Category: Algebra

Algebra Practice Problems – Free practice problems in algebraic skills such as factoring and simplifying expressions and solving equations. Each post has free exercises and answers.

Inverse Operations with Examples

Inverse Operations – Examples & Quiz

Welcome to the inverse operations with examples page.

Addition, subtraction, multiplication, and division are mathematical operations. 

Instructions: Please see the examples below before attempting this quiz. Then perform inverse operations to solve each of the following equations.

[WpProQuiz 19]

What are Inverse Operations?

You will need to understand inverse operations in order to solve algebra problems on your examination.

Subtraction is the inverse of addition and division is the inverse of multiplication.

An inverse operation changes a number or variable back to its original value. So, you reverse the operation when you perform inverse operations.

Inverse operations can be performed on numbers, variables, or algebraic terms. Now look at the examples that follow.

Inverse operations examples – Addition

The inverse of addition is subtraction.

Original value:  5

Original operation:  5 + 3 = 8

Inverse operation:  8 – 3 = 5

Inverse operations – Subtraction

The inverse of subtraction is addition

Original value:  8

Original operation:  8 – 6 = 2

Inverse operation:  2 + 6 = 8

Inverse operations – Multiplication

The inverse of multiplication is division.

Original value:  x

Original operation:  x × y = xy

Inverse operation:  xy ÷ y = x


Inverse operations – Division

The inverse of division is multiplication.

Original value:  5b

Original operation: 5b ÷ b = 5

Inverse operation:  5 × b = 5b

You may also want to view our posts on:

Order of operations – PEMDAS

Algebraic Equations

More algebra practice problems

Math Placement Test Download for Inverse Operations

Get the downloads

Order of Operations PEMDAS

What is the Order of Operations?

The phrase “order of operations” or “PEMDAS” refers to the order in which you need to perform multiplication, division, addition, subtraction, and exponential operations in an equation.

To see the free study questions, please go to the next section.

Order of Operations – PEMDAS

The phrase PEMDAS is sometimes used to remember the correct order to perform operations in an equation.

PEMDAS means parentheses, exponents, division, multiplication, addition, and subtraction.


So, when you see an equation, you need to perform the operations from left to right in this order:

1)    Parentheses – Do  the operations inside parentheses or brackets first.

2)   Exponents – Then perform the operations indicated by the exponents.

3)   Multiplication and Division – Then do the multiplication and division from left to right.

4)   Addition and Subtraction – Then do the addition and subtraction from left to right.

Some students remember the rule “PEMDAS” by using the phrase:

PEMDAS:  Please Excuse My Dear Aunt Sally

Let’s look at a couple of problems on PEMDAS or order of operations.

Order of Operations PEMDAS – Basic Example

Now study this example of how to carry out the order of operations “PEMDAS” rule.

Example:   1 + 64 ÷ 4 ÷ 2 × 5 – 3 = ?

Answer:   38

In this problem, we do not have any parentheses or exponents to deal with.

So, start by doing the multiplication and division first.

Multiplication and division are performed in the same step, so you need to divide and then multiply from left to right as shown below when you have division and multiplication together like in this example.

1 + 64 ÷ 4 ÷ 2 × 5 – 3 =

1 + (64 ÷ 4) ÷ 2 × 5 – 3 =

1 + 16 ÷ 2 × 5 – 3 =

1 + (16 ÷ 2) × 5 – 3 =

1 + 8 × 5 – 3 =

1 + (8 × 5) – 3 =

1 + 40 – 3

Then do the addition and subtraction from left to right.

1 + 40 – 3 =

41 – 3 = 38

Order of Operations PEMDAS – Advanced Example

Let’s look at a more difficult example of how to carry out the order of operations “PEMDAS” rule.

Example:   (4 + 5)2 + 9 ÷ 3 – 2 × 6 = ?

Answer:   72


Do the operations in the parentheses first.

(4 + 5)2 + 9 ÷ 3 – 2 × 6 =

(9)2 + 9 ÷ 3 – 2 × 6

Then do the exponential operation.

92 + 9 ÷ 3 – 2 × 6 =

(9 × 9) + 9 ÷ 3 – 2 × 6 =

81 + 9 ÷ 3 – 2 × 6

Next do the multiplication and division from left to right.

81 + 9 ÷ 3 – 2 × 6 =

81 + (9 ÷ 3) – 2 × 6 =

81 + 3 – 2 × 6 =

81 + 3 – (2 × 6) =

81 + 3 – 12

Then do the addition and subtraction from left to right.

81 + 3 – 12 =

(81 + 3) – 12 =

84 – 12 = 72

Order of Operations PEMDAS – Further Practice

If you found these exercises difficult, you should try our pre-algebra problems.

If you are testing in college math, you may also want to look at our other algebra study materials.

Quadratic Expression

What is the Quadratic Expression?

A quadratic expression is a statement in algebra that has a variable that is raised to the power of 2.

Please scroll to the bottom of the page for free practice exercises.

Quadratic Expression, Equation and Inequality

A quadratic expression should not be confused with a quadratic equation.

Quadratic equations contain an equals sign, while quadratic expressions do not.


Quadratic inequalities include the less than or greater than signs, while quadratic expressions do not.

ax2 + bx + c = 0 is an example of a quadratic equation.

ax2 + bx + c > 0 is an example of a quadratic inequality.

But quadratic inequalities and equations do include quadratic expressions.

Everything to the left of the equals sign or greater than sign in our examples above is a quadratic expression.

So, ax2 + bx + c = 0 is a quadratic expression.

Expanding Quadratics

Quadratics that contain terms in parentheses can be multiplied and expanded in order to make a quadratic expression.

If there is one set of parenthesis, the expression can be multiplied by using the distributive property of multiplication as shown below:

x(x + 2) =

(x × x) + (x × 2) =

x2 + 2x

The above expression is a quadratic expression because it contains x2.

If you have terms in two sets of parentheses, you can multiply using the FOIL method.

In order to use the FOIL method, you have to multiply the terms together in this order:  FIRST    OUTSIDE   INSIDE   LAST

Look at this example:

(x + 1)(2x + 3) =

 FIRST     OUTSIDE   INSIDE   LAST

(x × 2x) + (x × 3) + (1 × 2x) + (1 × 3) =

2x2 + 3x + 2x + 3 =

2x2 + 5x + 3

We explain the FOIL method in more depth in another post.

If you are not confident about how to solve algebra problems with the FOIL method, be sure to view that post.


Now you should try the exercises in the next section.

Quadratic Expression – Exercises

Instructions: Use multiplication in the problems below to expand and re-write them as quadratic expressions. Your answers should contain a variable raised to the power of two.

1)   x(x – 4)

2)   2x(4x + 6)

3)   –3x(x – 5)

4)   (x + 2)(x – 3)

5)   (3x – 3)(2x – 4)

Quadratic Expression – Answers

1)    x(x – 4) =

(x × x) + (x × –4) =

x2 – 4x

2)   2x(4x + 6) =

(2x × 4x) + (2x × 6) =

8x2 + 12x

3)  –3x(x – 5) =

(–3x × x) + (–3x × –5) =

–3x2 + 15x

4)   (x + 2)(x – 3) =

(x × x) + (x × –3) + (2 × x) + (2 × –3) =

x2 + –3x + 2x + –6 =

x2 – x – 6

5)   (3x – 3)(2x – 4) =

(3x × 2x) + (3x × –4) + (–3 × 2x) + (–3 × –4) =

6x2 + –12x + –6x + 12 =

6x2 – 18x + 12

You might also like to have a look at our posts on the quadratic formula and quadratic equations.

Quadratic Equation

What is the Quadratic Equation?

The quadratic equation can be written in different forms.

The standard quadratic equation is in the form:

ax2 + bx + c = 0

Please scroll to the next section of this page to see the free examples.


Standard Quadratic Equation

You will see several questions on quadratics on your algebra test.

A quadratic equation contains a variable such as the x2 term as the highest power.

In other words, an equation is quadratic only when it has a variable that is raised to the power of 2.

Let’s look at our equation again.

ax2 + bx + c = 0

In the equation above, a, b, and c are numbers.

a, b, and c can be fractions or decimals. They do not have to be whole numbers.

a, b, and c can also be imaginary numbers.

Other Forms of the Quadratic Equation

The standard quadratic equation can be re-written if the terms are re-arranged.

The standard quadratic equation can also be re-written if the b or c terms are equal to zero.

However, the a term can never equal zero in a quadratic equation.

We can write the standard form of the equation as follows:

x2 + bx + c = n (when a = 1) 

x2 + bx = n (when c = 1) 

x2 = n (when b = 0 and c = 0) 

Before trying the exercises at the bottom of the page, you should view our posts on quadratic forms and the FOIL method.

Quadratic Expressions

Quadratic equations must contain an equals sign.

A quadratic expression is the part of the quadratic equation without the equals sign.

So, ax2 + bx + c is an example of a quadratic expression.


In other words, on one side of the equation, a quadratic equation includes a quadratic expression.

You should also view our posts on the quadratic formula and quadratic expressions.

Solving Quadratic Equations

We look at how to solve quadratic equations in our post entitled “Solving Quadratic Equations.”

Click on the following links for:

More quadratics exercises

More algebra exercises

Quadratic Forms

What are Quadratic Forms?

Quadratic forms contain a variable such as the x2 term as their highest power.

There are various quadratic forms.

To view the free examples, please scroll down to the next section.

Quadratic Forms – Types

Type 1:   ax2 + bx + c = 0


The “standard” quadratic form is  ax2 + bx + c = 0

In the equation above, a, b, and c are positive or negative numbers.

a, b, and c can also be fractions or imaginary numbers.

So, 2x2 + 5x – 3 = 0 is an example of this quadratic form.

Note that the variable a can never be equal to zero in the quadratic form.

3x + 2 = o is not a quadratic because none of the terms are raised to the power or 2.

The standard quadratic form can be rewritten as other equations such as:

ax2 = bx + c

ax2 + bx = c

ax2 + c = bx

Type 2:   x2 + bx + c = 0

When a = 1, we can write the equation as follows:  x2 + bx + c = 0

So, x2 + 5x + 6 = 0 is an example of this quadratic form.

As mentioned above, the variable a can never be equal to zero in the quadratic form.

Type 3:  (ax + b)(x + c) = 0

The factored form of the equation is as follows: (ax + b)(x + c) = 0

An example of this form is: (x + 1)(x – 5) = 0

Type 4:  x2 + bx = 0

When c = 0, the quadratic form can be written as shown below:

x2 + bx = 0

x2 – 10x = 0 is an example of this quadratic form.

Type 5:   x2 = n


When b = 0 and c = 0, the quadratic form can be written as shown below:

x2 = 4

If you wish to solve for x in the above questions, you should also see our posts on the quadratic formula and solving quadratic equations.

Synthetic Division

What is Synthetic Division?

Synthetic division is used to divide polynomials by first degree binomials.

A first degree binomial is an expression that has a variable that is not raised to any power. So, x + 5 is an example of a first degree binomial.

Free exam sample questions are provided at the end of this page.



Synthetic Division – Explanation of Terminology

To perform synthetic division, you need to identify the coefficients and constants in the polynomial.

You will remember that a polynomial is an algebraic expression that has two or more terms.

Terms can contain either numbers or variables or both.

A coefficient is the number part of a term.

A constant is a number on its own. Constants are usually placed at the end of an algebraic expression.

Synthetic Division – Example

Consider the following polynomial:

x4 + 5x3 – 2x2 – 28x – 12

In the second term, which is 5x3, the coefficient is 5, for example.

Now look at this practice problem.

Example:  Divide the polynomial x4 + 5x3 – 2x2 – 28x – 12 by the first degree binomial x + 3. Use synthetic division to find your answer.

STEP 1: Identify the coefficients and constants.

Remember that if you have a variable without a number in front of it, the coefficient is 1.

x4 + 5x3 – 2x2 – 28x – 12 =

(1 × x4) + (5 × x3) – (2 × x2) – (28 × x) – 12 =

(1 × x4) + (5 × x3) + (–2 × x2) + (–28 × x) + (–12)

Our coefficients and constants are the numbers before each of the multiplication signs.

The coefficients are 1, 5, –2, and –28.

The constant is –12.

STEP 2: Identify the constant in the binomial and find its opposite.

We are going to divide by the binomial x + 3.

So, the constant in our binomial is 3.

The opposite of the constant is –3.

STEP 3: Write the problem out in synthetic division format, placing the opposite of the constant in an upside-down division sign.

Divide the numbers from step 1 by the opposite of the constant from step 2.

–3|    1    5    –2   –28   –12

STEP 4: Perform the synthetic division.

Bring down the 1 and multiply it by –3. Put that result under the 5. Then add 5 and –3 to get 2.

–3|  1     5    –2   –28   –12

     –3   –6      24      12

1     2

STEP 5: Repeat the operations from step 4 until you find the remainder.

–3|   1     5    –2   –28   –12

       –3    –6     24     12

1       2    –8     –4       0

Now try the practice problems in the next section.

Synthetic Division – Exercises

Instructions:  Perform synthetic division to find the remainder for the following problems. If you do not know how to perform synthetic division, please see the example above before completing the exercises.

1)   (x3 + 2x2 + x + 5) ÷ (x + 2)

2)   (x3 – 6x2 – 2x + 14) ÷ (x – 6)

3)   (x3 + x2 – x – 4) ÷ (x – 2)

4)   (x4 + 5x3 + 10x2 + 14x + 10) ÷ (x + 3)

5)   (x4 – 9x3 + 13x2 + 4x + 21) ÷ (x – 7)

Synthetic Division – Answers

1)   The remainder is 3.

Our question was: (x3 + 2x2 + x + 5) ÷ (x + 2)

The opposite of the constant in our binomial is –2.

Our coefficients and constant are:  1     2      1     5

Set up the synthetic division to solve as shown below.

–2|  1     2      1     5

     –2     0    –2 

 1     0      1      3

2)   The remainder is 2.

Our question was: (x3 – 6x2 – 2x + 14) ÷ (x – 6)

The opposite of the constant in our binomial is 6.

Our coefficients and constant are: 1   –6   –2    14

Set up the synthetic division to solve as shown below.

6|  1     –6   –2    14

       6      0   –12

1     0    –2       2

3)   The remainder is 6.

Our question was: (x3 + x2 – x – 4) ÷ (x – 2)

The opposite of the constant in our binomial is 2.

Our coefficients and constant are:  1     1    –1   –4

Set up the synthetic division to solve as shown below.

2|  1     1    –1    –4

      2      6     10

1     3      5       6

4)   The remainder is 4.

Our question was: (x4 + 5x3 + 10x2 + 14x + 10) ÷ (x + 3)

The opposite of the constant in our binomial is –3.

Our coefficients and constant are:   1     5    10   14    10

Set up the synthetic division to solve as shown below.

–3|  1     5     10    14    10

     –3   –6   –12   –6

1     2      4       2      4

5)   The remainder is 0.

Our question was: (x4 – 9x3 + 13x2 + 4x + 21) ÷ (x – 7)

The opposite of the constant in our binomial is 7.

Our coefficients and constant are:  1   –9    13    4     21

Set up the synthetic division to solve as shown below.

7|    1     –9     13     4     21

          7   –14   –7  –21

1     –2     –1    –3      0

You should also view our posts on the remainder theorem and factoring.

Perfect Squares

What are Perfect Squares?

Perfect squares are the result of multiplying a number or variable by itself.

For example, 4 and 9 are perfect squares, because they are the products of two times two and three times three, respectively.

2 × 2 = 4

3 × 3 = 9

To see the free examples, please go to the next section of this page.

Perfect Squares in Algebra

In algebra, perfect squares are the product of two equal variables.


For instance, a2 and b2 are examples of this, because they are the products of a times a and b times b, respectively.

a × a = a2

b × b = b2

Binomials are two term expressions.

So, the expression a2 + b2 is an example of a binomial.

One type of binomial is the difference between two perfect squares.

The expression a2 – b2 is an example of the difference between two perfect squares.

Factoring the Difference of Two Perfect Squares

If the two terms of a binomial are perfect squares and the terms are subtracted, the binomial can be factored.

The factored form is as follows:

a2 – b2 = (a + b)(a – b)

In other words, the variables in the first set of parentheses will be added to each other and the variables in the second set of parentheses will be subtracted from each other when factoring the difference between two perfect squares.


We can prove the form above by multiplying each of the terms and canceling out as shown below.

a2 – b2 =

(a + b)(a – b) =

(a × a) + (a × –b) + (b × a) + (b × –b) =

a2 + –ab + ab + –b2 =

a2 – ab + ab – b2 =

a2 – ab + ab – b2 =

a2 – b2

In order to perform the factoring, you will need to be acquainted with the laws of square roots.

You will also need to understand how to do the FOIL method on quadratics.

You may also want to try our other problems on factoring and algebra.

Greatest Common Factor with GCF Examples

Greatest Common Factor (GCF) Examples

This page will show problems on the greatest common factor with GCF examples.

To factor means to find two or more quantities whose product equals the original quantity.

GCF Quiz with Examples

Instructions: Try this GCF quiz, after studying the examples below. You need to find the greatest common factor of the following expressions.


[WpProQuiz 18]

How to Find the Greatest Common Factor

The greatest common factor of two numbers in the largest number that is a factor of both of them.

Consider the following algebraic expression:

30x2y3z + 18x3y2z2 + 24xy4z3

To find the greatest common factor, we need to follow these steps:

STEP 1:

Look for the common variables and exponents.

Variables: Each term has a factor of x, a factor of y, and a factor of z.

Exponents of x: The exponents on the x variables of each term are 2, 3, and 1. The common factor will be the variable with the smallest exponent. So, x1, simplified to x, is a common factor.

Exponents of y: The exponents on the y variables of each term are 3, 2, and 4. Remember that the common factor will be the variable with the smallest exponent. So, y2 is a common factor.

Variable z: The exponents on the z variables of each term are 1, 2, and 3. So, z is a common factor.

GCF for variables: Accordingly, the common factor for the variables is xy2z.

STEP 2:

Look at the coefficients. Remember that a coefficient is the number part of the term.

Our equation is: 30x2y3z + 18x3y2z2 + 24xy4z3

Coefficients: The coefficients are 30, 18, and 24. We need to find the greatest common factor for the coefficient part of the expression.

Dividing: To do this we have to find the largest number that each of these numbers (30, 18, and 24) can be divided by.

List the Common Factors: Each of these coefficients can be divided by 6. So each coefficient has a factor 6. There is no larger common factor.

30 = 6 × 5

18 = 6 × 3

24 = 6 × 4

STEP 3:

The common factor for the variables is xy2z and the common factor for the coefficients is 6.

Final GCF: The greatest common factor is 6xy2z.

STEP 4:

Re-write the expression in factored form, placing the GCF at the front of the new expression.

Work out the expression:

30x2y3z + 18x3y2z2 + 24xy4z3 =

(6 × 5)(x × x)(y2 × y)z + (6 × 3)(x × x2)y2(z × z) + (6 × 4)x(y2 × y2)(z × z2) =

6xy2z(6 × 5)(x × x)(y2 × y)z + (6 × 3)(x × x2)y2(z × z) + (6 × 4)x(y2 × y2)(z × z2) =

6xy2z(5 × x × y) + (3 × x × z) + (4 × y2 × z2) =

Factored Expression: 6xy2z(5xy + 3xz + 4y2z2)

Factoring – Further Practice

Before you go on to other topics, you should look at the following post:

Try our free algebra practice test

Math Placement Test Download with GCF Examples

Get the downloads

Inverse Functions

What Are Inverse Functions?

Functions are relationships between mathematical values.

Functions have input values, and an operation on each input value results in a single output value.

For some algebra problems, we want to work backwards from the output to discover what the input was.

When we work backwards to find the input, we use what is called an inverse function.


If you need to review functions before completing these exercises, see our post on Function Notation.

You can also try our Algebra Practice Test.

Inverse Functions – Example

An example of a function that has an inverse is:

ƒ(x) = 2x + 3.

The inverse of this function is written as follows:

f–1(x) = (x – 3) ÷ 2

In the notation for the inverse function above, you will notice that the  –1 exponent is given after the function.

The –1 exponent is a special notation used to indicate an inverse function.

Setting Up Inverse Functions

Let’s look at our functions again.

Our original function was:

ƒ(x) = 2x + 3

Our inverse function was:

f–1(x) = (x – 3) ÷ 2

To set up an inverse function, you usually need to perform four steps.

Step 1:

Replace ƒ(x) with y.

ƒ(x) = 2x + 3

y = 2x + 3

Step 2:

Then isolate x to one side of the equation.

y = 2x + 3

y – 3 = 2x + 3 – 3

y – 3 = 2x

(y – 3) ÷ 2 = x

x = (y – 3) ÷ 2

Step 3:

Then swap x and y.

x = (y – 3) ÷ 2

y = (x – 3) ÷ 2

Step 4:

Finally, replace y with the function f–1(x) to solve.

y = (x – 3) ÷ 2

f–1(x) = (x – 3) ÷ 2

Inverse Functions – Proofs

To prove an inverse function, put a value for x into the original function.

Let’s use x = 6 in our original function as an example.

ƒ(x) = 2x + 3

ƒ(x) = (2 × 6) + 3

ƒ(x) = 12 + 3

ƒ(x) = 15

Then take the result, or the output, from the original function and use it as in input in the inverse function.

f–1(x) = (x – 3) ÷ 2

f–1(x) = (15 – 3) ÷ 2

f–1(x) = 12 ÷ 2

f–1(x) = 6


So, our result or output from the inverse function is the same as our input in the original function.

Functions Without Inverses

Not every function have as inverse.

This is the case when two or more inputs to the function result in the same output.

For instance, consider the following function:  ƒ(x) = x2

When x = 2, the output is 4, but when x = –2, the output is also 4.

Therefore, the function ƒ(x) = x2 does not have an inverse.

A function will not have an inverse if any of the following are true:

  • The inverse function has an even-numbered exponent.
  • The inverse function has an absolute value symbol.
  • The graph of the function is a flat line.

Inverse Functions – Exercises

Find the inverses for the following functions.

If the function does not have an inverse, provide an explanation.

  1. ƒ(x) = 5x + 4
  2. ƒ(x) = 2x – 7
  3. ƒ(x) = x4
  4. ƒ(x) = (x + 2) ÷ (x – 3)
  5. ƒ(x) = | x – 3 |

Inverse Functions – Answers

Answer 1:

ƒ(x) = 5x + 4

Replace ƒ(x) with y.

y = 5x + 4

Then isolate x to one side of the equation.

y = 5x + 4

y  – 4 = 5x + 4 – 4

y – 4 = 5x

(y – 4) ÷ 5 = x

x = (y – 4) ÷ 5

Then swap x and y.

x = (y – 4) ÷ 5

y = (x – 4) ÷ 5

Finally, replace y with the function f–1(x) to solve.

y = (x – 4) ÷ 5

ƒ-1(x) = (x – 4) ÷ 5

Answer 2:

ƒ(x) = 2x – 7

Replace ƒ(x) with y.

y = 2x – 7

Then isolate x to one side of the equation.

y = 2x – 7

y + 7 = 2x – 7 + 7

y + 7 = 2x

(y + 7) ÷ 2 = x

x = (y + 7) ÷ 2

Then swap x and y.

x = (y + 7) ÷ 2

y = (x + 7) ÷ 2

Finally, replace y with the function f–1(x) to solve.

y = (x + 7) ÷ 2

ƒ-1(x) = (x + 7) ÷ 2

Answer 3:

ƒ(x) = x4

This function does not have an inverse because it has an even numbered exponent.

In other words, it is not a one-to-one function because there is more than one input for a single output.

For instance, the inputs –2 and 2 both result in the output of 16.

Answer 4:

ƒ(x) = (x + 2) ÷ (x – 3)

Replace ƒ(x) with y.

y = (x + 2) ÷ (x – 3)

Then isolate x to one side of the equation.

y = (x + 2) ÷ (x – 3)

y × (x – 3) = (x + 2)

xy – 3y = x + 2

xy – 3y  – x = 2

xy – 3y + 3y – x = 2 + 3y

xy  – x = 2 + 3y

x(y  – 1) = 2 + 3y

x = (2 + 3y) ÷ (y  – 1)

Then swap x and y.

x = (2 + 3y) ÷ (y  – 1)

y = (2 + 3x) ÷ (x  – 1)

y = (3x + 2) ÷ (x – 1)

Replace y with the function f–1(x) to solve.

y = (3x + 2) ÷ (x – 1)

ƒ-1(x) = (3x + 2) ÷ (x – 1)

Answer 5:

ƒ(x) = | x – 3 |

This function does not have an inverse because it has the absolute value symbol.

Like question 3 above, it is not a one-to-one function because there is more than one input for a single output.

For instance, the inputs x = 1 and x = 5 both result in the output of 2.

| 1 – 3 | = | –2 | = 2

| 5 – 3 | = | 2 | = 2

More Practice with Inverse Functions

If you are testing in trigonometry, you should also look at our practice problems on trig functions.

Function Notation Formula and Equation

Function Notation Formula: ƒ(x)

Any function notation formula or equation begins with the symbol ƒ(x). Function notation finds the value of y for a given operation on x.

In algebraic functions, the value of ƒ(x) = y. As an example,  the function ƒ(x) = 2x + 5 is the same as the equation y = 2x + 5.

Function Notation Quiz

Instructions: Try the quiz below on using function notation. You may also want to view the examples in the second part of the page.

[WpProQuiz 17] 





Function Notation Equation – Input and Output






Let’s look at our function again: ƒ(x) = 2x + 5


The value of x that we put into the function notation equation is called the input. The value of ƒ(x) = y is called the output. Input and output values can be positive or negative.


How to Solve Using Function Notation Formula


Example 1:


Place 1 in the function for the input value of x to solve the function ƒ(1):


ƒ(x) = 2x + 5


ƒ(x) = (2 × 1) + 5 = 2 + 5 = 7


So, in the above function notation and resulting equation, when x = 1, the output is 7.


Example 2:


When the input is x = 2, the output is 9:


ƒ(x) = 2x + 5


ƒ(x) = (2 × 2) + 5 = 4 + 5 = 9


So, using function notation, we get the equation from the input and then find the output.


Evaluating the Function


Working out the function notation formula from the inputs to find the outputs is called evaluating the function.


You might see the instructions: “Evaluate the Function” on your exam.


Inverse Functions


Sometimes we have the output and we want to work backwards to discover what the input was.


When we have to work backwards to find the input, we use what is called an inverse function.


Function Notation – Domain and Range


The set of all possible input values for any function notation formula is called the domain. The set of all possible output values for a function is called the range.


Sometimes the domain and range will be specified in the narrative details of the problem. For other exam questions, input and output values may be provided in a chart or table


Function Notation Formula – Advanced Problems


Sometimes you will have to solve one function and use the output from the first function as the input in the second function.


An example of an advanced function notation formula problem is as follows:


For the two functions f1(x) and f2(x), f1(x) = x + 5 and f2(x) = x2.


Step 1:


What is the value of f2(f1(2))?


To solve the problem, our first step is to find the output from the first function for the given input:


Our first function is (f1(2)), so the input for the first function is 2.


f1(x) = x + 5 = 2 + 5 = 7


Step 2:


The second step is to use the output from the first function as the input in the second function


f2(x) = x2 = 72 = 49


So, f2(f1(2)) = 49


Polynomial Functions


Polynomial functions are functions that can be written with coefficients, variables, and exponents. An example of a polynomial function is: f(x) = x2 + 3x + 7


You solve a polynomial function in the same way as any other function.


Solving Polynomial Functions


So, for example, consider the following question:


What is the value of f(x) = x2 + 3x + 7 when x = 3?


Substitute the value of 3 for x to solve the problem.


f(x) = x2 + 3x + 7


f(3) = 32 + (3 × 3) + 7


f(3) = 9 + 9 + 7


f(3) = 25


Trigonometric Functions


Sine, Cosine, and Tangent are the functions used in trigonometry.


If your entrance exam covers advanced math, you should look at our practice problems on trigonometry.


Free Algebra Practice Test


Function Notation Formula and Equation


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