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]]>Parabolas are the Graphs of Quadratic Equations.

There are 3 different forms of Quadratic Equations:

Standard Form: \boxed{ y = ax^2+bx+c }

Vertex Form: \boxed{ y = a(x-h)^2+k} . (h,k) = Vertex Coordinates.

Factored Form: \boxed{ y = a(x-r)(x-s)} . r and s = Zeros of the Parabola.

Standard Form: \boxed{ y = x^2+6x+8 } can be rewritten in as

Vertex Form: \boxed{ y = (x+3)^2-1} . It tells us that above Parabola has Vertex Coordinates = (3,-1) . The Process to convert from Standard Form to Vertex Form is called “Completing the Square” and can be done here with steps:

Solve A Quadratic Equation by Completing the Square

Factored Form: \boxed{ y = (x+4)(x+2)} . It tells us that above Parabola has Zeros -4 and -2 . The Process to convert from Standard Form to Factored Form is called “Factoring a Quadratic Equation” and can be done here with steps:

Factoring Quadratic Equations – Calculator

Or simply use the head menu when converting between the different Parabola Forms.

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]]> 1 – Step by Step Solver: Find the Range of a Parabola

2 – What is the Domain of a Parabola?

3 – How do I write the Range in Interval Notation?

4 – Example: Domain and Range of a Parabola

5 – Is there a Domain and Range Parabola Calculator?

Next, press the button to find the Domain and Range of a Parabola with Steps.

The reason for that is quadratic equations fall in the category of polynomials and thus don’t contain fractions, roots or radicals nor logarithms. Those 3 categories of functions have restrictions with regard to their domain.

Notice that the 3 is included in the Range so we use bracket [ in the Interval Notation. Since there is no upper bound for the Range we denote that with the Infinity Symbol which is always followed by the Open Interval symbol “)”. Think of Infinity as NOT being a concrete endpoint.

Again, the 5 is included in the Range so we use the bracket ] in the Interval Notation.

We are given the Standard Form

y=3x^2- 6x-2 .

Since ANY x value can be plugged into this equation (we have no fractions, radicals nor logarithms) we can conclude that the Domain is ‘all real numbers’. In Interval Notation: (-\infty , \infty)

To find the Range we first compute the x-coordinate of the vertex

h={ - b \over 2a}= { -(-6) \over (2*3)} = 1 .

Next, compute the y-coordinate of the vertex by plugging h=1 into the given equation:

k= 3*(1)^2-6(1)-2=-5 .

Therefore, the vertex is

(h,k)=(1,-5) .

The Range is simply all real numbers greater or equal to -5 or simply y>=-5.

Using Interval Notation: [-5 , \infty )

Explanation: This parabola opens to the top since the leading coefficient 3 is greater than 0. This implies that the vertex is a minimum and therefore the parabola takes on any value greater than 5. Since the Range contains all possible y- values the Range is [-5 , \infty)

Easy, wasn’t it?

Tip: When using the above Range of a Parabola Calculator for 3x^2-6x-2

we must enter the 3 coefficients a,b,c as a=3, b=-6 and c=-2.

Get it now? Try the above Range of a Parabola Calculator again if needed.

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]]>Just enter your Function and press the “Calculate Domain and Range” button. The Domain and Range will be displayed in a new window.

1) The Domain is defined as the set of all possible x-values that can be plugged into a function.

2) The Range of a function is defined as the set of all resulting y values.

1) The Domain is defined as the set of x-values that can be plugged into a function. Here, we can only plug in x-values greater or equal to 3 into the square root function avoiding the content of a square root to be negative.

Thus, domain is x>=3 .

Using Interval Notation we write: [3,\infty )

2) The Range of a function is defined as the set of all resulting y values. Here, the lowest y coordinate is y=0 achieved when x=3 is plugged in. The larger the x value plugged in the larger the y coordinate we obtain.

Thus, the range is y>=0 .

Using Interval Notation we write: [0,\infty )

1) The Domain are the x-values going left (from the smallest x-value) to right (to the largest x-value).

2) The Range are the y-values going from lowest (from the smallest y-value) to highest (to the largest y-value).

1) The Domain is all real numbers. Any number can be plugged into y=6.

2) The Range is just y=6. The lowest and highest y are both 6.

Domain and Range Calculator

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Enter your Numbers (separated by commas) & Press the Button.

The **Mean** in Mathematics is just another word for **Average**.

Mathematician denote the Mean using the symbol \overline{x} .

The Mean is found by adding up the given Numbers divided by the Number of Numbers given.

As a Formula:

\overline{x} = Sum of given Numbers / Number of Numbers = \frac{x1+x2+x3+..+xn}{n}

**Example:** We are given 2,4,9. Then, \overline{x} = \frac{2+4+9}{3}= \frac{15}{3} = 5

The Mean is one of 3 common ways to describe the center of a set of numbers besides the Median and the Mode.

For Instance, if 3 boys weigh 100, 110 and 150 pounds, then on Average (aka their Mean Weight) they weigh 360/3=120 Pounds. The 120 pounds can be seen as a representation of the weight of the 3 boys.

Notice that the Mean uses each boy’s weight. So if only boy is extra light or extra heavy it will have a significant impact on the Mean. On the contrary, the Median (described below) is designed to not get affected by unusual weights.

The Median is the Center in a sorted list of Numbers. In case of an odd number of numbers there will be a single center number. In case of an even number of numbers in the list the Median equals the average of the 2 center numbers.

**1. Example:** We are given 4,9,2. First we have to sort the list to get 2,4,9 . The Median is 2 because it is in the Center of this sorted list. We have an odd (3) numbers in our list.

**2. Example:** We are given 6,4,9,2. First we have to sort the list to get 2,4,6,9 . This list has an even number of numbers. Thus, the Median is 5 as the average of 4 and 6.

For Instance, if 3 boys weigh 100, 110 and 150 pounds, then their Median Weight is 110 Pounds.

The 110 pounds can be seen as a representation of the weight of the 3 boys.

Notice that the Median does not get affected by unusual weights.

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]]>**Standard Deviation of Sample ** s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^n (x - \overline{x})^2}

**Sample Standard Deviation** gives the average distance of your numbers to the mean of those numbers.

Example 1: A Standard Deviation of 0 means that the given set of numbers are the same since they don’t differ from their mean.

Example 2: A Standard Deviation of 1 means that the given set of numbers differ – on average – by 1 from their mean.

Bowling Example: A consistent bowler that bowls 110 and 90 games has a **lower **Standard Deviation than a bowler who bowls 50 and 150. While they each have a mean of 100 the 2. bowler scores varied much more from 100 than the first bowler.

Their Difference lies in the Denominators of their Formulas (for technical reasons):

When computing the Sample Standard Deviation we divide by n-1.

When computing the Population Standard Deviation we divide by n instead.

This is the formula for the Population Standard Deviation:

\sigma = \sqrt{ \frac{1}{n} \sum_{i=1}^n (x - \overline{x})^2}The 1 Standard Deviation rule refers to the **Empirical Rule** of Normal Distributions (aka Bell Curves). See image.

About 68% of the Data fall within 1 Standard Deviation of the Mean.

About 95% of the Data fall within 2 Standard Deviations of the Mean.

About 99.7% of the Data fall within 3 Standard Deviations of the Mean.

**Example:** About 95% of the US Population have an IQ between 80 and 120.

Reason: Mean IQ=100 and Standard Deviation=10. Thus, 95% Americans fall within 2 standard deviations, 2*10 = 20 of 100.

Note: This means that about 2.5% have an IQ higher than 120.

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Enter your Numbers (separated by commas) & Press the Button.

**Mean** = Average = \overline{x} = Sum of all Numbers / Number of Numbers in List.

**Median **= Center Number of Ordered List

**Mode **= The most frequent Number in List

**Range**=Largest – Smallest Number in List

** Variance of Population ** \sigma^2 = \frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2 , \quad \mu = Population Mean

**Standard Deviation of Population** \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2} , \enspace \mu = Population Mean

** Variance of Sample ** s^2= \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2 , \quad \overline{x} = Sample Mean

**Standard Deviation of Sample ** s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2} , \quad \overline{x} = Sample Mean

Let’s do an **Example with 7 numbers** (see right image)

a) To find the **Mean **we add up the 7 integers to get 80 and divide by 7 to get a Mean of 80/7 = 11.4 . Note: **Mean **is also called **Average**.

b) To find the **Median **we simply identify the center number to get a Median = 6. In case of 2 center numbers we will average them.

c) To find the **Mode **we simply identify the most common number to get a Mode = 1. Note: We may have 2 or more modes.

d) To find the **Range **we simply subtract the Minimum from the Maximum to get a **Range **= 42 – 1 =41.

e) **Outliers **are numbers that are way off.

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Next, press Solve to find the Variance.

Both Variance and Standard Deviation are Measures of Spread in Statistics.

The Variance is the Sum of the Squared Differences between the given Data and their Mean.

Taking the Square-Root of the Variance then gives the Standard Deviation. The Standard Deviation tells us by how much the data differ from the mean, the average distance from the mean. As formulas:

Say we are given 2,3,4 . Their mean is 3 since {2+3+4 \over 3} = 3 .

The Squared Differences between the data and mean are

(2-3)^2 + (3-3)^2 + (4-3)^2 = 1+0+1=2 .

Dividing that by the number of data points, here 3, yields s^2=2/3=0.666 as the variance.

Finally, take the Square Root of the Variance to find the Standard Deviation s= \surd (0.666)=0.81649658092 .

To find Variance we have to Square the Standard Deviation.

Reason: Standard Deviation is s , the Variance is s^2 .

To find Standard Deviation we have to take the Square Root of the Variance.

Reason: The Variance is s^2 , the Square of the Standard Deviation is s .

When computing Population Variance we divide by the Population Size N (as shown in the image above).

When computing Sample Variance we divide by the Sample Size N-1 instead (for technical reasons) .

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]]>Next, press the button to solve the equation with steps.

The Discriminant is the red part of the Quadratic Equation Solution Formula below:

What is the Discriminant of 2x^2+5x-3 ?

The coefficients are a=2, b=5 and c=-3.

We plug those into the formula for Discriminant D= b^2-4*a*c

to get D=5^2-4*2*(-3)

which simplifies to

25 - 8*(-3) = 25 + 24 = 49

implying the Discriminant is D=49.

Using the above Discriminant Calculator to solve 2x^2+5x-3=0 we must enter the coefficients a=2, b=5 and c=-3.

With steps we see the Discriminant is D=49 .

Get it now? Try the above Discriminant Calculator a few more times.

The Discriminant D= b^2-4*a*c is part of the Quadratic Equation, it is the part inside the square root.

The Discriminant ‘discriminates’ or ‘distinguishes’ 3 different types of solutions to the Quadratic Equation.

1) If the Discriminant D is greater than 0 then we can take the square root and we will have 2 real solutions.

2) If the Discriminant D is equal to 0 then we can take the square root of 0 and we will have 1 real solutions.

3) If the Discriminant D is less than 0 then we can take the square root of a negative number and we will have 2 complex solutions.

In the above Example we had a Discriminant D=49. Since it is a positive number we know that our Quadratic Equation 2x^2+5x-3=0 has 2 real (non-complex) Solutions. What you take from this is that a Discriminant tells you the type of solution our Quadratic Equation WITHOUT given you the actual Solutions. Those Solutions can be found using our handy Quadratic Equation Solver at: https://tinspireapps.com/OnlineCalculators/solve-quadratic-equation/

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]]>Enter the 3 coefficients of the quadratic equation in the above boxes.

Next, press the button to solve the equation with steps.

x^2+bx+c

In Factored Form it looks like this:

(x+r)*(x+s) where r,s are the 2 Zeros.

When distributing we get:

x^2+2 r s + r s

Matching the Coefficients

x^2+bx+c = x^2 + (r + s) x + r s

shows that the 2 Zeros r and s have to fulfill the 2 conditions:

1) r+s = b and

2) r*s = c

In Words:

1) r and s have to add to the value of the middle coefficient b.

2) r and s multiplied have to equal the constant coefficient c.

We first divide the entire equation by A to get:

x^2+(B/A)x+C/A = 0

Setting b=B/A and c=C/A we rewrite as

x^2+bx+c=0

The Factored Form looks like this:

(x+r)*(x+s) = 0 – r,s are the 2 Zeros.

Distributing terms we get

(x^2+2 r s + r s) = 0

We again Match the Coefficients:

x^2+bx+c = x^2 + (r + s) x + r s

It shows that the 2 Zeros r and s have to fulfill these 2 conditions:

1) r+s = b = B/A and

2) r s = c = C/A

In Words:

The 2 zeros r and s have to add to b = B/A.

And when multiplied equal c = C/A.

See below’s examples.

**1) Factor Quadratic Equations with Leading coefficient A = 1 **

We are to factor the Quadratic Equation

x^2- 6x+8 .

The 2 zeros when multiplied have to equal 8.

That could be 8 and 1 OR 4 and 2, and their negatives.

Additionally, they have to add to -6 which implies

the 2 zeros must be -4 and -2.

Therefore, the factored version is:

x^2- 6x+8 = (x-4)*(x-2) .

When asked to solve the Quadratic Equation

x^2- 6x+8=0 .

we use the above factored version and set each factor equal to 0:

Since x-4=0 we get x=4 ,

and since x-2=0 we get x=2 .

Thus, the 2 zeros are x=4 , x=2

Easy, wasn’t it?

Tip: When using the above Factor Quadratic Equation Solver to factor

x^2-6x+8 we must enter the 3 coefficients as

a=1, b=-6 and c=8.

** 2) Factor Quadratic Equations when A \ne 1 **

We are to factor the Quadratic Equation

2x^2- 12x+16 .

First divide by 2 to have a leading coefficient coefficient of A=1.

We get x^2- 6x+8 as we had in the above example.

Since

x^2- 6x+8 = (x-4)*(x-2)

we multiply by A=2 to get

2x^2- 12x+16 = 2*(x-4)*(x-2)

as the factored form.

Tip: When using the above Factor Quadratic Equation Solver to factor

2x^2-12x+16

we must enter the 3 coefficients a,b,c as

a=2, b=-12 and c=16.

This Video gives a great explanation on how to factor quadratic equations when the leading coefficient is not 1:

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]]>The post Vertex Form in 5 Minutes: <br> What is Vertex Form? How do I find it? appeared first on Mikes Calculators with Steps.

]]>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 .

Next, press the button to find the Vertex and Vertex Form with Steps.

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) .

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.

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) .

a

a

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

See the illustration below of the two possible Vertex options:

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.

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