Vertex

Vertex Form in 5 Minutes (Formulas, Solvers, Examples)

Published by
Mike Woodhill PhD

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 either
a 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 is
 y=a(x-h)²+k 
where (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.



Mike Woodhill PhD

As a long time instructor I would like you to benefit from my Step by Step Calculators. You may also schedule a live tutoring session to succeed in your class. 😉

Published by
Mike Woodhill PhD

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