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MATH 201 - Week 2
Ferenc Balogh
Concordia University
2008 Winter
Based on the textbook
J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson
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Lines (Chapter 1, Section 1.10)
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Slope of a Line
Equations of Lines
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Parallel and Perpendicular Lines
Slope as Rate of Change
What is a Function? (Chapter 2, Section 1)
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Point-Slope Form
Slope-Intercept Form
General Equation
Definition of Function
Evaluation of a Function
The Domain of a Function
Graphs of Functions (Chapter 2, Section 2)
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Graphing Functions
Graphing Piecewise Defined Functions
The Vertical Line Test
Equations That Define Functions
The slope gives the ”steepness” of a straight line as we move from
the left to the right.
It is defined as
rise
.
slope =
run
The Slope of a Line
If P(x1 , y1 ) and Q(x2 , y2 ) are on the
line l, then the slope of l is given by
m=
y2 − y1
x2 − x1
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The slope can be positive, negative or zero.
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Important: The slope of a vertical line is not defined!
Problem. Find the slope of the line that passes through the points
P(2, −1) and Q(4, 2).
Solution.
x1 = 2, y1 = −1, x2 = 4, y2 = 2.
m=
3
2 − (−1)
=
4−2
2
Question. Suppose that we are given a point P(x0 , y0 ) in the
plane and the slope m.
What is the equation of the line which passes through P and has
slope m?
Answer. The moving point Q(x, y ) is on the line if and only if
m=
y − y0
.
x − x0
The point-slope form
If m is the slope of a line and it passes through P(x0 , y0 ) then the
equation of the line is
m(x − x0 ) = y − y0 .
Problem. Find the equation of the line which passes through
P(1, 1) and has slope 2.
Solution. Using the point-slope form of
the equation,
2(x − 1) = y − 1
2x − 2 = y − 1
2x − y − 1 = 0
So the equation of the line is
2x − y − 1 = 0.
Problem. Find an equation of the line through the points
P(−2, 1) and Q(3, −1).
Solution. We calculate the slope of the
line first:
m=
−2
2
(−1) − 1
=
=− .
3 − (−2)
3+2
5
Now the point-slope form gives
2
− (x − (−2)) = y − 1
5
−2(x + 2) = 5(y − 1)
−2x − 4 = 5y − 5
0 = 2x + 5y − 1
So the equation of the line is 2x + 5y − 1 = 0.
Question. Suppose that we know the y -intercept of a line and its
slope m.
What is the equation of the line?
Answer. The line intersects the y -axis at the point P(0, b). So we
can use the point-slope form of the equation:
m(x − 0) = y − b.
This gives
The Slope-Intercept Form
If a line has y -intercept b its slope is m then
y = mx + b
is an equation of the line.
Problem. Find an equation of the line with slope 2 and
y -intercept −3.
Solution. The slope-intercept form of
the equation is
y = 2x − 3.
So the equation of the line is 2x − y − 3 = 0.
Equation of a horizontal line: y = a.
Equation of a vertical line: x = b.
Problem. Find equations of the
horizontal line and the vertical
line through the point R 2, 34 .
Solution. The horizontal line is given
by
3
y= .
4
The vertical line is given by
x = 2.
A linear equation is of the form
Ax + By + C = 0
where A, B, C are constants and A and B are not both 0.
Examples.
A non-vertical line has an equation of the form
mx − y + b = 0
(A = m, B = −1, C = b).
A vertical line has an equation of the form
x −a=0
(A = 1, B = 0, C = −a).
Conversely, the graph of every linear equation
Ax + By + C = 0
is a line:
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If B 6= 0 then the equation is
C
A
y =− x−
B
B
which is the equation of a non-vertical line with slope −A/B
and y -intercept −C /B.
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If B = 0 then A 6= 0 and the equation is
x =−
C
A
which is the equation of a vertical line.
General Equation of a Line
The graph of every linear equation
Ax + By + C = 0
(where A and B are not both 0) is a line.
Conversely, every line is a graph of a linear equation.
ALGEBRA
GEOMETRY
Problem. Sketch the graph of the linear equation x − 2y + 3 = 0.
Solution. The x-intercept of the line is
given by the equation
x −2·0+3 = 0
x
= −3
The y -intercept is given by
0 − 2y + 3 = 0
−2y
= −3
3
y =
2
The line is passing through the points (−3, 0) and 0, 32 .
Problem. Sketch the graph of the linear equation
5x + 2y − 1 = 0.
Solution. The second way is to use the
slope-intercept form:
5x + 2y − 1 = 0
2y
y
= −5x + 1
1
5
= − x+ .
2
2
Parallel Lines
Two non-vertical lines are parallel if and only if their slopes are the
same.
Note that two vertical lines are always parallel.
Perpendicular Lines
Two lines with slopes m1 and m2 are perpendicular if and only if
m1 m2 = −1.
Note that a horizontal line is always perpendicular to a vertical one.
Problem. Find an equation of the line through the point T (−1, 2)
which is parallel to the line 3x − 2y + 4 = 0.
Solution. The slope-intercept form of the equation of the given
line:
2y
y
= 3x + 4
3
=
x + 2.
2
The slope of this line is 32 . The parallel line has the same slope
therefore the point-slope form of its equation gives
3
(x − (−1)) = y − 2
2
3(x + 1) = 2(y − 2)
3x + 3 = 2y − 4
3x − 2y + 7 = 0.
Problem. Show that the points P(1, 0), Q(3, 1) and R(1, 5) are
the vertices of a right triangle.
Solution. The slope of the line passing
through P and Q is
m1 =
1−0
1
= .
3−1
2
The slope of the line passing through Q
and R is
m2 =
4
5−1
= − = −2.
1−3
2
The product of the slopes is
1
m1 m2 = (−2) = −1.
2
We conclude that the triangle with vertices P, Q and R is a right
triangle with right angle at Q.
Problem. Find an equation of the line that is perpendicular to the
line given by the equation 2x − y + 1 = 0 and passes through the
point P(−1, −2).
Solution. The slope-intercept form of the equation of the given
line:
y = 2x + 1
The slope m2 of the yet unknown line is given by 2m1 = −1 so
1
m2 = − .
2
The point-slope form provides the equation of the perpendicular
line:
1
− (x − (−1)) = y − (−2)
2
−(x + 1) = 2(y + 2)
−x − 1 = 2y + 4
0 = x + 2y + 5.
Example.
The water level in a lake is changing
according to the graph.
It is easy to see that the water level
changes linearly.
For example, it increases 4 inches in each 6 year-long period. In
other words, the rate of change is the ratio
4 inches
2 inch
=
.
6 years
3 year
The rate of change between two linearly related quantities is given
by the slope of the corresponding graph.
Problem. The tank of a car is full containing 40 litres of gas. After 572
kilometres of (monotonous) driving along the Trans-Canada Highway
from Winnipeg to Regina, the tank contains 3 litres only.
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What is the rate of change between the amount of gas consumed
and the distance driven?
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Find an equation which gives the amount of gas in terms of the
distance from Winnipeg (assuming linearity).
Solution. We can assume that the amount of gas is changing linearly in
terms of the distance (Trans-Canada highway).
The rate of change is given by the slope of that line:
m=
37 l
l
3 − 40 l
=−
≈ 0.065
572 − 0 km
572 km
km
Using the slope-intercept form, we have the equation
y =−
37
x + 40.
572
Examples.
temperature as a function of time
source: www.theweathernetwork.com
postage as a function of weight
source: www.canadapost.ca
Function
Let A and B denote two sets.
A function from the set A to the function B is a rule that assigns
to each element x in A exactly one element, called f (x) in B.
We write
f : A → B.
Some terminology for a function f : A → B
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f (x) is called the value of f at x or the image of x under f .
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The set A is called the domain of the function f .
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The range of the function f is the set of all possible values of
f (x) as x runs through the domain of f :
Range(f ) = {f (x) | x ∈ A}.
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The symbol that represents an element in the domain of f is
called an independent variable.
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The symbol that represents an element in the range of f is
called a dependent variable.
For example, if one writes
v = f (u),
then u is the independent variable, v is the dependent variable.
Problem. Let f (x) = 2x 2 − 4x +3. √
Calculate f (2), f (−3), f (0), f 53 , f ( 2), f (a), f (b + 2).
Solution.
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f (2) = 2 · 22 − 4 · 2 + 3 = 8 − 8 + 3 = 3
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f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33
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f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3
2
18
18−60+75
33
f 35 = 2 · 35 − 4 · 35 + 3 = 25
− 12
= 25
5 +3=
25
√
√ 2
√
√
√
f ( 2) = 2 · ( 2) − 4 · ( 2) + 3 = 4 − 4 2 + 3 = 7 − 4 2
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f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3
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f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =
2b 2 + 8b + 8 − 4b − 4 + 3 = 2b 2 + 4b + 7
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Problem.
Write down a formula for the postage to
be paid for a letter of type
other lettermail
as a function p of the weight w .
source: www.canadapost.ca
Solution. The domain of the function is the interval [0, 500] (the
unit is grams). In other words, w should be between 0 and 500.

1.15 if 0 ≤ w ≤ 100
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



1.92 if 100 < w ≤ 200
p(w ) =





2.65 if 200 < w ≤ 500
The domain of the function may be stated explicitly:
In the previous example, p was defined on the domain
0 ≤ w ≤ 500.
Convention. If the domain is given by an algebraic expression and
the domain is not stated explicitly, then the domain of the function
is the the set of all real numbers for which the expression is defined.
Examples.
√
1
h(x) =
r (x) = x − 1
x −2
The function h is given by an expression which is not defined at
x = 2 (the denumerator is 0 there). So the domain of h is
{x | x 6= 2}.
The square root is not defined for negative numbers: the domain
of g is therefore {x | x ≥ 1}.
Problem. Find the domain for each function.
f (x) =
x −2
x +3
g (s) =
p
4 − s2
h(r ) = √
r2
.
r −1
Solution.
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The denumerator of f cannot be zero therefore the domain of
f is {x | x 6= −3}.
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The expression 4 − s 2 in g is under a square root so it must
be nonnegative: the domain of g is {s | − 2 ≤ s ≤ 2}.
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The expression r − 1 in h is under a square root so it must be
nonnegative and it cannot be zero. The domain of h is
{r | r > 1}.
The Graph of a Function
Let f : A → R be a function on the domain A ⊆ R.
The graph of f is the set of points
{(x, f (x)) | x in A}
in the coordinate plane.
Example.
f (x) = 4x 3 + 2x 2 − 3x + 2
Examples.
A linear function is of the form
f (x) = mx + b
where m and b are some real numbers.
The graph of this linear function is the
line y = mx + b.
As a special case, if m = 0 we have a
constant function
f (x) = b
whose graph is the horizontal line
y = b.
Problem. Sketch the graph of the function
g (x) = (x − 1)2 − 1.
Solution.
We make a table of values first for some
specific values of x:
x
0
1
−1
2
−2
g (x)
0
−1
3
0
8
(x, g (x))
(0, 0)
(1, −1)
(−1, 3)
(2, 0)
(−2, 8)
Then we plot the points and try to join
them to obtain a curve.
Problem. Sketch the graph of the function
√
w (x) = x + 1.
Solution.
We make a table of values first for some
specific values of x:
(the domain of w is {x | x ≥ −1})
x
0
−1
1
2
3
w (x)
1
√0
√2
3
2
(x, w (x))
(0, 1)
(−1,
√0)
(1, √2)
(2, 3)
(3, 2)
Then we plot the points and try to join
them to obtain a curve.
Problem. Sketch the graph of the function
−x + 2 if x < 0
f (x) =
x 2 + 1 if x ≥ 0
Solution. We expect the behaviour of f to be different for x < 0
and x ≥ 0.
x
0
−1
−2
1
2
f (x)
1
3
4
2
5
(x, f (x))
(0, 1)
(−1, 3)
(−2, 4)
(1, 2)
(2, 5)
Then we plot the points and try to join
them to obtain a curve (be careful at the
point x = 0).
Vertical Line Test
A curve in the coordinate plane is the graph of a function if and
only if no vertical line intersects it more that once.
Examples.
graph of a function
NOT graph of a function
Problem. Does the equation
2y − x 2 − 3 = 0
define y as a function of x?
Solution. Yes, because we can solve for y and it gives a unique
solution in terms of the variable x:
2y − x 2 − 3 = 0
2y
y
= x2 + 3
1 2 3
x + .
=
2
2
In other words, y is given by the function
1
3
f (x) = x 2 + .
2
2
Problem. Does the equation
2y 2 − x 2 − 3 = 0
define y as a function of x?
Solution. No, because we can solve for y and but it gives two
possible solutions in terms of the variable x:
2y 2 − x 2 − 3 = 0
2y 2 = x 2 + 3
1 2 3
x +
y2 =
2r 2
1 2 3
y = ±
x + .
2
2
Assignments:
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Lines
p. 120, # 4, 16, 20, 28, 32
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What is a Function?
p. 155, # 4, 16, 30, 44, 48
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Graphs of Functions
p. 167, #8, 10, 28, 30, 32, 34, 36, 38, 44
The assignments are due to Wednesday, January 23.