The standard representation of the quadratic function looks like this:
The standard representation of the quadratic function looks like this:
When:
–
-
–
Let's see an example:
We are given the function:
What can we conclude?
therefore the parabola is smiling
therefore the function intersects the -axis at the point
Click here to learn more about the standard form of a quadratic function
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:
When:
- represents the value of the vertex.
- represents the value of the vertex.
For example:
In the function
The vertex of the parabola is:
Note-
The vertex formula is structured such that there is always a – before the , meaning vertex, but this does not necessarily mean that the vertex is negative.
If the parabola has a negative vertex, a will appear before the in the formula because times equals .
For example:
For example:
In the function
The vertex of the parabola is:
There is a before the in the formula, so it is .
Click here to learn more about the vertex form of a quadratic function
The product form shows multiplication between expressions. With the product form, we can easily determine the points of intersection of the function with the -axis.
The product form of the quadratic function looks like this:
where
and are the points of intersection of the parabola with the -axis.
As follows:
Let's see an example of the product form to understand better:
We can determine that:
The points of intersection with the -axis are:
Note - Since in the original template there is a minus before \(k\) and , we can infer that if there is a plus before one of them, it is negative, hence and not .
Click here to learn more about representing as a product of a quadratic function
Sometimes we encounter quadratic equations that are missing terms such as or and quadratic equations with denominators.
Quadratic equations with missing terms are quadratic equations where or are equal to .
When we have an incomplete equation where b=0:
We equate the constant term to the term with
and solve for . Note that a square root has two answers (negative and positive).
Let's see an example:
We equate the constant term to
and get:
When we have an incomplete equation where :
we can immediately determine that the parabola intersects the axis when
we will factor out the common factor and find the roots of the equation.
For example:
We factor out a common factor and get:
The factors that zero the equation are –
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
Find the standard representation of the following function
\( f(x)=(x-2)(x+5) \)
Determine the points of intersection of the function
\( y=(x-5)(x+5) \)
With the X
Create an algebraic expression based on the following parameters:
\( a=-1,b=-1,c=-1 \)
Create an algebraic expression based on the following parameters:
\( a=3,b=0,c=-3 \)
Create an algebraic expression based on the following parameters:
\( a=1,b=-1,c=1 \)
Find the standard representation of the following function
We will use the expanded distributive property.
(a+1)⋆(b+2) = ab+2a+b+2
Let's substitute the given values and solve:
(x-2)(x+5) =
x²-2x+5x+-2*5=
x²+3x-10
And that's the solution!
Determine the points of intersection of the function
With the X
In order to find the point of the intersection with the X-axis, we first need to establish that Y=0.
0 = (x-5)(x+5)
When we have an equation of this type, we know that one of these parentheses must be equal to 0, so we begin by checking the possible options.
x-5 = 0
x = 5
x+5 = 0
x = -5
That is, we have two points of intersection with the x-axis, when we discover their x points, and the y point is already known to us (0, as we placed it):
(5,0)(-5,0)
This is the solution!
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
\( a=-1,b=-8,c=0 \)
Create an algebraic expression based on the following parameters:
\( a=3,b=0,c=0 \)
Create an algebraic expression based on the following parameters:
\( a=2,b=4,c=8 \)
Create an algebraic expression based on the following parameters:
\( a=2,b=\frac{1}{2},c=4 \)
Create an algebraic expression based on the following parameters:
\( a=4,b=-16,c=0 \)
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
\( a=-1,b=1,c=0 \)
Create an algebraic expression based on the following parameters:
\( a=0,b=1,c=0 \)
Create an algebraic expression based on the following parameters:
\( a=-1,b=0,c=0 \)
Create an algebraic expression based on the following parameters:
\( a=3,b=6,c=9 \)
Create an algebraic expression based on the following parameters:
\( a=2,b=0,c=6 \)
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters:
Create an algebraic expression based on the following parameters: