Download U1B Linear Functions and Slope Review

UNIT 1B
Review of Linear
Functions and Slope
1
Slope of a Line
The term slope is often used to describe steepness or rate of change.
The pitch of a roof, the steepness of a ski run, the speed of a
car are all examples of slope. In each case, the slope is the
ratio of the rise to the run
In a coordinate system, we can determine the slope of any line
segment from its endpoints P1(x1, y1) and P2(x2, y2).
From P1 to P2:
• the rise is the difference in the y-coordinates:
y2 – y1 or ∆y
• the run is the difference in the x-coordinates:
x2 – x1 or ∆x
2
Slope of a Line
Let 𝑃1 𝑥1 , 𝑦1 and 𝑃2 𝑥2 , 𝑦2 be points on a non vertical line, L.
The slope of L is
𝒓𝒊𝒔𝒆 𝒚𝟐 − 𝒚𝟏
𝒎=
=
𝒓𝒖𝒏 𝒙𝟐 − 𝒙𝟏
3
Slope of a Line
A line that goes uphill as x increases has a positive slope..
A line that goes downhill as x increases has a negative slope.
4
Slope of a Line
A horizontal line has slope zero since all of its points have the same
y-coordinate.
P1(-1, 1)
P2(2, 1)
𝒎=
𝟏−𝟏
𝟎
= =𝟎
𝟐 − (−𝟏) 𝟑
For vertical lines, the slope is undefined since all of its points have
the same x-coordinate
𝟐 − (−𝟏) 𝟑
𝒎=
=
𝟏. 𝟓 − 𝟏. 𝟓 𝟎
P1 (1.5, 2)
division by 0 is undefined
P2 (1.5, -1)
5
Parallel Lines
Parallel lines form equal angles with the x-axis.
Hence, non-vertical parallel lines have the same slope.
m1 = m2
6
Perpendicular Lines
If two non-vertical lines L1 and L2 are perpendicular, their
slopes m1 and m2 satisfy m1m2 = – 1, so each slope is the
negative reciprocal of the other:
𝟑
Line 1 Slope = 𝟏 = 𝟑
3
1
× − = −1
1
3
Line 2 Slope = −
𝟏
𝟑
7
A(2, 1), B(5, 3)
𝟏 − 𝟑 −𝟐 𝟐
𝒎=
=
=
𝟐 − 𝟓 −𝟑 𝟑
Run = 3
Run = 3
Rise = 2
C(– 2, 2), D(1, 4)
Rise = 2
𝟐−𝟒
−𝟐 𝟐
𝒎=
=
=
−𝟐 − 𝟏 −𝟑 𝟑
Lines that are higher
on the right have
a positive slope.
Lines that have equal slopes are parallel
AB // CD and EF // GH
8
E(– 3 , 4), F(– 1 , – 2)
𝟒 − (−𝟐)
𝟔
𝒎=
=
= −𝟑
−𝟑 − (−𝟏) −𝟐
Run = – 1
Rise = 3
G(0, 5), H(1, 2)
Rise = – 6
𝟓−𝟐
𝟑
𝒎=
=
= −𝟑
𝟎 − 𝟏 −𝟏
Run = 2
Lines that are higher
on the left have
a negative slope.
Lines that have equal slopes are parallel
AB // CD and EF // GH
9
J(3 , –2), K(6 , – 4)
−𝟐 − (−𝟒)
𝟐
𝟐
𝒎=
=
=−
𝟑−𝟔
−𝟑
𝟑
L(4, –3), M(2, –6)
Run = –3
−𝟑 − (−𝟔) 𝟑
𝒎=
=
𝟒−𝟐
𝟐
Run = 2
Rise = 2
Rise = 3
𝟐 𝟑
− × = −𝟏
𝟑 𝟐
If the slopes of 2 lines are negative reciprocals
(product = – 1) they are perpendicular.
JK ┴ LM
10
R(– 2 , 4), S(5 , 4)
𝟒−𝟒
𝟎
𝒎=
=
=𝟎
−𝟐 − 𝟓 −𝟕
P(– 3, – 2 ), Q(– 3, 3)
−𝟐 − 𝟑
−𝟔
𝒎=
=
−𝟑 − (−𝟑)
𝟎
÷ by 0 is undefined
Horizontal lines will always have a slope of zero
Vertical lines will always have a slope which is undefined.
They are also perpendicular
11
QUESTION 9: Two students entered a car rally. During part of the
rally, they had to drive at a constant speed. The following graph
shows the distance traveled over a given time while traveling at this
constant speed.
P2
What is the slope of
the line and what
does it represent?
Distance (km)
P1
Time (hours)
Slope = (180 km – 60 km) = 60 km/h
(3 h – 1 h)
It represents the velocity of the car.
12
QUESTION 10
The pool at a fitness club is being drained. The graph shows the
number of kilolitres of water remaining after an elapsed time.
What
isthe
the
intercept
What
slope
of along
thealong
c)a)b)
What
isisthe
intercept
the
theand
vertical
axis
and
whatthis
line
what
does
it does
horizontal
axis
and
what
does this
intercept represent?
represent?
intercept
represent?
The horizontal intercept is (400, 0)
The vertical intercept is (0, 100)
It takes 400 min to drain the pool.
The pool has 100 kL in it
before it is drained.
Slope = (40 kL – 100kL)
(240 min – 0 min)
= – 0.25kL/min
The pool is draining at a rate of 0.25 kL/min
13
Equations of Lines
x
y
–4
3
–2
3
0
3
2
3
4
3
The horizontal line
through the point (2, 3)
has equation y = 3
x
y
2
–4
2
–2
2
0
2
2
2
4
The vertical line
through the point
(2, 3) has equation
x=2
The vertical line through the point (a, b) has equation x = a
since every x-coordinate on the line has the same value a.
The horizontal line through (a, b) has equation y = b since
every y-coordinate on the line has the same value b.
Finding Equations of
Vertical and Horizontal Lines
Horizontal Line is y = 8
Vertical Line is x = – 3
Y1 = 2x + 7
x
y
-3
1
0
7
(𝟎, 𝟕)
−𝟔
(−𝟑, 𝟏)
1 7
6
m=
2
=
30
3
y – intercept ( 0 , 7 )
−𝟑
Slope y-intercept form
y = mx + b
slope
y-intercept (0, b)
General Linear Equation
Although the general linear form helps in the quick
identification of lines, the slope-intercept form is the
one to enter into a calculator for graphing.
Ax + By = C
By = – Ax + C
y = – (A/B) x + C/B
Analyzing and Graphing a
General Linear Equation
Find the slope and y-intercept of the line 2𝑥 − 3𝑦 = 15
Rearrange for y
−𝟑𝒚 = −𝟐𝒙 + 𝟏𝟓
−𝟑
−𝟐
𝟏𝟓
𝒚 =
𝒙+
−𝟑
−𝟑
−𝟑
𝟐
𝒚= 𝒙−𝟓
𝟑
Slope is
𝟐
𝟑
y-intercept is (𝟎, −𝟓)
EXAMPLES
State the slopes and y-intercepts of the given linear functions.
1.
y = 4x
2.
y = 3x – 5
1
3. 𝑓 𝑥 = 𝑥 − 2
3
4
slope = m = _______
3
slope = m = _______
𝟏
𝟑
slope = m = _______
y -intercept ( 0 , 0 )
y -intercept ( 0 , - 5 )
y -intercept ( 0 ,, - 2 )
Find the slope and y-intercept of the following 3 lines
4. 2𝑥 − 3𝑦 = 15
𝟐
𝟑
slope = m = _______
y-intercept
𝟏
𝟐
slope = m = _______
y-intercept 𝟎,
𝟑
𝟐
𝟓
𝟑
slope = m = _______
𝟎,
𝟒
𝟑
𝟎, −𝟓
−𝟑𝒚 = −𝟐𝒙 + 𝟏𝟓
𝟐
𝒚= 𝒙−𝟓
𝟑
5. x + 2y = 3
𝟐𝒚 = −𝒙 + 𝟑
𝟏
𝟑
𝒚= 𝒙+
𝟐
𝟐
6. 5x – 3y = – 4
−𝟑𝒚 = −𝟓𝒙 − 𝟒
𝟓
𝟒
𝒚= 𝒙+
𝟑
𝟑
y-intercept
Example 7
𝟐
Find the equation in slope-intercept form for the line with slope and
𝟑
passes through the point (−𝟑, 𝟓)
Step 1: Solve for b using the point (−𝟑, 𝟓)
𝟐
𝟓 = −𝟑 + 𝒃
𝟑
b=7
Step 2: Find the equation
𝟐
𝟑
𝒚= 𝒙+𝟕
(𝟎, 𝟕)
(−𝟑, 𝟓)
Example 8
Find the equation in slope-intercept form for the line parallel to 𝒚 =
and through the point (10, -1)
Step 1: The slope of a parallel line will be
𝟐
𝒙
𝟓
+ 𝟐
𝟐
𝟓
Step 2: Solve for b using the point (𝟏𝟎, −𝟏)
𝟐
−𝟏 = 𝟏𝟎 + 𝒃
𝟓
𝒃 = −𝟓
Step 3: Find the equation
𝒚 =
𝟐
𝒙
𝟓
(𝟏𝟎, −𝟏)
–𝟓
(𝟎, −𝟓)
Example 9
Write the equation for the line through the point (– 1 , 2) that is
parallel to the line L: y = 3x – 4
Step 1: Slope of L is 3 so slope of any parallel line is also 3.
Step 2: Find b.
𝟐 = 𝟑(−𝟏) + 𝒃
𝒃 = 𝟓
Step 3: The equation of the line
parallel to
L: 𝒚 = 𝟑𝒙 – 𝟒 is
𝒚 = 𝟑𝒙 + 𝟓
Step 4: Graph on your calculator to check your work.
Use a square window.
Y1 = 3x – 4
Y2 = 3x + 5
(0, 5)
(0, – 4)
Example 10
Write the equation for the line that is perpendicular to
𝟐
𝒚 = 𝒙 + 𝟐 and passes through the point (10, – 1 )
𝟓
Step 1: The slope of a perpendicular line will be –
𝟓
𝟐
Step 2: Solve for b using the point (10, – 1)
𝟓
−𝟏 = – (𝟏𝟎) + 𝒃
Step 3: The equation of the line ┴ to
𝟐
𝟐
𝟓
𝒚 = 𝒙 + 𝟐
is 𝒚 = – 𝒙 + 𝟐𝟒
𝟓
𝟐
𝒃 = 𝟐𝟒
Step 4: Graph on your calculator to
check your work. Use a square window.
𝟐
Y1 = 𝒙 + 𝟐
𝟓
5
2
Y2 = – x + 24
Example 11
Write the equation for the line through the point (– 1, 2) that is
perpendicular to the line L: y = 3x – 4
Step 1: Slope of L is 3 so slope of any perpendicular line is −
𝟏
𝟑.
Step 2: Find b.
𝟏
𝟐 = − (−𝟏) + 𝒃
𝟑
𝟓
𝒃 =
𝟑
Step 3: Find the equation of the line perpendicular to L: y = 3x – 4
𝟏
𝟓
𝒚 =− 𝒙 +
𝟑
𝟑
Step 4: Graph on your calculator
to check your work.
Use a square window.
Y1 = 3x – 4
𝟏
𝟓
Y2 = − 𝒙 +
𝟑
𝟑
Example 12
Find the equation in slope-intercept form for the line that passes
through the points (7, −2) and (−5, 8).
Step 1: Find the slope
Step 2: Solve for b using either point
−𝟐 − 𝟖
−𝟏𝟎
𝟓
𝒎=
=
=−
𝟕 − (−𝟓)
𝟏𝟐
𝟔
𝟓
−𝟐 = − (𝟕) + 𝒃
𝟔
𝟐𝟑
𝒃 =
𝟔
𝟓
𝟖 = − (− 𝟓) + 𝒃
𝟔
𝒃 =
Step 3: Find the equation
𝒚 =
𝟓
− 𝒙
𝟔
𝟐𝟑
+
𝟔
(– 4, 8)
(7, – 2)
𝟐𝟑
𝟔
Example 13
Write the slope-intercept equation for the line through
(– 2, –1) and (5, 4).
−𝟏−𝟒
Slope = m =
−𝟐−𝟓
=
−𝟓
−𝟕
=
𝟓
𝟕
𝟓
– 𝟏 = (– 𝟐) + 𝒃
𝟕
𝒃 =
𝟑
𝟕
Equation for the line is
𝟓
𝟑
𝒚 = 𝒙 +
𝟕
𝟕
(5, 4)
(– 2, – 1)
UNIT 1B LESSON 3
SLOPE OF A SECANT
28
DEFINITION: A secant is a line that cuts a curve in two or more distinct points.
P2
P1
P2
P3
P1
SLOPE OF A SECANT
29
EXAMPLE
For the function y  x  1  3 draw the curve and the secant line
through the points A and C
2
X
Y
A
-1
1
B
0
-2
C
1
-3
D
2
-2
E
3
1
A(-1,1)
C(1, -3)
Equation of AC
𝒎=
−𝟑 − 𝟏
= −𝟐
𝟏 − (−𝟏)
𝟏 = −𝟐 −𝟏 + 𝒃
𝒃 = −𝟏
𝒚 = −𝟐𝒙 – 𝟏
𝒎<𝟎
30
EXAMPLE
For the function y  x  1  3 draw the curve and the secant line
through the points B and D.
2
X
Y
A
-1
1
B
0
-2
C
1
-3
D
2
-2
E
3
1
B(0,-2)
D(2, -2)
m=0
Equation of BD
m
2  (2)
0
02
−𝟐 = 𝟎(𝟎) + 𝒃
𝒃 = −𝟐
𝒚 = −𝟐
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EXAMPLE
For the function y  x  1  3 draw the curve and the secant lines
through the points B and E.
𝒎>𝟎
2
X
Y
A
-1
1
B
0
-2
C
1
-3
D
2
-2
E
3
1
E(3, 1)
B(0,-2)
Equation of BE
−𝟐 − 𝟏 −𝟏
𝒎=
=
=𝟏
𝟎−𝟑
−𝟏
−𝟐 = 𝟏(𝟎) + 𝒃
𝒃 = −𝟐
y=x–2
Notice: The slopes of secants on curves change as the points change.
Slopes may be positive, zero or negative.
32