Linear functions

Formulas for working with lines 

 
    y2 - y2
m =        
     x2 - x1
 Slope 

y = mx + b
 Slope-intercept form 
     
 y - y1 = m(x - x1)
Point-slope form 
     
  y = constant
Horizontal lines 
     
  x = constant
Vertical lines 
     
   m2 = m1
Parallel lines 

    -1
 m2 =    
     m1
Perpendicular lines

Example:     Determine the equation of a line passing through points (5,2) and (6,4). 
Let (x1,y1)=(5,2), (x2,y2)=(6,4) and compute the slope
m = (y2 - y1)/(x2 - x1) = (4-2)/(6-5) = 2
and the equation of the line
y - y1 = m(x - x1)
y - 2 = 2(x-5) OR 
y = 2x - 8
Example:     Determine the equation of the horizontal line passing through the point (5,2). 
The equation of a horizontal line is of the form y = constant.
The equation of this horizontal line is y = 2.
Example:     Determine the equation of the vertical line passing through the point (5,2). 
The equation of a vertical line is of the form x = constant.
The equation of this vertical line is x = 5.
Example:     Determine equation of a line parallel to the line 4x + 2y = 5
and passing through the point (1,2). 
Solving the equation 4x + 2y = 5 for y yields 
y = -2x + 5/2. The slope of this line is -2.  
Since we wish to find the equation of a parallel line 
we use the same slope of -2.  To find the equation 
of the parallel line substitute m = -2 and (x1,y1) = (1,2)
into the equation y - y1 = m(x - x1) to obtain
y - 2 = -2(x-1) OR y = -2x + 4.
Example:     Determine equation of a line perpendicular to the line 4x + 2y = 5
and passing through the point (1,2). 
Solving the equation 4x + 2y = 5 for y yields 
y = -2x + 5/2. The slope of this line is -2.  
Since we wish to find the equation of a perpendicular 
line we use the negative reciprocal m2 = -1/m1 = 1/2.  
To find the equation of the perpendicular line substitute 
m = 1/2 and (x1,y1) = (1,2) into the equation 
y - y1 = m(x - x1) to obtain y - 2 = 1/2(x-1) OR 
y = 1/2x + 3/2.

Linear functions

Linear functions have the form 
f(x) = mx + b
The graph of a linear function is a straight line with slope m and y-intercept b.
Linear function

Business applications 

Linear Cost, Revenue and Profit Model 
variable cost = (cost per item) × (number of items) 
cost = fixed cost + variable cost
revenue = (price per item) × (number of items) 
Compute the break-even point by setting R(x) = C(x).
Example:     A business has a monthly fixed cost of $100,000 and a production cost
of $14 per item. The product sells for $20 per item. Find the cost, revenue
and profit functions and determine the production level needed to break even. 
Let x = the number of items that are produced and sold.
C(x) = 14x + 100,000
R(x) = 20x
P(x) = R(x) - C(x) = 7x - 100,000
To compute the break even point set R(x) = C(x)
20x = 14x+100,000
7x = 100,000
 x = 14,285.7

Exercises 




(1) Compute the slope of the line passing through the 
    points (-3,6) and (2,-4).

     
-2 

     
-1/2 

     
0

     
1/2 


 



(2) Compute the equation of the line passing through the 
    points (-3,6) and (2,-4). 

     
y = 2x + 12 

     
y = -2x + 5 

     
y = -2x -12 

     
y = -2x 






 (3) The graph of f(x) = 2x - 3 could be

Linear function graph

     a) 

     
b) 

     
c) 

     
d) 


 



(4) Find the equation of the horizontal line passing through 
    the point (3,5). 

     
x = 3 

     
x = 5 

     
y = 3 

     
y = 5 


 



(5) Find the equation of the vertical line passing through 
    the point (3,5). 

     
x = 3 

     
x = 5 

     
y = 3 

     
y = 5 



 (6) A manufacturer has a monthly fixed cost of $200,000 and a 
    production cost of $20 per unit.  The product sells for $25
    per unit.  Determine the profit function. 

     
20x + 200,000 

     
20x - 200,000 

     
5x - 200,000 

     
5x + 200,000 





|  Table of Contents  |
|  Previous  |
|  Next  |