Algebraic Representation of a Function

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A function is a connection between an independent variable (X) (X) and a dependent variable (Y) (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) (Y) and the independent variable (X) (X) by means of an equation.

The following is the typical structure of a graphical representation:

  • Y=X+3 Y=X+3 , Y=2X5 Y=2X-5
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Is the given graph a function?

–7–7–7–6–6–6–5–5–5–4–4–4–3–3–3–2–2–2–1–1–1111222333444555666777–4–4–4–3–3–3–2–2–2–1–1–1111222333000

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For example, if the data is that every month, Daniel earns20.000 20.000 dollars.

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


If you are interested in this article you may also be interested in the following articles:

Graphical representation of a function

Notation of a function

Domain of a function

Indefinite integral

Assigning a numerical value to a function

Variation of a function

Increasing function

Decreasing function

Constant function

Growing and decreasing intervals of a function

Functions for seventh grade

In Tutorela you will find a variety of articles with interesting explanations about mathematics.


Exercises on algebraic representation of a function

Exercise 1

Task

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

Find the slope of the graph.

graph of the linear function passing through the points A(2,10) and B(-5,-4)

Solution

m=y2y1x2x1 m=\frac{y_2-y_1}{x_2-x_1}

Replace accordingly

x1=5,y1=4 x_1=-5,y_1=-4

x2=2,y2=10 x_2=2,y_2=10

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

147=2 \frac{14}{7}=2

Answer

2 2


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Exercise 2

Task

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

y=x+1 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 y=mx+b

Where m 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 y=-x+1

m=1 m=-1

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

Answer

m=1 m=-1


Exercise 3

Assignment

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

Solution

m=y2y1x2x1 m=\frac{y_2-y_1}{x_2-x_1}

Replace accordingly according to the data

x1=0,y1=4 x_1=0,y_1=4

x2=5,y2=6 x_2=-5,y_2=6

6450= \frac{6-4}{-5-0}=

25=25 \frac{2}{-5}=-\frac{2}{5}

Answer

25 -\frac{2}{5}


Do you know what the answer is?

Exercise 4

Task

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

Find the algebraic representation of the function

Solution

y=mx+b y=m\cdot x+b

m=3 m=-3

Replace accordingly

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

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

3=18+b -3=18+b

21=b -21=b

y=3x21 y=-3\cdot x-21

Answer

y=3x21 y=-3\cdot x-21


Exercise 5

Task

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

Find the slope of the graph

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

Solution

m=y2y1x2x1 m=\frac{y_2-y_1}{x_2-x_1}

Replace accordingly using the data

x1=0,y1=7 x_1=0,y_1=7

x2=8,y2=3 x_2=8,y_2=-3

3780= \frac{-3-7}{8-0}=

108=54 \frac{-10}{8}=-\frac{5}{4}

Answer

54 -\frac{5}{4}


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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 Y and the independent variable. X 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 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+4

y=x+1 y=-x+1

y=x5 y=x-5

y=x1 y=-x-1

We can observe that the variable X 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 X will have the exponent 2 2 , which will represent a parabola if it is graphed. Some examples are the following:

y=x2+4 y=x^2+4

y=x23 y=x^2-3

y=x2+5 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.

1 - How to get the algebraic representation of a table

Solution:

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

Let the points be

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

From where:

X1=3 X_1=-3 ,Y1=2 Y_1=-2

X2=1 X_2=-1 ,Y2=0 Y_2=0

Then we substitute in the formula of the slope

m=y2y1x2x1 m=\frac{y_2-y_1}{x_2-x_1}

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

m=0(2)1(3)=0+21+3 m=\frac{0-\left(-2\right)}{-1-\left(-3\right)}=\frac{0+2}{-1+3}

m=0+21+3=22=1 m=\frac{0+2}{-1+3}=\frac{2}{2}=1

Then m=1 m=1

Now taking the first point

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

We substitute in the equation of the line:

y=mx+b y=mx+b

2=13+b -2=1\cdot-3+b

2=3+b -2=-3+b

2+3=b -2+3=b

1=b 1=b

Knowing m m and b b

Substitute again in the equation of the line

y=mx+b y=mx+b

y=1x+1 y=1\cdot x+1

y=x+1 y=x+1

Result

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


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