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

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Test yourself on ways of representing the quadratic function!

einstein

Create an algebraic expression based on the following parameters:

\( a=-1,b=-1,c=-1 \)

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=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 #3

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 #4

Create an algebraic expression based on the following parameters:


a=1,b=8,c=0 a=-1,b=-8,c=0

Video Solution

Step-by-Step Solution

To solve this problem, we'll start by substituting the given parameters into the standard quadratic formula:

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

The problem gives us the values:

  • a=1 a = -1
  • b=8 b = -8
  • c=0 c = 0

This means we need to replace a a , b b , and c c in the formula:

y=(1)x2+(8)x+0 y = (-1)x^2 + (-8)x + 0

Simplifying this expression further:

  • The term with a a : (-1)x^2\) results in x2 -x^2 .
  • The term with b b : (-8)x\) simplifies to 8x -8x .
  • The term with c c : 0 0 contributes nothing to the expression, so it is omitted.

Thus, the final algebraic expression is:

y=x28x y = -x^2 - 8x

Therefore, the algebraic expression based on the given parameters is

x28x -x^2 - 8x .

Answer

x28x -x^2-8x

Exercise #5

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

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