### What is Vertex Form of a Quadratic Equation?

The Vertex Form is**f(x) = a(x-h)²+k**.

It is one of three ways to express a Quadratic Equation. The Vertex Form allows to read off the Vertex of a Parabola.

For Example,

**f(x) = 2(x-3)²+5**allows us to read off the Vertex

**(h,k) = (3,5)**

#### How does Vertex Form differ from Factored Form and Standard Form?

Each of the 3 Forms gives some information about its Graph, a Parabola.We will discuss each Form below in detail with examples. We also provide Calculators that allow you to convert your Quadratic Equations between the 3 Forms Step by Step.

1)

**f(x) = a(x-h)²+k**is called Vertex Form.

It gives us the Vertex Coordinates

**(h,k)**.

**Example: f(x) = 2(x-3)²+4 **

It tells us that the Graph of f(x) has Vertex Coordinates

**(3,4)**. The Vertex

**(3,4)**is a result of shifting the Standard Parabola

**f(x) = ax²**with its Vertex

**(0,0)**3 units right and 4 units up. See image below.

Since the leading coefficient

**a=2**is greater than 0 the parabola opens to the top which implies that this function’s Vertex

**(3,4)**is a minimum.

2) ** f(x) = ax²+bx+c ** is called **Standard Form.**

It tells us that the Graph of f(x) has a Y-intercept at **(0,c) **.

Example: **f(x) = 2x²+4x+5 **. Its graph has Y-Intercept 5 as shown in image below.

3) ** f(x) = a(x-r)(x-s) ** is called the **Factored Form. **

It tells us that the Graph of f(x) has the 2 Zeros **x=r , x=s **.
**Example: f(x)=2(x-3)(x-5) ** its graph has the 2 Zeros:

**x=3**and

**x=5**as shown in graph below.

### Two Ways to find the Vertex Form of a Quadratic Equation

Given the Standard Form of a Quadratic Equation **f(x)=ax²+bx+c **there is a quick and a longer way called “Complete The Square Method” to find the Vertex Form:

1) The *quick* way to find the Vertex Coordinates **(h,k) **uses the formula**h = -b/(2*a) **and **k = f(h) **

Once computed, **(h,k) **along with the leading coefficient **a **are plugged into **f(x) = a(x-h)²+k **.**Example: **Given the Standard Form **f(x)=1x²+8x+3 **then **h= -b/(2*a) = -8/(2*1) = -4 ** and thus **k = f(-4) = (-4)²+8(-4)+3 = 16-32+3 = -13 **

Replacing **(-4,-13) **along with a=1 yields the Vertex Form: **f(x) = 1(x+4)²-13 **.

See the image below: It shows the Vertex (-4,-13) of the Function **f(x)=1x²+8x+3** .

2) The Complete-The-Square Method finds the Vertex Coordinates **(h,k) **by converting the Standard Form **f(x)=ax²+bx+c **into Vertex Form **f(x) = a(x-h)²+k** in 3 Steps.**Example: **We redo the above example using the Complete the Square Method.

Step 1: Subtract 3 from **y=x²+8x+3 **to get **y-3=x²+8x **.

Step 2: To Complete the Square on **x²+8x **we must take half of 8 which is 4 and compute **(x+4)² = x²+8x+16 **.

Our function contains **x²+8x**. Since 16 is missing we simply add 16 to both sides:**y-3 + 16=x²+8x+16 **

We just Completed The Square! We can now simplify:**y+13=(x+4)² **

Step 3: Subtracting 13 yields the Vertex Form **y = (x+4)²-13 **.

Comparing this to the general Vertex Form **y = a(x-h)²+k **we can conclude that the given Quadratic Equation has the Vertex **(h,k) = (-4,-13) **.

### How do I find the Vertex Form when the Leading Coefficient a is not 1?

Say we want to find the Vertex of**y=2x²+8x+6**.

Using the above fast method we find h by computing

**h = -8/(2*2) = -2**.

We next plug in

**h=-2**to get

**k=2*(-2)²+8(-2)+6 = -2**.

Thus, the Vertex is

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

#### Let’s do the Complete the Square Method in 3 steps:

Step 1: We rewrite**y=2x²+8x+6**as

**y-6=2x²+8x = 2(x²+4x)**.

To Complete the Square on

**x²+4x**we must take half of 4 which is 2 and compute

**(x+2)² = x²+4x+4**

Step 2: Since 4 gets multiplied by 2 we must add 8 to both sides:

to get

**y-6 + 8=2(x²+4x+4)**. This Completes The Square.

We rewrite:

**y+2=2(x+2)²**

Step 3: Finally, subtracting 2 yields the Vertex Form

**f(x) = 2(x+2)²-2**.

Comparing this to the general Vertex Form

**f(x) = a(x-h)²+k**we can conclude that the given Parabola in has Vertex

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

### How do I find a Minimum or Maximum on the Graph of a Parabola?

Every Parabola has eithera

**Minimum**(opened to the top when leading coefficient a>0) or

a

**Maximum**(opened to the bottom when leading coefficient a<0).

The Vertex is that particular Point on the Graph of a Parabola.

See the illustration below of the two possible Vertex options:

### Example: How do I find the Vertex Form of a Quadratic Equation when the leading coefficient a is negative?

The two above methods work just fine: We are given the Standard Form **y = -3x² -6x-2 **.

Using the fast method, we first compute the x-coordinate of the Vertex **h= -b/(2a) = -(-6)/(2*-3) = 6/-6 = -1 **.

Next, compute the y-coordinate of the Vertex by plugging **h=-1** into the given equation: **k= -3*(-1)²-6*(-1)-2 = -3+6-2 = 1 **.

Therefore, the Vertex is **(h,k)=(-1,1) **.

Thus, the Vertex form of the Parabola is **y=-3*(x+1)²+1 **

Easy, wasn’t it?

Tip: When using our Vertex Form Calculator to solve **-3x²-6x-2=0** we must enter the 3 coefficients as **a=-3, b=-6** and **c=-2**.

Then, the calculator will find the vertex **(h,k)=(-1,1) **Step by Step.

Get it now? Try the Vertex Form Calculator below a few more times.

### How do I convert from Vertex to Standard Form?

The Vertex Form of a Parabola isy=a(x-h)²+kwhere

**(h,k)**are the Vertex Coordinates.

The Standard form of a Parabola is

y=ax²+bx+c

#### Let’s do an easy example first

Let**y=2(x-1)²-5**

we first apply the binomial formula to expand and get

**y=2(x²-2x+1)-5**

Next, we distribute the 2 to get

**y= 2x²-4x+2-5**

With 2-5=-3 we finally arrive at the Standard Form:

**y=2x²-4x-3**

In general, we obtain the Standard Form from the Vertex Form by using these 2 steps:

Given:

**y=a(x-h)²+k**

Step1: (Use Binomial Formula)

**y=a(x²-2hx+h²)+k**

Step2: (Distribute and Combine 2 like terms ah² and k)

**y=ax²-(2ah)x+(ah²+k)**

We created a separate page for you that teaches the Vertex to Standard Form Conversion. It also has a Solver that allows you convert you Vertex Form Equation into Standard Form. Visit our page here.