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Slopes of Tangent Lines 3 As the difference in the x values of points P and Q approaches ZERO we can express the slope of a tangent line as the following limit.
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UNIT 1B LESSON 7
USING LIMITS TOFIND TANGENTS
1
Slopes of Secant Lines
2
π=π (π+π) β π (π )
(π+π ) βπ
The slope of secant PQ is given by
πΈ ( ππ ,ππ )ππ πΈ (π+π , π (π+π))
π· ( ππ ,ππ )ππ π· (π , π (π ))
Slopes of Tangent Lines
3
As the difference in the x values of points P and Q approaches ZERO we can express the slope of a tangent line as the following limit.
π=π₯π’π¦πβπ
π (π+π )β π (π)(π+π )β π
We want to find the slope and the equation of any tangent line to the curve y = 2x2 + 4x β 1 using the general slope formula and having h (the change in x) approach 0.
4
m = lim [2(x + h)2 + 4(x + h) β 1] β [2x2 + 4x β 1] hβ0 (x + h) β x
m = lim [2x2 + 4xh + 2h2 + 4x + 4h β 1 β 2x2 β 4x + 1] hβ0 h
m = lim [2(x2 + 2xh + h2) + 4(x + h) β 1] β [2x2 + 4x β 1] hβ0 (x + h) β x
Lesson 7 EXAMPLE 1 Page 1
m = lim [ 4xh + 2h2 + 4h] hβ0 h
Lesson 7 EXAMPLE 1 (continued) Page 1
5
m = lim [4x + 2h + 4] hβ0
m = 4x + 4
The equation for the slope of any tangent line = m = 4x + 4
m = lim [ 4xh + 2h2 + 4h] hβ0 h
m = lim h(4x + 2h + 4) hβ0 h
m =[4x + 2(0) + 4]
Lesson 7 Page 1 conβt
6
The equation for the slope of any tangent line m = 4x + 4
slope of tangent line at 4(2) + 4 = 12
Equation of tangent line
This equation for the slope of any tangent line can be used for any x value
15 = 12(2) + b
b = β 9
y = 12x β 9
Point of tangency at (2,15)
Lesson 7 Page 1 conβt
7
The equation for the slope of any tangent line = m = 4x + 4 This equation for the slope of any tangent line can be used for any x value
slope of tangent line at 4(β1) + 4 = 0
Equation of tangent line
β 3 = 0(β 1) + b
b = β 3
y = β 3
Point of tangency at
Lesson 7 Page 1 conβt
8
The equation for the slope of any tangent line = m = 4x + 4
Equation of tangent line
This equation for the slope of any tangent line can be used for any x value
slope of tangent line at 4(β3) + 4 = β8
5 = β8(β3) + b
b = β19
y = β 8x β 19
Point of tangency at
9
Practice Question #1
Find the equation for the slope of the tangent line to the parabola y = 2x β x2
m = lim [2(x + h) β (x + h)2] β [2x β x2] hβ0 (x + h) β x
m = lim 2x + 2h β x2 β 2xh β h2 β 2x + x2 hβ0 h
m = lim 2h β 2xh β h2 hβ0 h
m = lim 2 β 2x β h = 2 β 2x β 0 = 2 β 2x hβ0
π
10
Practice Question #1a
Find the equation the tangent line to the parabola y = 2x β x2 when x = 2
π=π (π )β (π )π=π Point of tangency (2, 0)
Slope = m = 2 β 2x = 2 β 2(2) = β 2
0 = β 2(2) + b
b = 4
y = β2 x + 4
(π ,π)
11
Practice Question #1b
Find the equation the tangent line to the parabola y = 2x β x2 when x = β3
π=π(β3)β(β3)π=β 15 Point of tangency (-3, -15)
Slope = m = 2 β 2x = 2 β 2(β3 ) = 8
β15 = 8(β3) + b
b = 9
y = 8x + 9
(βπ ,βππ)
12
Practice Question #1c
Find the equation the tangent line to the parabola y = 2x β x2 when x = 0
π=π (0 ) β (0 )π=π Point of tangency (0, 0)
Slope = m = 2 β 2x = 2 β 2(0) = 2
0= β 2(0) + b
b = 0y = 2x
(π ,π)
13
Practice Question #2a
Find the equation for the slope of the tangent line to the parabola y = x2 + 4x β 1
m = lim [(x + h)2 + 4(x + h) β 1 ] β [x2 + 4x β 1] hβ0 (x + h) β x
m = lim x2+ 2xh + h2 + 4x + 4h β 1 β x2 β 4x + 1 hβ0 h
m = lim 2xh + h2 + 4h hβ0 h
m = lim 2x + h + 4 hβ0
Slope = m =2x + 0 + 4 = 2x + 4
π
14
Practice Question #2 aFind the equation of the tangent line to the parabola y = x2 + 4x β 1 when x = β3
y = (β3)2 + 4(β3) β 1 = β 4 Point of tangency (-3, β4)
Slope = m = 2x + 4 = 2(β3) + 4 = β2
β 4 = β2(β3) + b
b = β 10
y = β2x β 10
(βπ ,βπ)
15
Practice Question #2 bFind the equation of the tangent line to the parabola y = x2 + 4x β 1 when x = β2
y = (β2)2 + 4(β2) β 1 = β 5 Point of tangency (-2, β5)
Slope = m = 2x + 4 = 2(β2) + 4 = 0
β 5 = 0(β2) + b
b = β5
y = β5
(βπ ,βπ)
16
Practice Question #2 cFind the equation of the tangent line to the parabola y = x2 + 4x β 1 when x = 0
y = (0)2 + 4(0) β 1 = β 1 Point of tangency (0, β 1 )
Slope = m = 2x + 4 = 2(0) + 4 = 4
β 1 = 4(0) + b
b = β 1
y = 4x β 1
Consider this:
153
2731
15273
1527
31527
153
2731
15273
1527
31527
153
2731
15273
1527
31527
153
2731
15273
1527
31527
Lesson 7 Page 4
17
Working with Compound fractions
18
743
34
75
x
ππ Γ·π=ΒΏππΓπ
π=πππ
OR π(π)(π)
=πππ
ΒΏπ π+ππ Γ·πΒΏ
π π+ππ Γπ
π=ππ+πππ
ORππ+π(π)(π)
=π π+πππ
Lesson 7 Page 4 Example 2
19
Find the slope of the tangent to at the point where x = 3.π (π )=π βππ
continuedβ
π₯π’π¦πβπ
π+πβππ+π β πβπ
π(π+π ) βπ
π₯π’π¦πβπ
π (π+πβπ )π (π+π)
β(π+π) (πβπ )π(π+π)
π
π₯π’π¦πβπ
π (π+πβπ ) β(π+π)(πβπ)ππ (π+π )
so at x = 3 the slope of the tangent is 20
Lesson 7 Page 4 Example 2 conβt
π₯π’π¦πβπ
ππ+ππβπ πβ(ππβπ π+ππβππ)ππ (π+π )
π₯π’π¦πβπ
ππ+ππβπ πβππ+π πβππ+ππππ (π+π )
π₯π’π¦πβπ
ππππ (π+π )
π₯π’π¦πβπ
ππ (π+π )
ππ(π+π)
=πππ
πππ=
ππ
PRACTICE QUESTION 3Find the slope of the tangent toat the point where x = 5.
21continuedβ
π (π )=π+ππ
π₯π’π¦πβπ
π+π+ππ+π β π+π
π(π+π) βπ
π₯π’π¦πβπ
π (π+π+π )π (π+π )
β(π+π) (π+π )π (π+π)
π
π₯π’π¦πβπ
π (π+π+π )β(π+π)(π+π)ππ (π+π )
so at x = 5 the slope of the tangent is
22
Practice question 3 conβt
π₯π’π¦πβπ
ππ+ππ+πβ(ππ+π+ππ+π)ππ (π+π)
π₯π’π¦πβπ
ππ+ππ+πβ ππβπβ ππβπππ (π+π )
π₯π’π¦πβπ
βπππ (π+π )
βπ
βπππ =
βπππ
βππ(π+π)
=βπππ
π=π₯π’π¦πβπ
πβπ (π+π)π+π βπβπ π
ππ
π=π₯π’π¦πβπ
πβπ πβπππ+π βπβππ
ππ
π=π₯π’π¦πβπ
π (πβπ πβππ)π (π+π)
β(π+π )(πβπ π)
π (π+π)π
π=ππππβπ
( πβπππβπ ππ ) β(πβπ ππ+πβπ ππ)ππ (π+π)
continuedβ
Practice Question 4
Find the slope of the tangent to at the point where x = 1.
23
π (π )=πβπ ππ
π=ππππβπ
( πβπ ππβπ ππ ) β(πβπ ππ+πβπ ππ)ππ (π+π)
continuedβ
Practice Question 4 con`t
π=ππππβπ
βπππ (π+π)
π=ππππβπ
βππ (π+π)
=βπ
π (π+π)=
βπππ
24
π=βπππ =βπ
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