The Parabola y=ax2+bx+c y=ax^2+bx+c 

This function is a quadratic function and is called a parabola.

We will focus on two main types of parabolas: maximum and minimum parabolas.

Minimum Parabola

Also called smiling or happy.

A vertex is the minimum point of the function, where YY is the lowest.

We can identify that it is a minimum parabola if the aa equation is positive.

1b - We can identify that it is a minimum parabola if the equation a is positive


Maximum Parabola

Also called sad or crying.

A vertex is the maximum point of the function, where YY is the highest.

We can identify that it is a maximum parabola if the aa equation is negative.

2b - We can identify that it is a maximum parabola if the a equation is negative

To the parabola,

the vertex marks its highest point.

How do we find it?


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Test yourself on the quadratic function!

einstein

\( y=x^2+x+5 \)

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Find the vertex of the parabola

One of the following two methods can be chosen:


The first method: using the formula for the vertex of the parabola

X=b2aX=\frac{-b}{2a}

The value of XX that we receive will be replaced in the parabola function and we will obtain the value of YY relevant.


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The second method: using two symmetric points

The formula to find XX a vertex using two symmetric points is:

B3 - The formula to find X a vertex using two symmetric points

The vertex XX that we receive in the function to find the value of the vertex YY.

Now, we will move on to the points of intersection of the parabola with the XX and YY axes


Point of Intersection with the Axes

When we want to find the point of intersection with the XX axis:

We will set Y=0Y=0 in the quadratic equation and solve using a trinomial or the root formula.

We can find parabolas that are not zero and that do not have any point of intersection with the XX axis, or that have 11 or a maximum of 22.

When we want to find a point of intersection with the YY axis:

We will set X=0X=0 in the quadratic equation and find the solutions.

Wonderful. Now we will move on to the areas of increase and decrease of the quadratic function.


Do you know what the answer is?

Areas of Increase and Decrease

The areas of increase and decrease describe the XX where the parabola increases and where the parabola decreases.

The parabola changes its domain once, at the vertex.

Let's see this in the figure:

B4 - The areas of increase and decrease describe the X where the parabola increases decreases

When there is a graph:

We will examine what happens when the XXs are smaller than the vertex XX and what happens when the XXs are greater than the vertex XX.

When there is no graph:

  1. We will examine the equation of the function and determine based on the coefficient of X2X^2 whether it is a minimum or maximum function.
  2. Find the vertex XX according to the formula or by symmetric points.
  3. We will plot a graph according to the data we have found and clearly see the areas of increase and decrease.

Positive and Negative Domains

Positive domain: describes the XX where the graph of the parabola is above the XX axis, with a YY value positive.

Negative domain: describes the XX where the graph of the parabola is below the XX axis, with a negative YY value.

To find the domains of positivity and negativity, we will plot the graph of the parabola and ask:

At what XX values is the graph of the parabola above the XX axis, with a positive YY value? This will be the domain of positivity of the parabola.

At what XX values is the graph of the parabola below the XX axis, with a negative YY value? This will be the domain of negativity of the parabola.

Let's see this on the graph:

B5 - Positive domain and negative domain


We will find the points of intersection with the axes and mark them on the coordinate system.

  1. Find the vertex of the parabola and mark it on the coordinate system.
  2. We will understand if the parabola is a maximum or minimum (according to the coefficient aa) and will draw accordingly.

Examples and exercises with solutions of parabola

Exercise #1

y=x2+x+5 y=x^2+x+5

Video Solution

Step-by-Step Solution

To solve this problem, we will identify the parameters of the given quadratic function step-by-step:

  • Step 1: Define the problem statement: We have y=x2+x+5 y = x^2 + x + 5 .
  • Step 2: Identify the standard form of a quadratic function, which is y=ax2+bx+c y = ax^2 + bx + c .
  • Step 3: Compare the given quadratic expression with the standard form to identify the coefficients.

Now, let's analyze the quadratic function provided:

From the given expression y=x2+x+5 y = x^2 + x + 5 :
- The coefficient of x2 x^2 is 1 1 , so a=1 a = 1 .
- The coefficient of x x is 1 1 , so b=1 b = 1 .
- The constant term is 5 5 , so c=5 c = 5 .

Therefore, the parameters of the quadratic function are a=1 a = 1 , b=1 b = 1 , and c=5 c = 5 .

Consequently, the correct choice from the provided options is (a=1,b=1,c=5) (a = 1, b = 1, c = 5) .

Answer

a=1,b=1,c=5 a=1,b=1,c=5

Exercise #2

y=x2+x+5 y=-x^2+x+5

Video Solution

Step-by-Step Solution

To solve the problem of identifying the coefficients in the quadratic function y=x2+x+5 y = -x^2 + x + 5 , we follow these steps:

  • Step 1: Write down the general form of a quadratic equation: y=ax2+bx+c y = ax^2 + bx + c .

  • Step 2: Compare the given equation y=x2+x+5 y = -x^2 + x + 5 to the general form.

  • Step 3: Identify the value of each coefficient:

    • The coefficient of x2 x^2 is 1-1, so a=1 a = -1 .

    • The coefficient of x x is +1+1, so b=1 b = 1 .

    • The constant term is +5+5, so c=5 c = 5 .

Therefore, the parameters of the quadratic function are a=1 a = -1 , b=1 b = 1 , and c=5 c = 5 .

This matches choice 2, confirming the parameters in the quadratic function.

Final Answer: a=1,b=1,c=5 a=-1, b=1, c=5 .

Answer

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

Exercise #3

y=3x2+3x4 y=3x^2+3x-4

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Compare the given quadratic function with the standard form.
  • Step 2: Directly identify the coefficients a a , b b , and c c .
  • Step 3: Verify the correct choice from the provided options, if applicable.

Now, let's work through each step:
Step 1: The given quadratic function is y=3x2+3x4 y = 3x^2 + 3x - 4 . The standard form for a quadratic equation is y=ax2+bx+c y = ax^2 + bx + c .
Step 2: By comparing the given equation to the standard form, we can identify the coefficients:
- a=3 a = 3 , from the term 3x2 3x^2 .
- b=3 b = 3 , from the term 3x 3x .
- c=4 c = -4 , from the constant term 4-4.
Step 3: With these values, compare them to the given choices. The choice that matches these values is option 3: a=3,b=3,c=4 a = 3, b = 3, c = -4 .

Therefore, the solution to the problem is a=3,b=3,c=4 a = 3, b = 3, c = -4 .

Answer

a=3,b=3,c=4 a=3,b=3,c=-4

Exercise #4

y=4x2+3 y=-4x^2+3

Video Solution

Step-by-Step Solution

To solve this problem, we'll compare the given quadratic function with its standard form:

  • Step 1: Recognize the given function as y=4x2+3 y = -4x^2 + 3 .
  • Step 2: Write down the standard form of a quadratic function, which is y=ax2+bx+c y = ax^2 + bx + c .
  • Step 3: Match corresponding terms to identify a a , b b , and c c .

Now, let's work through these steps:

Step 1: The given function is y=4x2+3 y = -4x^2 + 3 .

Step 2: The standard form of a quadratic function is y=ax2+bx+c y = ax^2 + bx + c .

Step 3: By direct comparison:
- The coefficient of x2 x^2 in the given expression is 4-4. Therefore, a=4 a = -4 .
- There is no x x term in the given expression, which implies the coefficient b=0 b = 0 .
- The constant term in the given expression is 3 3 , indicating c=3 c = 3 .

Therefore, the solution is a=4 a = -4, b=0 b = 0, c=3 c = 3, which matches with choice 3.

Answer

a=4,b=0,c=3 a=-4,b=0,c=3

Exercise #5

y=3x24 y=-3x^2-4

Video Solution

Step-by-Step Solution

To solve this problem, we will follow these steps:

  • Step 1: Identify the form of a standard quadratic equation.
  • Step 2: Compare the given function with the quadratic standard form.
  • Step 3: Match the coefficients to the given answer choices.

Now, let's work through each step:
Step 1: The standard form of a quadratic function is y=ax2+bx+c y = ax^2 + bx + c .
Step 2: Given the function y=3x24 y = -3x^2 - 4 , we compare this with the standard form:

  • Coefficient a a is associated with x2 x^2 . Here, a=3 a = -3 .
  • Coefficient b b is associated with x x . Since there is no x x term, b=0 b = 0 .
  • The constant term c c is the standalone number, which is c=4 c = -4 .
Step 3: Given the coefficients a=3 a = -3 , b=0 b = 0 , and c=4 c = -4 , match these with the choices provided. The correct choice is Choice 4: a=3,b=0,c=4 a = -3, b = 0, c = -4 .

Therefore, the solution to the problem is a=3,b=0,c=4 a = -3, b = 0, c = -4 .

Answer

a=3,b=0,c=4 a=-3,b=0,c=-4

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