In Mathematics, the vertex formula helps to find the vertex coordinate of a parabola, when the graph crosses its axes of symmetry. Generally, the vertex point is represented by (h, k). We know that the standard equation of a parabola is y=ax²+bx+c. Here, if the coefficient of x² is positive, the vertex should be at the bottom of the U-shaped curve. If the coefficient of x² is negative, then the vertex should be at the top of the U-shaped curve. In this article, we are going to learn the standard form and vertex form of a parabola, vertex formula, and examples in detail.

The two vertex formulas to find the vertex is:

Formula 1: (h, k) = (-b/2a, -D/4a)

where,

• D is the denominator
• h,k are the coordinates of the vertex

Formula 2: x-coordinate of the vertex = -b / 2a

Vertex Form of Parabola

We know that the standard form of the parabola is y=ax²+bx+c.

Thus, the vertex form of a parabola is y = a(x-h)² + k.

Now, let us discuss the vertex formula in detail.

Vertex Formula

The vertex formula is used to find the vertex of a parabola. There are two ways to find the vertex of a parabola.

Vertex, (h, k) = (-b/2a, -D/4a)

Where “D” is the discriminant where D = b² – 4ac.

“h” and “k” are the coordinates of the vertex.

The above formula can also be written as follows:

The other method to find the vertex of a parabola is as follows:

We know that the x-coordinate of a vertex, (i.e) h is -b/2a.

Now, substitute the x-coordinate value in the given standard form of the parabola equation y=ax²+bx+c, we will get the y-coordinate of a vertex.

Solved Examples Using Vertex Formula

Example 1:

Find the vertex of a parabola, y=3x²+12x-12.

Solution:

Given parabola equation: y=3x²+12x-12.

The given parabola equation is of the standard form y=ax²+bx+c.

By comparing the given equation and standard form, we get

a = 3. b= 12, c = -12.

We know that the vertex formula is (-b/2a, -D/4a)

We know that D = b² – 4ac.

Therefore, D = (12)²-4(3)(-12)

D = 144+144

D = 288.

Now, substitute all the known values in the formula, we get

Vertex, (h, k) = ( (-12/2(3)), (-288/4(3))

(h, k) = (-12/6, -288/12)

(h. k) = (-2, -24)

The vertex (h, k) of the parabola y=3x²+12x-12 is (-2, -24).

Example 2: 

Find the vertex of the parabola y=3x²-6x+1.

Solution:

Given parabola equation: y = 3x²-6x+1.

The standard form of a parabola is  y=ax²+bx+c.

By comparing standard form and given parabola equations, we get a = 3, b=-6, c = 1.

We know that the formula to calculate the x-coordinate of a vertex is -b/2a.

Hence, h = -(-6)/2(3)

h = 6/6 = 1

Therefore, the x-coordinate of a vertex is 1.

Now, we need to find the y-coordinate of a vertex. (i.e.) k.

To get the value of the y-coordinate, substitute x =1 in the given equation y = 3x²-6x+1.

Hence, y-coordinate (k) = 3(1)²-6(1)+1

y-coordinate (k) = 3-6+1 = -2.

Hence, the coordinate of the vertex of a parabola (h, k) is (1, -2).

What is Vertex Formula?

The vertex formula helps to find the vertex coordinates of a parabola. The standard form of a parabola is y = ax² + bx + c. The vertex form of the parabola y = a(x – h)² + k. There are two ways in which we can determine the vertex(h, k). They are:

• (h, k) = (-b/2a, -D/4a), where D(discriminant) = b² – 4ac
• (h,k), where h = -b / 2a and evaluate y at h to find k.

Derivation of Vertex Formulas

Formula 1

We know that the standard form of a parabola is, y = ax² + bx + c. Let us convert it to the vertex form y = a(x – h)² + k by completing the squares.

Subtracting c from both sides:

y – c = ax² + bx

Taking “a” as the common factor:

y – c = a (x² + b/a x)

Here, half the coefficient of x is b/2a and its square is b²/4a². Adding and subtracting this on the right side (inside the parentheses):

y – c = a (x² + b/a x + b²/4a² – b²/4a²)

We can write x² + b/a x + b²/4a² as (x + b/2a)². Thus, the above equation becomes:

y – c = a ( (x + b/2a)² – b²/4a²)

Distributing “a” on the right side and adding “c” on both sides:

y = a (x + b/2a)² – b²/4a + c

y = a (x + b/2a)² – (b² – 4ac) / (4a)

Comparing this with y = a (x – h)² + k, we get:

h = -b/2a

k = -(b² – 4ac) / (4a)

We know that b² – 4ac is the discriminant (D).

Thus, the vertex formula is: (h, k) = (-b/2a, -D/4a) where D = b² – 4ac

Formula 2

If you feel difficult to memorize the above formula, you can just remember the formula for the x-coordinate of vertex and then just substitute it in the given equation y = ax² + bx + c to get the y-coordinate of the vertex.

x-coordinate of the vertex(h) = -b / 2a

Alternatively, if you do not want to use any of the above formulas to find the vertex, then you can just complete the square to convert y = ax² + bx + c of the form y = a(x – h)² + k manually and find the vertex (h, k).

Examples Using Vertex Formula

FAQs on Vertex Formula

The Vertex of a Parabola

The vertex of a parabola is the point where the parabola crosses its axis of symmetry.   If the coefficient of the x2

term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “ U ”-shape.  If the coefficient of the x2 term is negative, the vertex will be the highest point on the graph, the point at the top of the “ U”-shape.

The standard equation of a parabola is

y=ax2+bx+c

.But the equation for a parabola can also be written in “vertex form”:

y=a(x−h)2+k

In this equation, the vertex of the parabola is the point (h,k)

You can see how this relates to the standard equation by multiplying it out:

y=a(x−h)(x−h)+k

y=ax2−2ahx+ah2+k

. This means that in the standard form, y=ax2+bx+c , the expression −b/2a gives the x -coordinate of the vertex.

How to Convert From Standard Quadratic Form to Vertex Form

Most of the time when you’re asked to convert quadratic equations between different forms, you’ll be going from standard form (ax2+bx+c) to vertex form (a(x−h)2+k).

The process of converting your equation from standard quadratic to vertex form involves doing a set of steps called completing the square.

Let’s walk through an example of converting an equation from standard form to vertex form. We’ll start with the equation y=7×2+42x−3/14.

The first thing you’ll want to do is move the constant, or the term without an x or x2 next to it. In this case, our constant is −3/14. (We know it’s negative 3/14 because the standard quadratic equation is ax2+bx+c, not ax2+bx−c.) 

First, we’ll take that −3/14 and move it over to the left side of the equation:

y+3/14=7x²+42x

The next step is to factor out the 7 (the a value in the equation) from the right side, like so:

y+3/14=7(x2+6x)

Note that on the left side of equation, we made sure to include our a value, 7, in front of the space where our constant will go; this is because we’re not just adding the constant to the right side of the equation, but we’re multiplying the constant by whatever is on the outside of the parentheses. (If your a value is 1, you don’t need to worry about this.)

The next step is to complete the square. In this case, the square you’re completing is the equation inside of the parentheses—by adding a constant, you’re turning it into an equation that can be written as a square.

To calculate that new constant, take the value next to x (6, in this case), divide it by 2, and square it.

Last step: move the non-y value from the left side of the equation back over to the right side:

Congratulations! You’ve successfully converted your equation from standard quadratic to vertex form.

Now, most problems won’t just ask you to convert your equations from standard form to vertex form; they’ll want you to actually give the coordinates of the vertex of the parabola.

To avoid getting tricked by sign changes, let’s write out the general vertex form equation directly above the vertex form equation we just calculated:

Whew, that was a lot of shuffling numbers around! Fortunately, converting equations in the other direction (from vertex to standard form) is a lot simpler.

How to Convert From Vertex Form to Standard Form

Converting equations from their vertex form to the regular quadratic form is a much more straightforward process: all you need to do is multiply out the vertex form.

Parabola Vertex Form Practice: Sample Questions

To wrap up this exploration of vertex form, we have four example problems and explanations. See if you can solve the problems yourself before reading through the explanations!

Parabola Vertex Form Practice: Solutions

#1: What is the vertex form of the quadratic equation x2+2.6x+1.2?

Start by separating out the non-x variable onto the other side of the equation:

y−1.2=x2+2.6x

Since our a (as in ax2+bx+c) in the original equation is equal to 1, we don’t need to factor it out of the right side here (although if you want, you can write y−1.2=1(x2+2.6x)).

Next, divide the x coefficient (2.6) by 2 and square it, then add the resulting number to both sides of the equation:

#2: Convert the equation 7y=91×2−112 into vertex form. What is the vertex?

When converting an equation into vertex form, you want the y have a coefficient of 1, so the first thing we’re going to do is divide both sides of this equation by 7:

Now, there are a couple of ways to go from here. The sneaky way is to use the fact that there’s already a square written into the vertex form equation to our advantage.

First, we’ll move the constant over to the left side of the equation:

Vertex formula

For the vertex form of the parabola, y = a(x – h)² + k, the coordinates (h, k) of the vertex are,

(h, k) = (-b/2a, -D/4a)

where,

a is the coefficient of x²,

b is the coefficient of x,

D = b² – 4ac is the discriminant of the standard form y = ax² + bx + c.

Derivation

Suppose we have a parabola with standard equation as, y = ax² + bx + c.

This can be written as,

y – c = ax² + bx

y – c = a (x² + bx/a)

Adding and subtracting b²/4a² on the RHS, we get

y – c = a (x² + bx/a + b²/4a² – b²/4a²)

y – c = a ((x + b/2a)² – b²/4a²)

y – c = a (x + b/2a)² – b²/4a

y = a (x + b/2a)² – b²/4a + c

y = a (x + b/2a)² – (b²/4a – c)

y = a (x + b/2a)² – (b² – 4ac)/4a

We know, D = b² – 4ac, so the equation becomes,

y = a (x + b/2a)² – D/4a

Comparing the above equation with the vertex form y = a(x – h)² + k, we get

h = -b/2a and k = -D/4a

This derives the formula for coordinates of the vertex of a parabola.

Sample Problems

Problem 1. Find the coordinates of the vertex for the parabola y = 2x² + 4x – 4.

Solution:

We have the equation as, y = 2x² + 4x – 4.

Here, a = 2, b = 4 and c = -4.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² – 4ac.

D = (4)² – 4 (2) (-4)

= 16 + 32

= 48

So, x – coordinate of vertex = -4/2(2) = -4/4 = -1.

y – coordinate of vertex = -48/4(2) = -48/8 = -6

Hence, the vertex of the parabola is (-1, -6).

Problem 2. Find the coordinates of the vertex for the parabola y = 3x² + 5x – 2.

Solution:

We have the equation as, y = 3x² + 5x – 2.

Here, a = 3, b = 5 and c = -2.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² – 4ac.

D = (5)² – 4 (3) (-2)

= 25 + 24

= 49

So, x – coordinate of vertex = -5/2(3) = -5/6

y – coordinate of vertex = -49/4(3) = -49/12

Hence, the vertex of the parabola is (-5/6, -49/12).

Problem 3. Find the coordinates of the vertex for the parabola y = 3x² – 6x + 1.

Solution:

We have the equation as, y = 3x² – 6x + 1.

Here, a = 3, b = -6 and c = 1.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² – 4ac.

D = (-6)² – 4 (3) (1)

= 36 – 12

= 24

So, x – coordinate of vertex = 6/2(3) = 6/6 = 1

y – coordinate of vertex = -24/4(3) = -24/12 = -2

Hence, the vertex of the parabola is (1, -2).

Problem 4. Find the coordinates of the vertex for the parabola y = 3x² + 8x – 8.

Solution:

We have the equation as, y = 3x² + 8x – 8.

Here, a = 3, b = 8 and c = -8.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² – 4ac.

D = (8)² – 4 (3) (-8)

= 64 + 96

= 160

So, x – coordinate of vertex = -8/2(3) = -8/6 = -4/3

y – coordinate of vertex = -160/4(3) = -160/12 = -40/3

Hence, the vertex of the parabola is (-4/3, -40/3).

Problem 5. Find the coordinates of the vertex for the parabola y = 6x² + 12x + 4.

Solution:

We have the equation as, y = 6x² + 12x + 4.

Here, a = 6, b = 12 and c = 4.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² – 4ac.

D = (12)² – 4 (6) (4)

= 144 – 96

= 48

So, x – coordinate of vertex = -12/2(6) = -12/12 = -1

y – coordinate of vertex = -48/4(6) = -48/24 = -2

Hence, the vertex of the parabola is (-1, -2).

Problem 6. Find the coordinates of the vertex for the parabola y = x² + 7x – 5.

Solution:

We have the equation as, y = x² + 7x – 5.

Here, a = 1, b = 7 and c = -5.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b² – 4ac.

D = (7)² – 4 (1) (-5)

= 49 + 20

= 69

So, x – coordinate of vertex = -7/2(1) = -7/2

y – coordinate of vertex = -69/4(1) = -69/4

Hence, the vertex of the parabola is (-7/2, -69/4).

Problem 7. Find the coordinates of the vertex for the parabola y = 2x² + 10x – 3.

Solution:

We have the equation as, y = x2 + 7x – 5.

Here, a = 1, b = 7 and c = -5.

Now, it is known that the coordinates of the vertex are given by, (-b/2a, -D/4a) where D = b2 – 4ac.

D = (7)2 – 4 (1) (-5)

= 49 + 20

= 69

So, x – coordinate of vertex = -7/2(1) = -7/2

y – coordinate of vertex = -69/4(1) = -69/4

Hence, the vertex of the parabola is (-7/2, -69/4).

Finding the X Coordinate of the Vertex

Now, let’s look at an example where we use the vertex formula and a table of values to graph a function.

Example 1 – Finding the Vertex

The vertex, also known as your maximum point, is (-1, 4.5). The zeros of the function are: (-4,0) and (2,0). These are the points where the parabola crosses the x-axis.

Ok, ready to try one on your own?

Practice Problem

Given the following function:

h(x) = 1/2x² – x +2

• Predict whether the parabola will open up or down.
• Find the x coordinate of the vertex using the vertex formula.
• Create a table of values and graph the parabola.
• Identify the zeros of function.

Solution

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