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 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.

**First step:** We will find the $X$ of the vertex according to the formula $x=\frac{(-b)}{2a}$

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

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

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

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

The following function has been graphed below.

\( f(x)=-x^2+5x+6 \)

Calculate point C.

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:

$y=ax^2+bx+c$

**Reminder: **

$a$ positive –> happy-faced parabola

$a$ negative –> sad-faced parabola

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

To find the vertex of the parabola, we must solve for the value of its $X$ and its $Y$.

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

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

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

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

Test your knowledge

Question 1

The following function has been graphed below.

\( f(x)=x^2-6x+8 \)

Calculate point B.

Question 2

The following function has been graphed below:

\( f(x)=x^2-6x \)

Calculate point C.

Question 3

The following function has been graphed below:

\( f(x)=x^2-8x+16 \)

Calculate point C.

Here is the following equation of the parabola –>

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

Find the vertex of the parabola.

**Solution:**

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

$b=2$

$a=5$

We will obtain:

$x=\frac{-20}{2 \times 5}$

$x=\frac{-20}{10}$

$x=-2$

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

in the original parabola equation.

We will obtain:

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

$y=5 \times 4-40+4$

$y=20-40+4$

$y=-16$

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

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

To find the vertex of the parabola in this way, we must first find the points of intersection of the parabola with the $X$ axis.

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

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

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

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

Do you know what the answer is?

Question 1

Find the vertex of the parabola

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

Question 2

Find the vertex of the parabola

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

Question 3

Find the vertex of the parabola

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

Here is the following parabola equation ->

$f(x)=x^2-8x+12$

Find the vertex of the parabola.

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

$x^2-8x+12=0$

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

$x_{1,2} = {-8 \pm \sqrt{8^2-4 \times 1 \times 12} \over 2 \times 1}$

$x_{1,2} = {-8 \pm 4 \over 2}$

$X_1=6$

$X_2=2$

**We will obtain:**

$X={(6+2)\over2}=4$

**We will obtain:**

$Y=4^2-8 \times 4+12$

$Y=-4$

The vertex of the parabola is:

$(4,-4)$

As you can see, the second method seems to be quite longer.

However, if you already have $2$ points of intersection of the parabola with the $X$ 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 $Y$ of the vertex by placing the data in the original equation.

The following function has been graphed below.

$f(x)=-x^2+5x+6$

Calculate point C.

To solve the question, let's recall the formula for finding the vertex of a parabola:

Let's substitute the known data into the formula:

-5/2(-1)=-5/-2=2.5

In other words, the x-coordinate of the vertex of the parabola is found when the X value equals 2.5,

Now let's substitute this into the parabola equation and find the Y value

-(2.5)²+5*2.5+6= 12.25

Therefore, the coordinates of the vertex of the parabola are (2.5,12.25).

$(2\frac{1}{2},12\frac{1}{4})$

The following function has been graphed below.

$f(x)=x^2-6x+8$

Calculate point B.

$(3,-1)$

The following function has been graphed below:

$f(x)=x^2-6x$

Calculate point C.

$(3,-9)$

The following function has been graphed below:

$f(x)=x^2-8x+16$

Calculate point C.

$(4,0)$

Find the vertex of the parabola

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

$(-1,-1)$

Check your understanding

Question 1

Find the vertex of the parabola

\( y=x^2+3 \)

Question 2

Find the vertex of the parabola

\( y=x^2-6 \)

Question 3

Find the vertex of the parabola

\( y=x^2 \)