Ways to represent a quadratic function

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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

Start practice

Test yourself on ways of representing the quadratic function!

Create an algebraic expression based on the following parameters:

\( a=2,b=2,c=2 \)

Practice more now

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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

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