Create an algebraic expression based on the following parameters: $$ a=\frac{1}{2},b=\frac{1}{2},c=\frac{1}{2} $$

$$ \frac{x^2}{2}+\frac{x}{2}+\frac{1}{2} $$

$$ \frac{x^2}{2}+2x+\frac{1}{2} $$

$$ \frac{x^2}{2}-\frac{x}{2}-\frac{1}{4} $$

Quadratic Function Representations Practice Problems

Understanding Ways of Representing the Quadratic Function

Complete explanation with examples

Ways to represent a quadratic function

Standard representation

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

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

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

Representation as a product

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

Other different representations

Different representations 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 –
$X=0,16$

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

Detailed explanation

Practice Ways of Representing the Quadratic Function

Test your knowledge with 97 quizzes

Create an algebraic expression based on the following parameters:

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

Examples with solutions for Ways of Representing the Quadratic Function

Step-by-step solutions included

Exercise #1

Create an algebraic expression based on the following parameters:

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

Step-by-Step Solution

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

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

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

Combine these terms to form the full expression:

$y = -x^2 - x - 1$

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

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

Answer:

$-x^2-x-1$

Video Solution

Exercise #2

Create an algebraic expression based on the following parameters:

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

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 $a = -1$ , $b = -2$ , and $c = -5$ .
Step 2: Use the standard quadratic equation format $y = ax^2 + bx + c$ .
Step 3: Substitute the known values into the equation:

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

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

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

Answer:

$-x^2-2x-5$

Video Solution

Exercise #3

Create an algebraic expression based on the following parameters:

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

Step-by-Step Solution

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

Step 1: Recognize the given parameters: $a = 4$ , $b = 2$ , and $c = 5$ .
Step 2: Acknowledge that the standard form for a quadratic expression is $y = ax^2 + bx + c$ .
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 $a = 4$ , $b = 2$ , and $c = 5$ .
Step 2: Our basis is the quadratic form $y = ax^2 + bx + c$ .
Step 3: Substituting the given values, we find:

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

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

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

Answer:

$4x^2+2x+5$

Video Solution

Exercise #4

Create an algebraic expression based on the following parameters:

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

Step-by-Step Solution

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

$y = ax^2 + bx + c$

Given the values:

$a = -1$
$b = 1$
$c = 0$

We substitute these into the formula:

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

Simplifying the expression gives:

$y = -x^2 + x$

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

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

Answer:

$-x^2+x$

Video Solution

Exercise #5

Create an algebraic expression based on the following parameters:

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

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 $y = ax^2 + bx + c$ .
Substitute $a = 1$ , $b = 1$ , and $c = 0$ into the equation.

Substituting these values, the expression becomes:

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

This simplifies to:

$y = x^2 + x$ .

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

Answer:

$x^2+x$

Video Solution

Frequently Asked Questions

Everything you need to know about Ways of Representing the Quadratic Function

What are the three main ways to represent a quadratic function?

+

The three main representations are: 1) Standard form: y=ax²+bx+c, 2) Vertex form: y=a(x-p)²+c, and 3) Product form: y=(x-t)(x-k). Each form highlights different properties of the parabola.

How do you find the vertex from vertex form y=a(x-p)²+c?

+

The vertex is at point (p,c). Remember there's always a minus sign before p in the formula, so if you see y=2(x+3)²+5, the vertex is at (-3,5) because -(+3) = -3.

What does the coefficient 'a' tell you about a quadratic function?

+

The coefficient 'a' determines: if a>0, the parabola opens upward (smiling); if a<0, it opens downward (sad). The larger |a| is, the narrower the parabola becomes.

How do you find x-intercepts from product form y=(x-t)(x-k)?

+

The x-intercepts are at points (t,0) and (k,0). For example, if y=(x-6)(x+5), the x-intercepts are at (6,0) and (-5,0). Note the sign change due to the minus in the formula.

What happens when b=0 in a quadratic equation?

+

When b=0, you have an incomplete quadratic like x²-16=0. Solve by isolating x²: x²=16, so x=±4. The parabola is symmetric about the y-axis.

What happens when c=0 in a quadratic equation?

+

When c=0, the parabola passes through the origin (0,0). Factor out the common factor: x²-16x=0 becomes x(x-16)=0, giving solutions x=0 and x=16.

How do you solve quadratic equations with fractions?

+

Find the common denominator and multiply each term to eliminate fractions. Then solve the resulting quadratic equation normally using factoring, completing the square, or the quadratic formula.

Which quadratic form is best for finding different properties?

+

Use standard form y=ax²+bx+c for y-intercept (c value), vertex form y=a(x-p)²+c for vertex location (p,c), and product form y=(x-t)(x-k) for x-intercepts (t,0) and (k,0).

More Ways of Representing the Quadratic Function Questions

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Identify the positive and negative domain Linking function properties to its representation Transition between different representations of a function With fractions Determine the algebraic representation of the following descriptions Finding Increasing or Decreasing Domains Identify the positive and negative domain Transition between different representations of a function Using roots With fractions Identify the positive and negative domain Linking function properties to its representation Transition between different representations of a function Using roots With fractions

Quadratic Function Representations Practice Problems

Understanding Ways of Representing the Quadratic Function

Ways to represent a quadratic function

Standard representation

Vertex presentation

Representation as a product

Other different representations

Different representations of a quadratic function –

Quadratic equations with denominators – fractions

Practice Ways of Representing the Quadratic Function

Examples with solutions for Ways of Representing the Quadratic Function

Frequently Asked Questions

What are the three main ways to represent a quadratic function?

How do you find the vertex from vertex form y=a(x-p)²+c?

What does the coefficient 'a' tell you about a quadratic function?

How do you find x-intercepts from product form y=(x-t)(x-k)?

What happens when b=0 in a quadratic equation?

What happens when c=0 in a quadratic equation?

How do you solve quadratic equations with fractions?

Which quadratic form is best for finding different properties?

More Ways of Representing the Quadratic Function Questions

Product Representation

Vertex Representation

Standard Representation

Practice by Question Type

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