Vertex of a parabola

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Vertex of the Parabola

The vertex of the parabola indicates the highest or maximum point of a sad-faced parabola, and the lowest or minimum point of a happy-faced parabola.

The first method to find the vertex of the parabola: (with formula)

First step: We will find the XX of the vertex according to the formula x=(βˆ’b)2ax=\frac{(-b)}{2a}

Second step: We will place the XX of the vertex we have found into the original parabola equation to find the YY of the vertex.


Second method to find the vertex of the parabola: according to 2 points of intersection with the X-axis and use of symmetry

First step: Find two points of intersection of the parabola with the XX axis using the quadratic formula.

Second step: Find the XX of the vertex: the point that is exactly between two points of intersection. The calculation will be done through the average of two XXs of the intersection points.

Third step: Place the XX of the vertex we have found into the original parabola equation to solve for the YY of the vertex.

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Test yourself on the vertex of the parabola!

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Find the vertex of the parabola

\( y=(x+1)^2 \)

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Vertex of the parabola

In this article, we will study the vertex of the parabola and discover easy ways to find it without too much effort.
The vertex of the parabola marks the highest point of a sad-faced parabola and, the lowest point of a happy-faced parabola.
Let's remember the equation of the parabola:
y=ax2+bx+cy=ax^2+bx+c

Reminder:Β 
aa Β positive –> happy-faced parabola
aa Β negative –> sad-faced parabola Β 

The notation of the parabola's vertex is as follows:Β (YΒ vertex,XΒ vertex)(Y~vertex, X~vertex)

Vertex of the parabola


Ways to Find the Vertex of the Parabola

First method - With formula

To find the vertex of the parabola, we must solve for the value of its XX and its YY.

To find the value of the XX of the vertex:
We will use the following formula: x=(βˆ’b)2ax=\frac{(-b)}{2a}

To find the value of the YY of the vertex:
We will place the value of the XX we have found into the original parabola equation and obtain the YY of the vertex.


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Let's look at an example

Here is the following equation of the parabola –>
y=5x2+20x+4y=5x^2+20x+4
Find the vertex of the parabola.

Solution:
To find the XX of the vertex we will place it in the formula x=(βˆ’b)2ax=\frac{(-b)}{2a}
b=2b=2
a=5a=5
We will obtain:
x=βˆ’202Γ—5x=\frac{-20}{2 \times 5}
x=βˆ’2010x=\frac{-20}{10}
x=βˆ’2x=-2

To find the YY of the vertex we will place the XX of the vertex we have found: βˆ’2-2
in the original parabola equation.
We will obtain:
y=5Γ—(βˆ’2)2+20Γ—(βˆ’2)+4y=5 \times (-2)^2+20 \times (-2)+4
y=5Γ—4βˆ’40+4y=5 \times 4-40+4
y=20βˆ’40+4y=20-40+4
y=βˆ’16y=-16

The vertex of the parabola is (βˆ’2,βˆ’16)(-2,-16)

Note: The fact of having obtained a parabola vertex with negative numbers does not mean that the parabola is a sad face parabola.


Second method: Find the points of intersection with the X-axis and use the points of symmetry in the quadratic formula.

To find the vertex of the parabola in this way, we must first find the points of intersection of the parabola with the XX axis.
To do this, we will set Y=0Y=0 in the original parabola equation, solve the quadratic equation with the help of the quadratic formula, and obtain two values of XX.
Reminder: The quadratic formula to solve a quadratic equation is: x=βˆ’bΒ±b2βˆ’4ac2ax = {-b \pm \sqrt{b^2-4ac} \over 2a}

Then, we will find the point that is exactly between the two XX values we obtained, and that will be the XX of the vertex.
To find the midpoint, we will calculate the average of the XXs.

After finding the XX of the vertex, we will place it in the original parabola equation and obtain the YY of the vertex.

Do you know what the answer is?

Let's look at an example

Here is the following parabola equation ->
f(x)=x2βˆ’8x+12f(x)=x^2-8x+12
Find the vertex of the parabola.

Solution

First step

First step: Find the points of intersection with the XX axis

We will set Y=0Y=0
We will obtain:
x2βˆ’8x+12=0x^2-8x+12=0
We will solve the quadratic equation by placing the data in the quadratic formula and we will obtain:

x1,2=βˆ’8Β±82βˆ’4Γ—1Γ—122Γ—1x_{1,2} = {-8 \pm \sqrt{8^2-4 \times 1 \times 12} \over 2 \times 1}

x1,2=βˆ’8Β±42x_{1,2} = {-8 \pm 4 \over 2}

X1=6X_1=6
X2=2X_2=2

Second step

Second step: Calculating the mean we will find the point that is exactly between the two intersection points –> XX of the vertex.

We will obtain:
X=(6+2)2=4X={(6+2)\over2}=4

Third step

Third step: We will place the XX of the vertex we obtained into the original parabola equation and find the YY of the vertex.

We will obtain:

Y=42βˆ’8Γ—4+12Y=4^2-8 \times 4+12
Y=βˆ’4Y=-4

The vertex of the parabola is:
(4,βˆ’4)(4,-4)


When is it convenient to use the second method?

As you can see, the second method seems to be quite longer.
However, if you already have 22 points of intersection of the parabola with the XX axis, it is advisable to use this method, find the point that is exactly between them by calculating the average and continue looking for the YY of the vertex by placing the data in the original equation.


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