Linear functions and Straight Lines
Linear Functions
• The equation f(x) = mx+b m and b are real
numbers is the equation of a linear
function. The domain is the set of all real
numbers. The graph of a linear function is
a straight line. Some examples of graphs
will follow in the next few slides.
1
f ( x)  x  8
4
f(x)= -2x+3
More examples
f ( x)  2
Graphing
• Graph
3
f ( x)  x  2
4
• using a table of
values for x and y
x y
-4
0
4
8
Solution:
Graphing using intercepts
• Graph 5x+6y = 30 using the x and y
intercepts:
• 1. Set x = 0 and solve for y
•
5(0) + 6y = 30
•
y=5.
• 2. Now, let y =0 and solve for x:
•
5x + 6(0) = 30, x = 6
• 3. Plot the two ordered pairs (0, 5) and (6,0) and connect the points
with a straight line.
solution
Special cases
• 1. The graph of x=k is the graph of a
vertical line k units from the y-axis.
• 2. The graph of y=k is the graph of the
horizontal line k units from the x-axis.
Some examples:
• 1. Graph x=-7
• 2. Graph y = 3
solutions
X=7
Y=3
Slope of a line
• Slope of a line:
• = rise
run
m=
 x1, y1 
y2  y1
x2  x1
Rise
 x2 , y2 
run
Slope-intercept form
• The equation y  mx  b
• is called the slope-intercept form of an
equation of a line .
• The letter m represents the slope and b
represents the y intercept.
Find the slope and intercept from
an equation of a line
• 1. Find the slope and y
intercept of the line
whose equation is
• 5x – 2y = 10
Solution: Solve the equation for y in
terms of x. Identify the coefficient
of x as the slope and the yintercept is the constant term.
Therefore: the slope is 5/2 and the y
intercept is -5
-
5 x  2 y  10
2 y  5 x  10
5 x 10 5
y
  x5
2 2 2
Point-slope form
• The point- slope form of
the equation of a line is
y  y1 
as follows:
m( x  x1 )
• It is derived from the
definition of the slope of a
line:
y y
2
1
x2  x1
m
Examples
• Find the equation of the line through the
points (-5, 7) and (4, 16) :
• Solution:
 5, 7 
(4,16)
16  7
9
m
 1
4  (5) 9
y  16  1( x  4)  y  x  4  16  x  12
Applications
• Office equipment was purchased for $20,000 and will
have a scrap value of $2,000 after 10 years. If its value
is depreciated linearly , find the linear equation that
relates value (V) in dollars to time (t) in years:
Solution: when t = 0 , V = 20,000 and when t = 10, V = 2,000. Thus, we
have two ordered pairs (0, 20,000) and (10, 2000). We find the
slope of the line using the slope formula. The y-intercept is already
known (when t = 0, V = 20,000, so the y-intercept is 20,000) . The
slope is : (2000-20,000)/(10 – 0) = -1,800.
Therefore, our equation is V(t)=
- 1,800t + 20,000