$y=-x+1$

$y=x-5$

$y=-x-1$

We can observe that the variable $X$ has as exponent one, and if we graph these functions we will always get a straight line, so they represent a linear function.

How is the algebraic representation of a quadratic function?

A quadratic function can be seen as an equation where the variable $X$ will have the exponent $2$, which will represent a parabola if it is graphed. Some examples are the following:

$y=x^2+4$

$y=x^2-3$

$y=-x^2+5$

How to get the algebraic representation of a table?

A function can be represented verbally, algebraically, in a table of values and graphically, therefore, we can go from one representation to another, in this case we are going to study how to go from a table to an algebraic representation, for such a case let's see the following example:

Example

Assignment

From the following table find its algebraic representation.

Solution:

From the previous table we are going to take any two points, in order to get its slope:

Let the points be

$A=\left(-3,-2\right)$,$B=\left(-1, 0\right)$

From where:

$X_1=-3$,$Y_1=-2$

$X_2=-1$,$Y_2=0$

Then we substitute in the formula of the slope

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{0-\left(-2\right)}{-1-\left(-3\right)}$

$m=\frac{0-\left(-2\right)}{-1-\left(-3\right)}=\frac{0+2}{-1+3}$

$m=\frac{0+2}{-1+3}=\frac{2}{2}=1$

Then $m=1$

Now taking the first point

$A=\left(-3,-2\right)$

We substitute in the equation of the line:

$y=mx+b$

$-2=1\cdot-3+b$

$-2=-3+b$

$-2+3=b$

$1=b$

Knowing $m$ and $b$

Substitute again in the equation of the line

$y=mx+b$

$y=1\cdot x+1$

$y=x+1$

Result

The algebraic expression is $y=x+1$.

Do you think you will be able to solve it?

## Examples with solutions for Algebraic Representation of a Function

### Exercise #1

Is the given graph a function?

### Step-by-Step Solution

It is important to remember that a function is an equation that assigns to each element in domain X one and only one element in range Y

Let's note that in the graph:

$f(0)=2,f(0)=-2$

In other words, there are two values for the same number.

Therefore, the graph is not a function.

No

### Exercise #2

Which of the following equations corresponds to the function represented in the graph?

### Step-by-Step Solution

Use the formula for finding slope:

$m=\frac{y_2-y_1}{x_2-x_1}$

We take the points:

$(0,-2),(-2,0)$

$m=\frac{-2-0}{0-(-2)}=$

$\frac{-2}{0+2}=$

$\frac{-2}{2}=-1$

We substitute the point and slope into the line equation:

$y=mx+b$

$0=-1\times(-2)+b$

$0=2+b$

We combine like terms:

$0+(-2)=b$

$-2=b$

Therefore, the equation will be:

$y=-x-2$

$y=-x-2$

### Exercise #3

Which of the following equations corresponds to the function represented in the table?

### Step-by-Step Solution

We will use the formula for finding slope:

$m=\frac{y_2-y_1}{x_2-x_1}$

Let's take the points:

$(-1,4),(3,8)$

$m=\frac{8-4}{3-(-1)}=$

$\frac{8-4}{3+1}=$

$\frac{4}{4}=1$

We'll substitute the point and slope into the line equation:

$y=mx+b$

$8=1\times3+b$

$8=3+b$

Let's combine like terms:

$8-3=b$

$5=b$

Therefore, the equation will be:

$y=x+5$

$y=x+5$

### Exercise #4

Is the given graph a function?

No

### Exercise #5

Is the given graph a function?