Examples Functions in Math

Functions in math deals with finding unknown variable from the given expression with the help of known values. In algebraic expression the variable are represented in alphabetic letters. Functions in math, the numbers are consider as constants. Algebraic expression deals with real number, complex number, and polynomials. In algebraic expression several identities to find the x values by using this we can easily find the algebraic expression of the particular function. The example function in math  may include the function of p(x), q(x),… to find the x value of the functions.

Examples function in math:

Q(y) = 4y2+12y + 40. In this equation we need to find the variable of Q(2) functions in math her y is 2.

Problems using examples functions in math

Examples functions in math

Problem 1: Examples functions in math using p(y) = y2 +2y +4

p(y) = y2 +2y +4 find the f(6).

Solution :

Given the function of p(y) there is y value is given function in math

p(y) = y2 +2y +4 find the p(6)

The value of x is 2 is given

p(6) = 62 +2*6 +4

p(6) = 36 +12 +4 In this step 6 square is 36 it is calculate and 2*6 is 12 be added

p(6) = 52.

The functions in math p(6) = y2 +2y +4 find the p(6) is 52.

Problem 2 Examples functions in math using q(x) = x2 +2x +40 find the q(3).

q(x) = x2 +2x +40 find the q(3).

Solution :

Given the examples functions in math of q(x) there is x value is given functions in math

q(x) = x2 +2x +40 find the f(3)

The value of x is 2 is given

q(3) = 32 +2*3 +40

q(3) = 9 + 6 +40 In this step 3 square is 9 it is calculate with 2*3 is 6 be added to 40 to find the example function in math q(3).

q(3) = 55.

The examples functions in math q(3) = x2 +2x +40 find the q(3) is 55

Example functions in math using cubic equation

Problems1: examples functions in math using f(x) = x3 +2x2 + 2x + 4 to find the f(3).

F(x) = x3 +2x2 + 2x + 4 find the functions in math f(3).

Solution

Given the function in math of  f(x) there is x value is given as 3. Find example function in math.

f(x) = x3 +2x2 + 2x + 4 find the f(3).

Here the value of x is given as 3

f(3) = 33 + 2*32 + 2*3 +4

f(3) = 27 +18+ +6 +4 In this step 3cube is calculated  as 27 and  3square is 9.

f(3) = 55.

The math functions of f(x) = x3 +2x2 + 2x + 4 find the f(3) = 55.

Problems 2: Examples functions in math using q(x) = x3 +2x2 + 4x + 25 to find the q(3).

q(x) = x3 +2x2 + 4x + 25 find the functions in math q(3).

q(x) = x3 +2x2 + 4x + 25 find the math function in q(3).

Solution Given the examples functions in math of  q(x) there is x value is given as 3. Find example function in math.

q(x) = x3 +2x2 + 2x + 25 find the f(3).

Here the value of x is given as 3

q(3) = 33 + 2*32 + 2*3 +25

q(3) = 27 +18+ 6 + 25 In this step 3cube is calculated  as 27 and  3 square is 9 multiplied with 2 and add 25 to find the example function in math of q(3).

q(3) = 76 .

The examples functions in math of q(x) = x3 +2x2 + 2x + 25  is the q(3) = 76.

Length of Line Segment

The line segment is the straight line and it has two points. They are starting point and ending point. The starting point is in the starting place of the line and the ending point is in the ending place of the line. The length of line segment is the distance between the starting point and ending point of a line.

Diagram and formula - Length of line segment:

The formula to find the length of the line segment with two points (x1,y1) and (x2, y2) is

Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Example problems - Length of line segment:

Find the length of line segment and the points are (1,1) and (2, 2).

Solution:

Given , (1,1) and (2, 2).

Let us take (1 ,1 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-1)^2 + (2-1)^2)`

=`sqrt(1^2 + 1^2)`

= `sqrt(1 + 1)`

= `sqrt(2)`

Find the length of line segment and the points are (1,1) and (3, 3).

Solution:

Given , (1,1) and (3, 3).

Let us take (1 ,1 ) as (x1, y1) and (3, 3) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((3-1)^2 + (3-1)^2)`

= `sqrt(2^2 + 2^2)`

= `sqrt(4+4)`

= `sqrt(8)` .

= `sqrt(4 * 2)`

= `sqrt(4)` `xx` `sqrt(2)`

= 2 `xx` `sqrt(2)` .

Find the length of line segment and the points are (3,3) and (2, 2).

Solution:

Given , (3,3) and (2, 2).

Let us take (3 ,3 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-3)^2 +(2-3)^2)`

= `sqrt((-1)^2 + (-1)^2)`

= `sqrt(1 + 1)`

= `sqrt(2)` .

Find the length of line segment and the points are (2,2) and (5, 5).

Solution:

Given , (2,2) and (5, 5).

Let us take (2 ,2 ) as (x1, y1) and (5, 5) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((5-2)^2 + (5-2)^2)`

= `sqrt(3^2 + 3^2)`

= `sqrt(9+9)`

= `sqrt(18)`

= `sqrt(9 * 2)`

= `sqrt(9)` `xx` `sqrt(2)`

= 3 `xx` `sqrt(2)`

Mean Difference Standard Deviation

Definition of mean difference:

Mean difference is defined as the measure of the difference between the given data set and mean value.For finding the mean difference first have to find the mean,

Formula for finding the mean,

`barx = (sumx) / n`

Using the mean value mean difference have to be found Formula for mean difference is,

`x-barx`

Definition of Standard Deviation:

Standard Deviation is the determination of describing the variability and spread of the Data set in the given total values in data set. It is used to take the measurement for the average of numbers in the given Data set. Standard Deviation given by the square root for the summation of the total squared mean difference and it is divided by the total number of values minus one.

Formula for standard deviation,

S =` sqrt(((sum(x - barx))) / (n-1))`

Steps for calculating mean difference and standard deviation:

Get the mean for the given n numbers in the given data set.
Get mean difference of each given numbers in the Data set from the mean.
Take Square for all each deviations. It is called as the squared mean deviation.
Calculate the summation for standard  mean deviations.
Now apply the Standard Deviation formula for finding the Standard deviation form the mean.


Mean difference standard Deviation - Example Problems:

Mean difference standard Deviation - Problem1:

Calculate the mean difference and standard deviation in the following data set.56, 52, 54, 57, 58.

Solution:

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (56+ 52+ 54+ 57+ 58) / 5`        

`barx = 277 / 5`

`barx = 55.4`

Mean difference is given below

x                             `(x - barx)`

56                         56 - 55.4 =   0.6

52                         52 - 55.4 =  -3.4

54                        54 - 55.4 =   -1.4

57                        57 - 55.4 =    1.6

58                        58 - 55.4 =    2.6


Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.6)^2+(-3.4)^2+(-1.4)^2+(1.6)^2+(2.6)^2 )/ (5-1))`

S = `sqrt(23.2 / 4)`

S = `sqrt(5.8)`

S = 2.40831892

Mean difference standard Deviation - Problem 2:

Calculate the mean difference and standard Deviation for the given data set. 23, 25, 24, 26.

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (23+ 25+ 24+ 26) /4`        

`barx = 98 / 4`

`barx = 24.5`

Mean difference is given below

x                                  `(x-barx)`

23                        23 - 24.5 =   0.5

25                        25 - 24.5 = - 0.5

24                        24 - 24.5 =   1.5

26                        26 - 24.5 = - 1.5

Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.5)^2+(-0.5)^2+(1.5)^2+(-1.5)^2)/ (4-1))`

S = `sqrt(5 /3)`

S = `sqrt(1.66666667)`



S = 1.29099445

Finding Orthocenter

Orthocentre is point of intersection of all the three altitudes of a triangle.

The altitude of a triangle is the line perpendicular from one vertex to the opposite side.

Steps to find the orthocentre of triangle ABC algebrically :-

1) Find the equation of any two altitudes of the triangle

The slope of the line joining the opposite side BC  is calculated ,

Then,  the perpendicular slope is then calculated.

slope of a line is `(y2 - y1 )/(x2-x1)` =  m ,

perpendicular slope is  `(-1)/(m)`

Given are vertex A and slope of line,

we plug the values in y = mx +b we get the y intercept b

The equation of the altitude is y = mx + b

Same way we find the equation of the other altitude

2) Find point of intersection of the two altitudes is the orthocentre of the triangle.

Steps to find the orthocentre of a triangle graphically:

1) First we plot all the points of the triangle

2) We draw two altitudes of the trianlge

3) The point of intersection of the altitudes is the Orthocentre

Examples of Solving Orthocentre

Find the orthocentre of the of the triangle ABC whose vertices are A(-2,-1) , B ( -1,-4) and C (0,-5) .

Solution:    {i) Equation of altitude AD which is perpendicular to BC

x 2 = -1, y2 = -4 , x 1 = 0 , y1 = -5

slope of a line BC is  `(-4 +5)/(-1-0)` =  `(1)/(-1)`  =  -1

slope of line AD is `(-1)/(-1)` =  1

Line AD passes through A(-2,-1) and slope m = 1,

Equation of line is y = 1x +b

we plug in x = -2, y = -1,  we get,

-1 = -2 + b

-1 +2  = b

b = 1

Equation of AD is   y = x + 1 ------------------> 1
(ii) Equation of Altitude BE perpendicular to AC

x2 = 0, y2 = -5, x1 = -2, y1 = -1

slope of AC =  `(-5+1)/(0+2)` = `(-4)/(2)`  =  -2

Perpendicular slope is  `(-1)/(-2)`  =  `1/(1/2)`  = slope of BE

Equation of BE is y = `(1)/(2)` x + b

B (-1,-4)  Plug in x = -1, y = -4              ,

-4   = -1/2 +b

-4 + 1/2 = b ,  b = 7/2

Altitude BE is        y = `(1)/(2)` x + `(7)/(2)`  ---------------> 2

(iii) Point of intersection of AD , BE

Solve 1 and 2,    x +1  = `(1)/(2)` x + `(7)/(2)`

we get (-9,-8)

Solving Orthocentre of Different Types of Triangles using Graphs

Orthocentre of a right angled triangle is the vertex which is the right angle.

Orthocentre of a an obtuse angled triangle is  outside the triangle.

For an equilateral triangle , the orthocenter lies on the perpendicular bisector of each side of the triangle.

Some graphs to illustrate the above facts:-

Does Geometry use Integers

The geometry generally use the integers and the variables. The answer for the question "Does geometry use integers?" is yes, the geometry uses the integers.For specific purpose of using the geometry with integers, the separate topic available as algebraic geometry. The integers are used for to represent the co-ordinates in vertex, some equations of the line in the 2d geometry. The examples and practice problems are given below for the question "Does geometry use integers or not?".

some examples to explain "does geometry use integers"

• Consider the given points  (4,4), (2,3), (8,3), (6, 2 ), (5, 1) and plot them in the graph. And also we have to denote the quadrant in which each of the point lies.

Solution:

The plotted points in the coordinate plane are shown in the graph. All the co-ordinates of the points are in positive so all will be in the first quadrant itself. And also that  the co-ordinates of the points are not having the zero term, they are all having only the non-zero values. Therefore the points are not lie on the x-axis and the y-axis.

• Consider the points (−3, −4) and (−9, 11) and find the horizontal and the vertical distances between them.

Solution:

For the points (−3, −4) and (−9, 11) The horizontal distance between the two points is a distance between the point corresponding to x coordinates−3 and −9 on the number line x- axis; i.e., (−3) − (−9) = 9 − 3 = 6. and the vertical distances between the two points is the distance between the points corresponding to y co-ordinates −4 and 11 on the number line y axis; i.e., (11) − (−4) = 15.

some more problems to explain "does geometry use integers"

• Consider the line passing through (5,6) and (15,9) and state whether the line is rising up or falling down find the slope.

Solution:

Take (5,6) as (x₁, y₁) and (15, 9) as (x₂, y₂). Then the slope of the line is

m = y₂-y₁ / x₂-x₁

  = `(9-6)/(15-5)`

  = `3/10`

The slope is a positive(+ve) number and so the line is rising up. Here the geometry is used to determine(find) the direction of the lines using the integers.

• Consider the line passing through (−16, 29) and (40, −6) and state whether the line is rising up or falling down and find the slope.
  

Solution:

The slope of the line is

m = y₂-y₁ / x₂-x₁

  = `(-6-29)/(40-(-16))`

  = `-35/56` 

  = `-5/8`

Here m is a negative number, the line will be falling down

In this problem the geometry is used to determine the direction of the lines using the integers.

Isosceles Triangle

Isosceles triangle is a three sided closed polygon with two equal sides and two equal angles.

Propeties of an isosceles triangle:

The legs are equal in the length and the third side is the base of the triangle.
The angles formed in the base side are equal in measures.i.e. they they are congruent to each other.
The angle formed by two legs of the triangle is vertex angle.

geometry isoscles triangle

Formula related related to isosceles triangle:

1. Formula to find the vertex angle of the isosceles triangle is

Sum of the angles in triangle is 180 degrees.

Here the base angle is equal.

a+a+ vertex angle= 180

Vertex angle = 180-2a.

2.Formula for the area of the isosceles triangle is `1/2 (BH)` square.units.

3.Formula for the perimeter of the isosceles triangle is sum of three sides.

Model problems to isosceles triangle

Pro 1: Find the vertex angle of the triangle whose base angle is 70 degrees?

Sol: Step 1:   Let the base angle be 70 degrees

We know that the base angles in the isosceles traingle is equal

Step 2:  Sum of the angles in triangle is 180 degrees.

Angle 1+ Angle 2+ Angle 3 = 180 degrees

x + x + vertex angle = 180

Step 3:   Vertex angle= 180- (140)

Vertex angle is 40 degrees

Vertex angle of the geometry isosceles triangle is 40 degrees.

Pro 2. Find the area of the isosceles triangle whose base side is 10cm and the height of the triangle is 3 cm.

Sol:  Step 1:   Base side, B= 10cm

Height of the triangle, H= 3cm

Step 2:  Formula for the area of the isosceles triangle is 1/2 (BH) square.units.

= `(1/2) (10 xx3)`

= `(30/2)`

= 15cm2

Area of the isosceles triangle = 15cm2

Pro 3: Find the perimeter of the isosceles triangle whose base side is 10cm and leg side is 5cm?

Sol: Step 1: We know that leg sides are equal in length.

Perimeter of the isosceles triangle  = (10+5+5)

= 20 cm

Perimeter of the isosceles triangle is 20cm

Geometry Reflections

In geometric, the determination of reflection of a point in a line translation is performed only through a given point or line. In reflection of a point in a line, the basic ideas of reflection are the transformation of a point to a reflected point that is the equal length of the opposite side of a  line. The transformation of  reflections in two right angular axes produce a rotation of straight angle (180°), that is a half turn.

Reflection of a point:

Reflection:

The reflection is the "flip or mirror image" of it over a line.  So, the actual and mirror images are identical to each other.

The two very common reflections of polygons are given by

Horizontal reflections.

Vertical reflections.

The colored vertices used for each of the triangle.  The line of reflection is halfway from both red points, blue points, and green points. The line of reflection is directly in the center of both points.

Reflection of  a point in a line to describe the X-axis is consider the mirror line or axis of reflection. Therefore, we change the given geometric object point  into x = x  and y = -y.

geometry reflections

Reflection of  a point in a line to describe the Y-axis is consider the mirror line or axis of reflection.. Therefore, we change the given geometric object point  into x = -x  and y = y.

geometry reflections

Reflection Example:

We take a shape of triangle, Learn horizontal reflection assists, the transformation of the geometric triangle of co-ordinates are P, Q, and R. It is transformed by using the horizontal reflection, that is the horizontal line y = x.

Solution:

Step 1: To draw the triangle object from the given co-ordinates are P, Q, and R in the X-Y plane.

geometry reflections

Step 2:  To draw the horizontal line or mirror line of that triangle PQR, and draw the perpendicular line to the horizontal line or mirror line from each co-ordinates of  given triangular object. And then measure the distance of each perpendicular lines are y1, y2, y3.

geometry reflections

Step 3:  To plot the mirrored points are P', Q', R', the same vertical distance from the horizontal line or mirror line.

geometry reflections

Step 4:  To join the plotted points P', Q', R'. We get the horizontally reflected triangular object.

geometry reflections

Step 5:   This is required horizontal reflections of triangular object.

Horizontal reflection of X- axis:

Horizontal reflection assist that concepts to describe the x-axis are considering the mirror line. Then change all the coordinates of the given geometric object into the x = x and y = -y. For example,

geometry reflections