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

How does it differ from Factored Form and Standard Form?

There are 3 ways to express Quadratic Equations.

Each one gives some information about the Graph of a Parabola:

1) \boxed{ f(x) = a(x-h)^2+k } This is called **Vertex Form.**

It tells us that the Graph of f(x) has Vertex Coordinates (h,k) .

**2 Shifts of a Parabola:** Shifting the Standard Parabola f(x) = ax^2

h units right and k units up results in f(x) = a(x-h)^2+k .

That also means: the original Vertex was shifted

from (0,0) to its new position at (h,k) . **Example: f(x) = 2(x-3)^2+4 **

It tells us that the Graph of f(x) has Vertex Coordinates (3,4) . So the Standard Parabola f(x) = ax^2 and its Vertex (0,0) was shifted 3 units right and 4 units up.

Since the 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) \boxed{ f(x) = ax^2+bx+c } This is called **Standard Form.**

It tells us that the Graph of f(x) has a Y-intercept at x=c .**Example: f(x) = 2x^2+4x+5 **

It tells us that the Graph of f(x) has Y-Intercept 5.

Note: We find its Zeros using the famous Quadratic Equations.

3) \boxed{ f(x) = a(x-r)(x-s)} This is called **Factored Form.**

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

It tells us that the Graph of f(x) has the 2 Zeros: x=3 and x=5 .

### Solution with Steps

Enter the 3 coefficients a,b,c of the Quadratic Equation in the above 3 boxes.Next, press the button to find the Vertex and Vertex Form with Steps.

### How do I find the Vertex Form of a Quadratic Equation?

Given the Standard Form of a Quadratic Equation f(x)=ax^2+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 \over 2*a} and k=f({-b \over 2a })

Once computed, (h,k) along with the leading coefficient a are plugged into f(x) = a(x-h)^2+k .**Example: **f(x)=1x^2+8x+3 then

h= {-b \over 2*a} = {-8 \over 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:

f(x) = 1(x+4)^2-13 .

2) The Complete-The-Square Method finds the Vertex Coordinates (h,k) by converting the Standard Form f(x)=ax^2+bx+c

into f(x) = a(x-h)^2+k format in 3 Steps.**Example: **We redo the above Example using the Complete the Square Method in 3 steps.

1) We subtract 3 from y=x^2+8x+3 to get y-3=x^2+8x .

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

2) Since the 16 is missing we simply add 16 to both sides:

y-3 + 16=x^2+8x+16

We just Completed The Square! We now simplify:

y+13=(x+4)^2

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=2x^2+8x+6 .Using the above fast method we find h by computing h= {-8 \over 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). Simple enough.

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

1) We rewrite y=2x^2+8x+6 as y-6=2x^2+8x = 2(x^2+4x) .

To Complete the Square on x^2+4x we must take half of 4 which is 2 and compute (x+2)^2 = x^2+4x+4

2) We therefore add 8 to both sides (as 2*4=8 to the right side):

to get y-6 + 8=2(x^2+4x+4) . This Completes The Square. We rewrite:

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

3) 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 Vertex Form is

(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?

We are given the Quadratic Equation y=-3x^2- 6x-2 .

First, compute the x-coordinate of the Vertex

h={ - b \over 2a}= { -(-6)\over (2*-3)} = { 6\over -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 the above Vertex Form Calculator to solve -3x^2-6x-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 above Vertex Form Calculator a few more times.