The vertex of the parabola - Examples, Exercises and Solutions

πPractice the vertex of the parabola

5 Questions

The quadratic function

The vertex of the parabola

Practice

Add a new subject

## 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 $X$X of the vertex according to the formula $x=\frac{(-b)}{2a}$x=2a(βb)β

**Second step:** We will place the $X$X of the vertex we have found into the original parabola equation to find the $Y$Y 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 $X$X axis using the quadratic formula.

**Second step:** Find the $X$X of the vertex: the point that is exactly between two points of intersection. The calculation will be done through the average of two $X$Xs of the intersection points.

**Third step:** Place the $X$X of the vertex we have found into the original parabola equation to solve for the $Y$Y of the vertex.

Detailed explanation

Start practice

## Test yourself on the vertex of the parabola!

Find the vertex of the parabola

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

Practice more now

## 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:

$y=ax^2+bx+c$y=ax2+bx+c

**Reminder:**

$a$a positive β> happy-faced parabola

$a$a negative β> sad-faced parabola

The notation of the parabola's vertex is as follows:$(Y~vertex, X~vertex)$(Yvertex,Xvertex)

## 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 $X$X and its $Y$Y.

To find the value of the $X$X of the vertex:

We will use the following formula: $x=\frac{(-b)}{2a}$x=2a(βb)β

To find the value of the $Y$Y of the vertex:

We will place the value of the $X$X we have found into the original parabola equation and obtain the $Y$Y of the vertex.

Join Over 30,000 Students Excelling in Math!

Endless Practice, Expert Guidance - Elevate Your Math Skills Today

Test your knowledge

Question 1

Find the vertex of the parabola

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

Question 2

Find the vertex of the parabola

\( y=(x-3)^2-1 \)

Question 3

Find the vertex of the parabola

\( y=x^2+3 \)

### Let's look at an example

Here is the following equation of the parabola β>

$y=5x^2+20x+4$y=5x2+20x+4

Find the vertex of the parabola.

**Solution:**

To find the $X$X of the vertex we will place it in the formula $x=\frac{(-b)}{2a}$x=2a(βb)β

$b=2$b=2

$a=5$a=5

We will obtain:

$x=\frac{-20}{2 \times 5}$x=2Γ5β20β

$x=\frac{-20}{10}$x=10β20β

$x=-2$x=β2

To find the $Y$Y of the vertex we will place the $X$X of the vertex we have found: $-2$β2

in the original parabola equation.

We will obtain:

$y=5 \times (-2)^2+20 \times (-2)+4$y=5Γ(β2)2+20Γ(β2)+4

$y=5 \times 4-40+4$y=5Γ4β40+4

$y=20-40+4$y=20β40+4

$y=-16$y=β16

The vertex of the parabola is $(-2,-16)$(β2,β16)

**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 $X$X axis.

To do this, we will set $Y=0$Y=0 in the original parabola equation, solve the quadratic equation with the help of the quadratic formula, and obtain two values of $X$X.**Reminder:** The quadratic formula to solve a quadratic equation is: $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$x=2aβbΒ±b2β4acββ

Then, we will find the point that is exactly between the two $X$X values we obtained, and that will be the $X$X of the vertex.

To find the midpoint, we will calculate the average of the $X$Xs.

After finding the $X$X of the vertex, we will place it in the original parabola equation and obtain the $Y$Y of the vertex.

Do you know what the answer is?

Question 1

Find the vertex of the parabola

\( y=x^2 \)

Question 2

Find the vertex of the parabola

\( y=x^2-6 \)

Question 3

Find the vertex of the parabola

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

### Let's look at an example

Here is the following parabola equation ->

$f(x)=x^2-8x+12$f(x)=x2β8x+12

Find the vertex of the parabola.

#### Solution

##### First step

#### First step: Find the points of intersection with the $X$X axis

We will set $Y=0$Y=0**We will obtain:**

$x^2-8x+12=0$x2β8x+12=0

We will solve the quadratic equation by placing the data in the quadratic formula and we will obtain:

$x_{1,2} = {-8 \pm \sqrt{8^2-4 \times 1 \times 12} \over 2 \times 1}$x1,2β=2Γ1β8Β±82β4Γ1Γ12ββ

$x_{1,2} = {-8 \pm 4 \over 2}$x1,2β=2β8Β±4β

$X_1=6$X1β=6

$X_2=2$X2β=2

##### Second step

#### Second step: Calculating the mean we will find the point that is exactly between the two intersection points β> $X$X of the vertex.

**We will obtain:**

$X={(6+2)\over2}=4$X=2(6+2)β=4

##### Third step

#### Third step: We will place the $X$X of the vertex we obtained into the original parabola equation and find the $Y$Y of the vertex.

**We will obtain:**

$Y=4^2-8 \times 4+12$Y=42β8Γ4+12

$Y=-4$Y=β4

The vertex of the parabola is:

$(4,-4)$(4,β4)

#### 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 $2$2 points of intersection of the parabola with the $X$X 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 $Y$Y of the vertex by placing the data in the original equation.

## Examples and exercises with solutions on the vertex of the parabola

### Exercise #1

Find the vertex of the parabola

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

### Video Solution

### Answer

$(-1,0)$(β1,0)

### Exercise #2

Find the vertex of the parabola

$y=(x-1)^2-1$y=(xβ1)2β1

### Video Solution

### Answer

$(1,-1)$(1,β1)

### Exercise #3

Find the vertex of the parabola

$y=(x-3)^2-1$y=(xβ3)2β1

### Video Solution

### Answer

$(3,-1)$(3,β1)

### Exercise #4

Find the vertex of the parabola

$y=x^2+3$y=x2+3

### Video Solution

### Answer

$(0,3)$(0,3)

### Exercise #5

Find the vertex of the parabola

$y=x^2$y=x2

### Video Solution

### Answer

$(0,0)$(0,0)

Check your understanding

Question 1

Find the vertex of the parabola

\( y=(x-3)^2 \)

Question 2

Find the vertex of the parabola

\( y=(x-3)^2+6 \)

Question 3

Find the vertex of the parabola

\( y=(x-7)-7 \)

Start practice

π Click here to practice