Ways to represent a quadratic function

Standard representation

The standard representation of the quadratic function looks like this:

y=ax2+bx+cy=ax^2+bx+c

When:

aa

  • The coefficient of X2X^2
  • determines whether the parabola will be a maximum or minimum parabola (sad or smiling) and how open or closed it will be.
  • aa must be different from 00.
  • If aa is positive – the parabola is a minimum parabola – smiling
  • If aa is negative – the parabola is a maximum parabola – sad
  • The larger aa is – the narrower the function will be, and vice versa.

bb -

  • The coefficient of XX
  • can be any negative or positive number

   c    c

  • The free term
  • can be any positive or negative number and determines the intersection point with the Y  axis


Let's see an example:

We are given the function:
y=5x2+2x+7y=5x^2+2x+7
What can we conclude?
A=5A=5 therefore the parabola is smiling
C=7C=7 therefore the function intersects the YY-axis at the point (0,7)(0,7)

Click here to learn more about the standard form of a quadratic function

Vertex presentation

The vertex form of a quadratic function allows us to identify its vertex directly from the function!

The vertex form of a quadratic function is:
Y=a(Xp)2+cY=a(X-p)^2+c

When:
PP - represents the XX value of the vertex.
CC - represents the YY value of the vertex.

For example:
In the function
Y=6(X5)2+2Y=6(X-5)^2+2

The vertex of the parabola is: 
(2,5)(2,5)

Note-
The vertex formula is structured such that there is always a – before the PP, meaning XX vertex, but this does not necessarily mean that the XX vertex is negative.
If the parabola has a negative XX vertex, a ++ will appear before the PP in the formula because times equals ++.

For example:
For example:
In the function
Y=6(X+3)2+8Y=6(X+3)^2+8

The vertex of the parabola is:
(3,8)(-3,8)
There is a ++ before the 33 in the formula, so it is 3-3.

Click here to learn more about the vertex form of a quadratic function

Representation as a product

The product form shows multiplication between 22 expressions. With the product form, we can easily determine the points of intersection of the function with the XX-axis.
The product form of the quadratic function looks like this:
y=(xt)(xk)y=(x-t)*(x-k)
where
tt and kk are the 22 points of intersection of the parabola with the XX-axis.
As follows: (t,0)(k,0)(t,0) (k,0)
Let's see an example of the product form to understand better:
y=(x6)(x+5)y=(x-6)*(x+5)
We can determine that:
The points of intersection with the XX-axis are:
(6,0) (6,0) 
 (5,0) ( -5,0)
Note - Since in the original template there is a minus before \(k\) and tt, we can infer that if there is a plus before one of them, it is negative, hence 5-5 and not 55.

Click here to learn more about representing as a product of a quadratic function

Other different representations

Different representations of a quadratic function –

Sometimes we encounter quadratic equations that are missing terms such as BXBX or CC and quadratic equations with denominators.

Quadratic equations with missing terms are quadratic equations where cc or bb are equal to 00.

When we have an incomplete equation where b=0:
We equate the constant term to the term with x2x^2
and solve for XX. Note that a square root has two answers (negative and positive).
Let's see an example:
x216=0x^2-16=0
We equate the constant term to x2x^2
and get:
X2=16X^2=16
X=4,4X=4,-4

When we have an incomplete equation where c=0c=0:
we can immediately determine that the parabola intersects the YY axis when Y=0Y=0
we will factor out the common factor and find the roots of the equation.

For example:
x216x=0x^2-16x=0

We factor out a common factor and get:
X(X16)=0X(X-16)=0
The factors that zero the equation are –
X=0,16X=0,16

Quadratic equations with denominators – fractions

Sometimes we encounter quadratic equations with a fraction (numerator and denominator). To read the function more accurately, we need to get rid of the fraction.
To solve quadratic equations with denominators-
find the common denominator, multiply each term, and obtain an equation without a fraction. Then solve it completely normally and find the solutions.


Click here to learn more about different representations of a quadratic function

Practice Ways of Representing the Quadratic Function

Examples with solutions for Ways of Representing the Quadratic Function

Exercise #1

Create an algebraic expression based on the following parameters:

a=1,b=1,c=1 a=-1,b=-1,c=-1

Video Solution

Step-by-Step Solution

The goal is to express the quadratic equation y=ax2+bx+c y = ax^2 + bx + c using the given parameters a=1 a = -1 , b=1 b = -1 , and c=1 c = -1 .

First, substitute the values of a a , b b , and c c into the standard form:

  • Substituting a=1 a = -1 , the term becomes x2 -x^2 .
  • Substituting b=1 b = -1 , the term becomes x -x .
  • Substituting c=1 c = -1 , the term remains 1-1.

Combine these terms to form the full expression:


y=x2x1 y = -x^2 - x - 1

Therefore, the algebraic expression for the parameters a=1 a = -1 , b=1 b = -1 , and c=1 c = -1 is: x2x1 -x^2 - x - 1 .

Comparing with the given choices, the correct choice is option 4: x2x1 -x^2-x-1

Answer

x2x1 -x^2-x-1

Exercise #2

Create an algebraic expression based on the following parameters:

a=1,b=2,c=5 a=-1,b=-2,c=-5

Video Solution

Step-by-Step Solution

To create the algebraic expression for the quadratic function given the parameters, we follow these steps:

  • Step 1: Identify the values to substitute into the equation. Here, we have a=1 a = -1 , b=2 b = -2 , and c=5 c = -5 .
  • Step 2: Use the standard quadratic equation format y=ax2+bx+c y = ax^2 + bx + c .
  • Step 3: Substitute the known values into the equation:

Substituting these values, we get:
y=(1)x2+(2)x+(5) y = (-1)x^2 + (-2)x + (-5)

Simplify this expression:
This simplifies to x22x5-x^2 - 2x - 5.

Therefore, the algebraic expression is x22x5 -x^2 - 2x - 5 .

Answer

x22x5 -x^2-2x-5

Exercise #3

Create an algebraic expression based on the following parameters:

a=4,b=2,c=5 a=4,b=2,c=5

Video Solution

Step-by-Step Solution

To derive the algebraic expression based on the parameters given, we follow these steps:

  • Step 1: Recognize the given parameters: a=4 a = 4 , b=2 b = 2 , and c=5 c = 5 .
  • Step 2: Acknowledge that the standard form for a quadratic expression is y=ax2+bx+c y = ax^2 + bx + c .
  • Step 3: Substitute the given parameter values into this quadratic expression.

Now, let's implement these steps to form the quadratic expression:
Step 1: The given parameters are a=4 a = 4 , b=2 b = 2 , and c=5 c = 5 .
Step 2: Our basis is the quadratic form y=ax2+bx+c y = ax^2 + bx + c .
Step 3: Substituting the given values, we find:

y=4x2+2x+5 y = 4x^2 + 2x + 5

This substitution provides us with the quadratic expression y=4x2+2x+5 y = 4x^2 + 2x + 5 , fulfilling the problem's requirements.

Therefore, the correct algebraic expression is 4x2+2x+5 4x^2 + 2x + 5 .

Answer

4x2+2x+5 4x^2+2x+5

Exercise #4

Create an algebraic expression based on the following parameters:

a=1,b=1,c=0 a=-1,b=1,c=0

Video Solution

Step-by-Step Solution

To determine the algebraic expression, we start with the standard quadratic function:

y=ax2+bx+c y = ax^2 + bx + c

Given the values:

  • a=1 a = -1
  • b=1 b = 1
  • c=0 c = 0

We substitute these into the formula:

y=(1)x2+1x+0 y = (-1)x^2 + 1x + 0

Simplifying the expression gives:

y=x2+x y = -x^2 + x

Thus, the algebraic expression, when these parameters are substituted, is:

The solution to the problem is x2+x \boxed{-x^2 + x} .

Answer

x2+x -x^2+x

Exercise #5

Create an algebraic expression based on the following parameters:

a=1,b=1,c=0 a=1,b=1,c=0

Video Solution

Step-by-Step Solution

To determine the algebraic expression, we will substitute the given parameters into the standard form of the quadratic function:

  • The standard quadratic form is y=ax2+bx+c y = ax^2 + bx + c .
  • Substitute a=1 a = 1 , b=1 b = 1 , and c=0 c = 0 into the equation.

Substituting these values, the expression becomes:

y=1x2+1x+0 y = 1 \cdot x^2 + 1 \cdot x + 0 .

This simplifies to:

y=x2+x y = x^2 + x .

Therefore, the algebraic expression, based on the given parameters, is x2+x x^2 + x .

Answer

x2+x x^2+x

Exercise #6

Create an algebraic expression based on the following parameters:

a=3,b=0,c=3 a=3,b=0,c=-3

Video Solution

Step-by-Step Solution

To solve the problem of creating an algebraic expression with the given parameters, we will proceed as follows:

  • Step 1: Identify the given coefficients for the quadratic function, which are a=3 a = 3 , b=0 b = 0 , and c=3 c = -3 .
  • Step 2: Substitute these values into the standard quadratic expression y=ax2+bx+c y = ax^2 + bx + c .

Through substitution, the expression becomes:

y=3x2+0x3 y = 3x^2 + 0x - 3

We can further simplify this expression:

y=3x23 y = 3x^2 - 3

Thus, the algebraic expression with the given parameters is y=3x23 y = 3x^2 - 3 .

The correct answer corresponds to choice number 1: 3x23 3x^2-3 .

Therefore, the solution to the problem is

y=3x23 y = 3x^2 - 3

Answer

3x23 3x^2-3

Exercise #7

Create an algebraic expression based on the following parameters:


a=1,b=1,c=1 a=1,b=-1,c=1

Video Solution

Step-by-Step Solution

To solve this problem, let's form the algebraic expression using the standard quadratic formula:

y=ax2+bx+c y = ax^2 + bx + c

Given are the values:
a=1 a = 1 ,
b=1 b = -1 ,
c=1 c = 1 .

Substituting these values into the formula, we have:
y=1x2+(1)x+1 y = 1 \cdot x^2 + (-1) \cdot x + 1

This simplifies to:
y=x2x+1 y = x^2 - x + 1

Thus, the algebraic expression is x2x+1\boldsymbol{x^2 - x + 1}.

The correct choice from the given options is:

Choice 3: x2x+1 x^2-x+1

Answer

x2x+1 x^2-x+1

Exercise #8

Create an algebraic expression based on the following parameters:

a=1,b=16,c=64 a=1,b=16,c=64

Video Solution

Step-by-Step Solution

To solve this problem, let's proceed with the construction of the quadratic expression:

  • Step 1: Recognize the standard form of a quadratic expression, which is ax2+bx+c ax^2 + bx + c .
  • Step 2: Substitute the given values into this formula:
    • a=1 a = 1
    • b=16 b = 16
    • c=64 c = 64
    Plugging in these values, we determine the expression to be 1x2+16x+64 1x^2 + 16x + 64 , which simplifies to x2+16x+64 x^2 + 16x + 64 .

Thus, the algebraic expression we derive from these parameters is the quadratic expression:

x2+16x+64 x^2 + 16x + 64

This matches the correct choice provided in the given multiple-choice options.

Answer

x2+16x+64 x^2+16x+64

Exercise #9

Create an algebraic expression based on the following parameters:

a=4,b=16,c=0 a=4,b=-16,c=0

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Formulate using the standard quadratic expression template.
  • Step 2: Substitute the given parameters.
  • Step 3: Simplify the resultant expression.

Now, let's work through each step:

Step 1: We use the standard form of a quadratic expression, which is ax2+bx+c ax^2 + bx + c .

Step 2: Substitute the values a=4 a = 4 , b=16 b = -16 , and c=0 c = 0 into this template:

ax2+bx+c4x216x+0 ax^2 + bx + c \rightarrow 4x^2 - 16x + 0

Step 3: Simplify the expression:

The expression simplifies to 4x216x 4x^2 - 16x .

Thus, the algebraic expression based on the given parameters is 4x216x 4x^2 - 16x .

Checking against the answer choices, the correct choice is: 4x216x 4x^2 - 16x .

Answer

4x216x 4x^2-16x

Exercise #10

Create an algebraic expression based on the following parameters:

a=12,b=12,c=12 a=\frac{1}{2},b=\frac{1}{2},c=\frac{1}{2}

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify and substitute the values of a a , b b , and c c into the equation y=ax2+bx+c y = ax^2 + bx + c .
  • Step 2: Simplify the equation to obtain the required expression.
  • Step 3: Compare the simplified expression with the provided multiple-choice answers.

Let's work through each step:

Step 1: The given coefficients are a=12 a = \frac{1}{2} , b=12 b = \frac{1}{2} , and c=12 c = \frac{1}{2} . Substitute these values into the standard quadratic form y=ax2+bx+c y = ax^2 + bx + c :

y=12x2+12x+12 y = \frac{1}{2}x^2 + \frac{1}{2}x + \frac{1}{2}

Step 2: The expression is already simplified. The coefficients are correctly substituted, and no further simplification is needed:

y=x22+x2+12 y = \frac{x^2}{2} + \frac{x}{2} + \frac{1}{2}

Step 3: Compare this expression to the provided multiple-choice options. The correct match is:

Choice 1: x22+x2+12 \frac{x^2}{2} + \frac{x}{2} + \frac{1}{2}

Therefore, the algebraic expression is x22+x2+12 \frac{x^2}{2} + \frac{x}{2} + \frac{1}{2} .

Answer

x22+x2+12 \frac{x^2}{2}+\frac{x}{2}+\frac{1}{2}

Exercise #11

Create an algebraic expression based on the following parameters:

a=5,b=3,c=4 a=5,b=3,c=-4

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify the general form of the quadratic expression.
  • Step 2: Substitute the given values a=5a = 5, b=3b = 3, and c=4c = -4 into the quadratic form.
  • Step 3: Write down the resultant expression.

Now, let's work through each step:
Step 1: The general form of a quadratic expression is ax2+bx+cax^2 + bx + c.
Step 2: We are given a=5a = 5, b=3b = 3, and c=4c = -4. Substituting these into the expression, we get:

5x2+3x45x^2 + 3x - 4

Therefore, the solution to the problem is 5x2+3x45x^2 + 3x - 4.

Answer

5x2+3x4 5x^2+3x-4

Exercise #12

Create an algebraic expression based on the following parameters:

a=1,b=2,c=0 a=1,b=2,c=0

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify the given information
  • Step 2: Apply the standard quadratic formula
  • Step 3: Perform the substitution

Now, let's work through each step:
Step 1: We are given a=1a = 1, b=2b = 2, and c=0c = 0.
Step 2: We'll use the formula y=ax2+bx+cy = ax^2 + bx + c to form our expression.
Step 3: By substituting the given values, we get:

  • Replace aa with 1: 1x2=x21x^2 = x^2.
  • Replace bb with 2: 2x=2x2x = 2x.
  • Replace cc with 0: since c=0c = 0, it does not contribute to the expression, so we can omit this term.

Therefore, we combine the terms to form the expression: x2+2xx^2 + 2x.

The correct answer choice based on our derived expression is: x2+2xx^2 + 2x.

Answer

x2+2x x^2+2x

Exercise #13

Create an algebraic expression based on the following parameters:

a=1,b=1,c=3 a=1,b=-1,c=3

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Identify the provided coefficients for the expression of the form ax2+bx+c ax^2 + bx + c .
  • Step 2: Substitute the provided values into the expression.

Now, let's perform these steps:

Step 1: The problem provides us with the coefficients a=1 a = 1 , b=1 b = -1 , and c=3 c = 3 for a quadratic expression ax2+bx+c ax^2 + bx + c .

Step 2: Substitute these values into the quadratic expression:

a=1 a = 1 : Multiply x2 x^2 by 1 1 , resulting in x2 x^2 .

b=1 b = -1 : Multiply x x by 1-1, resulting in x-x.

c=3 c = 3 : The constant term is 3 3 .

Thus, the algebraic expression is:

x2x+3 x^2 - x + 3 .

Comparing this result to the given choices, we find that this expression matches choice 3.

Therefore, the solution to the problem is x2x+3 x^2 - x + 3 .

Answer

x2x+3 x^2-x+3

Exercise #14

Create an algebraic expression based on the following parameters:

a=3,b=0,c=0 a=3,b=0,c=0

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Substitute the given values a=3 a = 3 , b=0 b = 0 , and c=0 c = 0 into the quadratic function formula y=ax2+bx+c y = ax^2 + bx + c .
  • Step 2: Simplify the expression.

Let's execute these steps:

Step 1: Substitute the values into the formula:
y=3x2+0x+0 y = 3x^2 + 0x + 0

Step 2: Simplify the expression:
Eliminate the terms with zero coefficients to get:
y=3x2 y = 3x^2

Thus, the algebraic expression for the quadratic function with a=3 a = 3 , b=0 b = 0 , and c=0 c = 0 is 3x2 3x^2 .

Therefore, the correct choice from the options provided is choice 1: 3x2 3x^2

Answer

3x2 3x^2

Exercise #15

Create an algebraic expression based on the following parameters:

a=2,b=12,c=4 a=2,b=\frac{1}{2},c=4

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow the steps outlined:

  • Step 1: Identify the given values for the quadratic function's parameters: a=2 a = 2 , b=12 b = \frac{1}{2} , and c=4 c = 4 .
  • Step 2: Apply these values to the standard quadratic form y=ax2+bx+c y = ax^2 + bx + c .
  • Step 3: Substitute the values to construct the algebraic expression.

Now, let's proceed with these steps:

Given the standard form of a quadratic expression y=ax2+bx+c y = ax^2 + bx + c :

Substituting the values, we obtain:

y=2x2+12x+4 y = 2x^2 + \frac{1}{2}x + 4

Therefore, the correct algebraic expression for the quadratic function is 2x2+12x+4 2x^2 + \frac{1}{2}x + 4 .

Answer

2x2+12x+4 2x^2+\frac{1}{2}x+4