Ways of Representing the Quadratic Function - Examples, Exercises and Solutions

The standard representation of the quadratic function looks like this:

$y=ax^2+bx+c$

When:

$a$ –

• The coefficient of $X^2$
• determines whether the parabola will be a maximum or minimum parabola (sad or smiling) and how open or closed it will be.
• $a$ must be different from $0$.
• If $a$ is positive – the parabola is a minimum parabola – smiling
• If $a$ is negative – the parabola is a maximum parabola – sad
• The larger $a$ is – the narrower the function will be, and vice versa.

$b$ -

• The coefficient of $X$
• can be any negative or positive number

$c$ –

• 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:
$y=5x^2+2x+7$
What can we conclude?
$A=5$ therefore the parabola is smiling
$C=7$ therefore the function intersects the $Y$-axis at the point $(0,7)$

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:
$Y=a(X-p)^2+c$

When:
$P$ - represents the $X$ value of the vertex.
$C$ - represents the $Y$ value of the vertex.

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

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

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

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

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

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

The product form shows multiplication between $2$ expressions. With the product form, we can easily determine the points of intersection of the function with the $X$-axis.
The product form of the quadratic function looks like this:
$y=(x-t)*(x-k)$
where
$t$ and $k$ are the $2$ points of intersection of the parabola with the $X$-axis.
As follows: $(t,0) (k,0)$
Let's see an example of the product form to understand better:
$y=(x-6)*(x+5)$
We can determine that:
The points of intersection with the $X$-axis are:
$(6,0)$
$( -5,0)$
Note - Since in the original template there is a minus before \(k\) and $t$, we can infer that if there is a plus before one of them, it is negative, hence $-5$ and not $5$.

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 $BX$ or $C$ and quadratic equations with denominators.

Quadratic equations with missing terms are quadratic equations where $c$ or $b$ are equal to $0$.

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

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

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

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

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