Slope of lines
A line joining two points P (x1 , y1 ) and Q(x2 , y2 ) has slope
Elementary Functions
m :=
Part 2, Polynomials
Lecture 2.0a, A Review of Linear Functions
Sam Houston State University
(If a line joins two distinct points P (x0 , y1 ) and Q(x0 , y2 ) in which the
x-coordinates are the same, then the line is the vertical line x = x0 and
we say the slope is “infinite” or “undefined.”)
2013
Elementary Functions
(1)
In the geometry of the Cartesian plane any two points determine a unique
line, so the slope of a line is a natural and important property of a line.
Dr. Ken W. Smith
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y2 − y1
∆y
=
x2 − x1
∆x
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Slope of lines
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Elementary Functions
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Equations for lines
Suppose we have a point P (x1 , y1 ) on a line of slope m. Then given any
other point Q(x, y) on the line,
Given two points P (x0 , y1 ) and Q(x0 , y2 ) on a line
the difference of y-values, written ∆y := y2 − y1 is the rise between P
and Q.
m=
The difference of x-values, written ∆x := x2 − x1 is called the run.
Thus the slope of a line is m :=
∆y
, the ratio of the rise to the run.
∆x
y − y1
.
x − x1
If we solve for y (as is our custom), then
The slope of a line identifies how quickly a line rises or falls.
It describes the number of units one rises as one moves one unit to the
right.
y − y1 = m(x − x1 )
(2)
y = m(x − x1 ) + y1
(3)
and so
For example, if the line has slope m = 4 and goes through the point
P (10, 100) then we know that
when x = 10 + 1 = 11, y = 100 + 4 = 104
Either of these equations (equation 2 or equation 3) will be called the
“point-slope” form for a line, since it is created out of a single point and
the slope.
when x = 11 + 1, y = 104 + 4 = 108
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when x = 13, y = 112
Elementary Functions
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Elementary Functions
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Linear Functions
Linear Functions
A favorite form for the equation of a line is the slope-intercept form.
We began by discussing linear equations y = a1 x + a0 .
Another equation for a line has form Ax + By = C where A, B and C are
constants associated with the line.
This form is called symmetric because it does not attempt to single out a
special variable (y) and express the equation in terms of that special
variable.
It is easy (isn’t it?) to see that a1 represents the slope of this line.
(Increase x by 1. What happens to y?)
The value a0 represents the y-intercept since it is the value of y when x is
zero.
The constants a1 and a0 are more popularly replaced by m and b and we
speak of the equation
y = mx + b
(4)
as the slope-intercept form for a line.
(It is not clear why we use the letter “m” for slope, or b for the
y-intercept.)
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Elementary Functions
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Linear Functions
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Elementary Functions
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Linear Functions
Worked exercises. Find the equation for the line of slope 4 passing
through the point (6, 20).
1 Put your answer in point-slope form.
2 Put your answer in slope-intercept form.
3 Put your answer in symmetric form.
1
2
3
In point-slope form we first write the formula for the slope using two
points, an arbitrary point (x, y) and the point (6, 20). So our answer
begin with 4 = y−20
x−6 . Clear denominators to get 4(x − 6) = y − 20.
If y = 4x + b is a line of slope 4 and (6, 20) is on the line then
20 = 4(6) + b =⇒ b = −4. Answer: y = 4x − 4.
To create the “symmetric” form for our line start with a previous
equations
4(x − 6) = y − 20
... and use algebra to put all variables on the left...
4x − 24 = y − 20
4x − y − 24 = −20
4x −Functions
y=4.
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Elementary
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In the next presentation, we continue to look at slopes of lines and
introduce the average rate of change (ARC) of a function.
(END)
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Elementary Functions
2013
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More exercises on slope
We continue to study the slope of a line, working through several
examples.
Find the equation for the line passing through the points (3, 8) and
(6, 20). Put your answer in point-slope form.
Elementary Functions
Part 2, Polynomials
Lecture 2.0b, Slope and ARC
Solution. First we find the slope of the line passing through (3, 8) and
(6, 20).
20 − 8
12
It is
=
= 4.
6−3
3
Dr. Ken W. Smith
Since the line goes through the point (6, 20) and has slope 4 then the
slope is
y − 20
4=
.
x−6
Multiplying through by the denominator gives
Sam Houston State University
2013
4(x − 6) = y − 20.
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Elementary Functions
2013
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Working some examples on slope
Elementary Functions
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Practice with slope
A related problem Find the x-intercept of the line passing through the
points (3, 8) and (6, 20).
The line y = f (x) has slope 4 and passes through the point
(12400, 999900).
Find f (12402).
Solution. Set y = 0 in the solution from the previous problem:
Let’s keep this as simple as possible.
4(x − 6) = 0 − 20
(x − 6) =
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Solution. A line of slope 4 has the property that every step to the right
creates a rise of 4.
−20
4
So 2 steps to the right (from x = 12400 to x = 12402) creates a rise of 8.
x − 6 = −5
Since we began at height 999900 we must end at 999900 + 8 = 999908.
x = 1.
Notice that we did not need to deal with any equations. Merely
understanding the meaning of slope was enough to quickly do this
problem!
The x-intercept is
(1, 0) .
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Elementary Functions
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Elementary Functions
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Practicing with slope
Practicing with slope
The graph of y = f (x) is a straight line of slope 12.
If f (gazillion) = google then what is f (gazillion + 3)?
The graph of y = f (x) is a straight line of slope 12.
Solution. f (gazillion + 3) = google + (3)(12) = google +36 .
If f (gazillion) = google then what is f (gazillion + 3)?
Easy!
Solution. f (gazillion + 3) = google + (3)(12) = google +36 .
That’s it!
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Elementary Functions
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The average rate of change of a function
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Elementary Functions
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The Average Rate of Change
Suppose two points P and Q are on the graph y = f (x) of a function f .
Since f is a function, then by the vertical line test, these two points
cannot have the same x-value.
The ARC of the function f (x) from x to x + h is the slope between the
points P (x, f (x)) and Q(x + h, f (x + h)).
Concentrate on the point P and write its coordinates as (x, f (x)).
The other point, Q has x-coordinate x + h for some value of h.
(So h is the “run” between the points P and Q.)
The coordinates of Q are Q(x + h, f (x + h)).
The slope of the line joining P to Q is
ARC :=
The slope of the line joining P to Q is
m :=
f (x + h) − f (x)
f (x + h) − f (x)
∆y
=
=
.
∆x
(x + h) − x
h
∆y
f (x + h) − f (x)
f (x + h) − f (x)
.
=
=
∆x
(x + h) − x
h
(6)
(5)
The ARC is a critical concept in calculus.
(x)
This expression, f (x+h)−f
, is the difference quotient, the slope of the
h
line connecting the points P (x, f (x)) and Q(x + h, f (x + h)).
It is the average rate of change (ARC) of the function f (x) between the
points P and Q.
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Elementary Functions
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Elementary Functions
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The ARC
The ARC
Let’s do an example or two.
Find the average rate of change of f (x) = x2 as x varies from x = 2 to
x = 5.
Worked Examples. Consider the quadratic function f (x) = x2 . Find the
difference quotient for this function.
Solution. The slope of the line through (2, f (2)) and (5, f (5))
is the slope between (2, 4) and (5, 25)
Solution. We compute
and this is equal to
f (x + h) − f (x)
(x + h)2 − x2
x2 + 2xh + h2 − x2
2xh + h2
=
=
=
= 2x + h .
h
h
h
h
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Elementary Functions
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Linear Functions
In the next presentation, we explore “quadratic” functions.
(END)
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Elementary Functions
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21
25 − 4
=
= 7.
3
5−2
Elementary Functions
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