linear function presentaion

y  ax  b



Linear Function

By: Avi Meshulam



August 21, 2009



Linear Function



1



by: Avi Meshulam



Methodology

• What is a linear function?



• Graphic & Algebraic description

• Variables & Parameters • The gradient



• Intersection with the axis

• The linear equation • Steps in solving a linear equation



• Parallel & Perpendicular lines

• Distances & Areas

August 21, 2009 Linear Function 2



What is a linear function?



by: Avi Meshulam



• Linear Function - a function which describes a straight line graphically or algebraically.



August 21, 2009



Linear Function



3



by: Avi Meshulam



Graphic Expression

• Representation of a linear equation by drawing it on a set of 2 axis (X,Y).

– For example: a graphic expression of



the function y=2x-3



August 21, 2009



Linear Function



4



by: Avi Meshulam



Algebraic Expression

• Describing the linear function by using an algebraic equation. • The general equation of the linear function:



Y=ax+b



– For example: the function y=2x+5.



August 21, 2009



Linear Function



5



by: Avi Meshulam



Variables & Parameters

• Let’s look at the at the general equation of the linear function:



• Variables: x,y. • Parameters: a,b.



Y=ax+b



• To learn about the meaning of the variables & parameters go to the next slide…



August 21, 2009



Linear Function



6



Variables & Parameters

(continuation)



by: Avi Meshulam



• The variables (x,y) present the x & y values of all the points which fulfill the equation. • Let’s take for example the equation y=x+7 and the point )2,9(. Is that point on the equation’s graph? • In order to answer this question, we will place the X value of the point )2( in the equation. If the answer will be the y value of the point )9(, then that point is on the equation’s graph.

August 21, 2009 Linear Function 7



Variables & Parameters

(continuation)

Y=2+7=9 • Let’s place:



by: Avi Meshulam



• We have found out that the point )2,9( fulfills the equation and therefore is a part of it.



August 21, 2009



Linear Function



8



Variables & Parameters

(continuation)

• parameters:

– a – the function’s gradient. – b – intersection point with y axis.



by: Avi Meshulam



• Further details about parameters will be given later…



August 21, 2009



Linear Function



9



Variables & Parameters

(summary)



by: Avi Meshulam



X value of the point



Y=ax+b

Y value of the point The line’s gradient intersection point with y axis



August 21, 2009



Linear Function



10



by: Avi Meshulam



The gradient

• The gradient (a) – the difference in y axis compared with the difference in x axis between 2 points.

– The gradient is also marked sometimes by the letter m. – Let’s look for example on the equation y=3x-5

• The equation’s gradient is 3.



August 21, 2009



Linear Function



11



by: Avi Meshulam



Finding the gradient

• The equation for finding the gradient is:



y y2  y1 m  x x2  x1



(x1,y1)



(x2,y2) ∆y ∆x



August 21, 2009



Linear Function



12



The gradient (continuation)

42 2 m  2 3 2 1

(2,2) ∆x



by: Avi Meshulam



• An example: let’s find the gradient between the following points: )2,2( ,)3,4(



(3,4) ∆y



August 21, 2009



Linear Function



13



by: Avi Meshulam



The gradient (continuation)

• Examples for different types of gradients:

– A positive gradient (a>0)



– A negative gradient )a<0)



August 21, 2009



Linear Function



14



by: Avi Meshulam



The gradient (continuation)

• Examples for different types of gradients (continuation):

– gradient 0 (y=b( ,(a=0)



– Undefined gradient (x=b( ,)a∞(



August 21, 2009



Linear Function



15



by: Avi Meshulam



Intersection points

• Intersection point with x axis • Intersection point with y axis

(0, y1) (x1,0)



y=0 (x1,0). (x=0) (0, y1(.



August 21, 2009



Linear Function



16



Intersection points

(continuation)



by: Avi Meshulam



• An example :let’s find the intersection points of the equation y=2x-3 with the axis:



Y=0

x=0



0=2x-3

y=2*0-3



2x=3

y=0-3



x=1.5

Y=-3



August 21, 2009



Linear Function



17



The linear equation

– )x1,y1( – a point on the equation.

– m – the equation’s gradient.



by: Avi Meshulam



• In order to find the linear equation, we will usually use y  y1  m( x  x1 ) the next equation:



• For example :a given point on a linear equation is )-2,5 (. The line’s gradient is -1. What is the linear equation? • We’ll place in the equation: y  5  1( x  ( 2)) y  5  1( x  2) y  5  x  2 y  x  2  5 y  x  3

August 21, 2009 Linear Function 18



Steps in solving a linear equation



by: Avi Meshulam



• We will implement the steps in solving a linear equation on the following example: 5  12  3

x4



• step I – range of definition – Since it is forbidden to divide by 0 ,the range of definition is: x  4  0



x4

August 21, 2009 Linear Function 19



Steps in solving a linear equation (continuation)

• step II – common denominator



by: Avi Meshulam



– Multiply the equation by a common number in order to cancel the denominator.



5  12  3 / ( x  4) x4 5  12( x  4)  3( x  4)



• step III – opening the brackets

– solve the multiplication in the brackets.



5 12x  48  3x 12

August 21, 2009 Linear Function 20



Steps in solving a linear equation (continuation)

• step IV – moving members



by: Avi Meshulam



– Gather the variables at the left member of the equation and the numbers at the right member. 12 x  3x  12  5  48 15 x  55



• step V – divide by the coefficient of X

15 x  55 / :15 55 x 15 x  3.67

August 21, 2009 Linear Function 21



by: Avi Meshulam



Parallel & perpendicular lines

• Parallel lines – lines which have the same gradient.

– For example: Y=2x+3 Y=2x-7



• Perpendicular lines – fulfill the following equation:



m1  m2  1

– For example: Y=2x+3



1 y   x3 2

August 21, 2009



 1 2      1  2

Linear Function 22



Parallel & perpendicular lines

(continuation)



by: Avi Meshulam



• A question :on a set of axis the following points are given A(-3,2) , B(1,6) .From point B coming out line BC to the X axis, so line AB creates with it a right angle. .point C is the intersection point of BC with the X axis .From point A coming down a vertical to the X axis and intersects it in point D.

– Find the linear equation of AB. – Find the gradient of BC. – Find the x,y values of point C.



– Find the area of quadrangle ABCD.

August 21, 2009 Linear Function 23



Parallel & perpendicular lines

• A solution:



by: Avi Meshulam



(continuation)



– In order to find a linear equation we need a point and a gradient . 2 points are given (A-‫ ו‬B) .We will use them to find the gradient .



2  6 4  1 3  1 4 – We place the gradient which we have just found and one of the points (in this case I choose B) in the general linear equation in order to find the equation of line AB. m



y  6  1( x  1) y  6  x 1 y  x5

August 21, 2009 Linear Function

A(-3,2)



B(1,6)



24



Parallel & perpendicular lines

• solution:



by: Avi Meshulam



(continuation)



– It is given that BC is perpendicular to AB. So, in order to find the gradient of BC ,we use the following equation :



m1  m2  1 1 m2  1 1 m2   1 1

B(1,6) A(-3,2)



August 21, 2009



Linear Function



25



Parallel & perpendicular lines

• Solution:



by: Avi Meshulam



(continuation)



– point C is the intersection point of BC with the X axis . therefore ,in order to find point C ,we will first find the linear equation of BC ,by placing point B and the gradient we found before in the following equation:



y  6  1( x  1) y  6  x 1 y  x  7

August 21, 2009 Linear Function

A(-3,2) A(-3,2)



B(1,6) B(1,6)



26



Parallel & perpendicular lines

(continuation)

• solution:



by: Avi Meshulam



– In the next step we will find point C by placing in the linear equation of BC, which we have solved before, the value y=0.



y  x  7 x  7  0  x  7 x7

A(-3,2)



B(1,6)



C(7,0)



August 21, 2009



Linear Function



27



Parallel & perpendicular lines

• solution:



by: Avi Meshulam



(continuation)



– In order to find the area of ABCD, we need to find the values of point D .it is given that a perpendicular from point A intersects the X axis in point D .meaning, points A and D have the same X value (3) .Since point D is on the X axis ,it’s values are D(-3,0).



– The area of a quadrangle is given by multiplying the base of the quadrangle by its height. The base is DC and the height is the y value of point B.

B(1,6)



DC=7-(-3)=10



A(-3,2) A(-3,2) C(7,0) C(7,0) D(-3,0)



S ABCD  10  6  60

August 21, 2009 Linear Function



28



by: Avi Meshulam



Distances & Areas

• Calculating distances – we distinct between two cases:

– Calculating distance between two points which are on the same side of the origin of axis ) 0,0).



AB=X2-X1

x1



A(x1,0)



B(x2,0)



x2



August 21, 2009



Linear Function



29



Distances & Areas (continuation)

• Calculating distances



by: Avi Meshulam



– Calculating a distance between 2 points which are on different sides of the origin of axis)0,0(.



AB=X2-X1



A(x1,0)



B(x2,0)



x1



x2



August 21, 2009



Linear Function



30



Distances & Areas (continuation)

• Calculating distances



by: Avi Meshulam



– An example :calculate the distance between the 2 points (0,2( ,)0,-3(.



AB=y1-y2

A(0,2)



AB=2-(-3)=2+3=5



y1



y2

B(0,-3)



August 21, 2009



Linear Function



31



Distances & Areas (continuation)

• Calculating areas:



by: Avi Meshulam



– Calculating a triangle’s area is by the equation :(base ah multiplied by height divided by 2(.



SV 



The base BC: AB=X3-X2 The height: h=Y1 The area of triangle ABC:

B(x2,0)



2

A(x1,y1)



C(x3,0)



SV ABC



( x3  x2 ) y1  2



August 21, 2009



Linear Function



32



Distances & Areas (continuation)

• Calculating areas:



by: Avi Meshulam



– An example :the next 2 functions are given :y=3x+4 ,y=-0.5x+2 . find the area of triangle ABC.



step I :match the function to the graph



A(x1,y1)



B(x2,0)



C(x3,0)



August 21, 2009



Linear Function



33



• Calculating areas:



Distances & Areas (continuation)



by: Avi Meshulam



– step II :find the values of points A,B,C.

• point A is the intersection point between the 2 functions ,so ,in order to find that point, we compare the equations.



3x  4  0.5 x  2 3x  0.5 x  2  4 3.5 x  2 2 4 x  3.5 7

12 5 2  4 y  3      4    4  1  4  2 7 7 7  7

August 21, 2009 Linear Function



2 A(x  4 ,y )  A   1 , 21   7 7

B(x2,0) C(x3,0)



34



• Calculating areas:



Distances & Areas (continuation)



by: Avi Meshulam



– step II :find the values of points A,B,C.

• points B and C are the intersection points of both functions with X axis (y=0).

point B:



3x  4  0 3 x  4 x 4 1  1 3 3



 4 2 A  , 2   7 7

B(x2,0) C(x3,0)



C (4, 0)



point C:



1 x20 2 1  x  2 / ( 2) 2 x4 



 1  B  1 , 0   3 



August 21, 2009



Linear Function



35



• Calculating areas:



Distances & Areas (continuation)



by: Avi Meshulam



Finding the height:



– step III :find the base & height of the triangle. 1 1 • finding base (BC):  1 BC  4   1   4  1  5 3 3  3



2 h2 7



- step IV :calculating the area:



 4 2 A  , 2   7 7



C (4, 0)



1 2 16 16 5 2  16  8 128 2 16 SV ABC  3 7  3 7   6 2 2 3  7  2 21 21

1



 1  B  1 , 0   3 



August 21, 2009



Linear Function



36



by: Avi Meshulam



August 21, 2009



Linear Function



37