Upload
ronald-bond
View
228
Download
0
Embed Size (px)
DESCRIPTION
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.
Citation preview
UNIT 1B LESSON 7
USING LIMITS TOFIND TANGENTS
1
Slopes of Secant Lines
2
𝒎=𝒇 (𝒙+𝒉) − 𝒇 (𝒙 )
(𝒙+𝒉 ) −𝒙
The slope of secant PQ is given by
𝑸 ( 𝒙𝟐 ,𝒚𝟐 )𝒐𝒓 𝑸 (𝒙+𝒉 , 𝒇 (𝒙+𝒉))
𝑷 ( 𝒙𝟏 ,𝒚𝟏 )𝒐𝒓 𝑷 (𝒙 , 𝒇 (𝒙 ))
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.
𝒎=𝐥𝐢𝐦𝒉→𝟎
𝒇 (𝒙+𝒉 )− 𝒇 (𝒙)(𝒙+𝒉 )− 𝒙
We want to find the slope and the equation of any tangent line to the curve y = 2x2 + 4x – 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) – 1] – [2x2 + 4x – 1] h→0 (x + h) – x
m = lim [2x2 + 4xh + 2h2 + 4x + 4h – 1 – 2x2 – 4x + 1] h→0 h
m = lim [2(x2 + 2xh + h2) + 4(x + h) – 1] – [2x2 + 4x – 1] h→0 (x + h) – x
Lesson 7 EXAMPLE 1 Page 1
m = lim [ 4xh + 2h2 + 4h] h→0 h
Lesson 7 EXAMPLE 1 (continued) Page 1
5
m = lim [4x + 2h + 4] h→0
m = 4x + 4
The equation for the slope of any tangent line = m = 4x + 4
m = lim [ 4xh + 2h2 + 4h] h→0 h
m = lim h(4x + 2h + 4) h→0 h
m =[4x + 2(0) + 4]
Lesson 7 Page 1 con’t
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 = – 9
y = 12x – 9
Point of tangency at (2,15)
Lesson 7 Page 1 con’t
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(–1) + 4 = 0
Equation of tangent line
– 3 = 0(– 1) + b
b = – 3
y = – 3
Point of tangency at
Lesson 7 Page 1 con’t
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(–3) + 4 = –8
5 = –8(–3) + b
b = –19
y = – 8x – 19
Point of tangency at
9
Practice Question #1
Find the equation for the slope of the tangent line to the parabola y = 2x – x2
m = lim [2(x + h) – (x + h)2] – [2x – x2] h→0 (x + h) – x
m = lim 2x + 2h – x2 – 2xh – h2 – 2x + x2 h→0 h
m = lim 2h – 2xh – h2 h→0 h
m = lim 2 – 2x – h = 2 – 2x – 0 = 2 – 2x h→0
𝒉
10
Practice Question #1a
Find the equation the tangent line to the parabola y = 2x – x2 when x = 2
𝒚=𝟐 (𝟐 )− (𝟐 )𝟐=𝟎 Point of tangency (2, 0)
Slope = m = 2 – 2x = 2 – 2(2) = – 2
0 = – 2(2) + b
b = 4
y = –2 x + 4
(𝟐 ,𝟎)
11
Practice Question #1b
Find the equation the tangent line to the parabola y = 2x – x2 when x = –3
𝒚=𝟐(–3)−(–3)𝟐=– 15 Point of tangency (-3, -15)
Slope = m = 2 – 2x = 2 – 2(–3 ) = 8
–15 = 8(–3) + b
b = 9
y = 8x + 9
(−𝟑 ,−𝟏𝟓)
12
Practice Question #1c
Find the equation the tangent line to the parabola y = 2x – x2 when x = 0
𝒚=𝟐 (0 ) − (0 )𝟐=𝟎 Point of tangency (0, 0)
Slope = m = 2 – 2x = 2 – 2(0) = 2
0= – 2(0) + b
b = 0y = 2x
(𝟐 ,𝟎)
13
Practice Question #2a
Find the equation for the slope of the tangent line to the parabola y = x2 + 4x – 1
m = lim [(x + h)2 + 4(x + h) – 1 ] – [x2 + 4x – 1] h→0 (x + h) – x
m = lim x2+ 2xh + h2 + 4x + 4h – 1 – x2 – 4x + 1 h→0 h
m = lim 2xh + h2 + 4h h→0 h
m = lim 2x + h + 4 h→0
Slope = m =2x + 0 + 4 = 2x + 4
𝒉
14
Practice Question #2 aFind the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = –3
y = (–3)2 + 4(–3) – 1 = – 4 Point of tangency (-3, –4)
Slope = m = 2x + 4 = 2(–3) + 4 = –2
– 4 = –2(–3) + b
b = – 10
y = –2x – 10
(−𝟑 ,−𝟒)
15
Practice Question #2 bFind the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = –2
y = (–2)2 + 4(–2) – 1 = – 5 Point of tangency (-2, –5)
Slope = m = 2x + 4 = 2(–2) + 4 = 0
– 5 = 0(–2) + b
b = –5
y = –5
(−𝟐 ,−𝟓)
16
Practice Question #2 cFind the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = 0
y = (0)2 + 4(0) – 1 = – 1 Point of tangency (0, – 1 )
Slope = m = 2x + 4 = 2(0) + 4 = 4
– 1 = 4(0) + b
b = – 1
y = 4x – 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
𝟑𝟒 ÷𝟕=¿𝟑𝟒×𝟏
𝟕=𝟑𝟐𝟖
OR 𝟑(𝟒)(𝟕)
=𝟑𝟐𝟖
¿𝟓 𝒙+𝟕𝟒 ÷𝟑¿
𝟓 𝒙+𝟕𝟒 ×𝟏
𝟑=𝟓𝒙+𝟕𝟏𝟐
OR𝟓𝒙+𝟕(𝟒)(𝟑)
=𝟓 𝒙+𝟕𝟏𝟐
Lesson 7 Page 4 Example 2
19
Find the slope of the tangent to at the point where x = 3.𝒇 (𝒙 )=𝒙 −𝟓𝒙
continued→
𝐥𝐢𝐦𝒉→𝟎
𝒙+𝒉−𝟓𝒙+𝒉 − 𝒙−𝟓
𝒙(𝒙+𝒉 ) −𝒙
𝐥𝐢𝐦𝒉→𝟎
𝒙 (𝒙+𝒉−𝟓 )𝒙 (𝒙+𝒉)
−(𝒙+𝒉) (𝒙−𝟓 )𝒙(𝒙+𝒉)
𝒉
𝐥𝐢𝐦𝒉→𝟎
𝒙 (𝒙+𝒉−𝟓 ) −(𝒙+𝒉)(𝒙−𝟓)𝒉𝒙 (𝒙+𝒉 )
so at x = 3 the slope of the tangent is 20
Lesson 7 Page 4 Example 2 con’t
𝐥𝐢𝐦𝒉→𝟎
𝒙𝟐+𝒙𝒉−𝟓 𝒙−(𝒙𝟐−𝟓 𝒙+𝒙𝒉−𝟓𝒉)𝒉𝒙 (𝒙+𝒉 )
𝐥𝐢𝐦𝒉→𝟎
𝒙𝟐+𝒙𝒉−𝟓 𝒙−𝒙𝟐+𝟓 𝒙−𝒙𝒉+𝟓𝒉𝒉𝒙 (𝒙+𝒉 )
𝐥𝐢𝐦𝒉→𝟎
𝟓𝒉𝒉𝒙 (𝒙+𝒉 )
𝐥𝐢𝐦𝒉→𝟎
𝟓𝒙 (𝒙+𝒉 )
𝟓𝒙(𝒙+𝟎)
=𝟓𝒙𝟐
𝟓𝟑𝟐=
𝟓𝟗
PRACTICE QUESTION 3Find the slope of the tangent toat the point where x = 5.
21continued→
𝒇 (𝒙 )=𝒙+𝟏𝒙
𝐥𝐢𝐦𝒉→𝟎
𝒙+𝒉+𝟏𝒙+𝒉 − 𝒙+𝟏
𝒙(𝒙+𝒉) −𝒙
𝐥𝐢𝐦𝒉→𝟎
𝒙 (𝒙+𝒉+𝟏 )𝒙 (𝒙+𝒉 )
−(𝒙+𝒉) (𝒙+𝟏 )𝒙 (𝒙+𝒉)
𝒉
𝐥𝐢𝐦𝒉→𝟎
𝒙 (𝒙+𝒉+𝟏 )−(𝒙+𝒉)(𝒙+𝟏)𝒉𝒙 (𝒙+𝒉 )
so at x = 5 the slope of the tangent is
22
Practice question 3 con’t
𝐥𝐢𝐦𝒉→𝟎
𝒙𝟐+𝒙𝒉+𝒙−(𝒙𝟐+𝒙+𝒙𝒉+𝒉)𝒉𝒙 (𝒙+𝒉)
𝐥𝐢𝐦𝒉→𝟎
𝒙𝟐+𝒙𝒉+𝒙− 𝒙𝟐−𝒙− 𝒙𝒉−𝒉𝒉𝒙 (𝒙+𝒉 )
𝐥𝐢𝐦𝒉→𝟎
−𝒉𝒉𝒙 (𝒙+𝒉 )
−𝟏
−𝟏𝟓𝟐 =
−𝟏𝟐𝟓
−𝟏𝒙(𝒙+𝟎)
=−𝟏𝒙𝟐
𝒎=𝐥𝐢𝐦𝒉→𝟎
𝟏−𝟐 (𝒙+𝒉)𝒙+𝒉 −𝟏−𝟐 𝒙
𝒙𝒉
𝒎=𝐥𝐢𝐦𝒉→𝟎
𝟏−𝟐 𝒙−𝟐𝒉𝒙+𝒉 −𝟏−𝟐𝒙
𝒙𝒉
𝒎=𝐥𝐢𝐦𝒉→𝟎
𝒙 (𝟏−𝟐 𝒙−𝟐𝒉)𝒙 (𝒙+𝒉)
−(𝒙+𝒉 )(𝟏−𝟐 𝒙)
𝒙 (𝒙+𝒉)𝒉
𝒎=𝒍𝒊𝒎𝒉→𝟎
( 𝒙−𝟐𝒙𝟐−𝟐 𝒙𝒉 ) −(𝒙−𝟐 𝒙𝟐+𝒉−𝟐 𝒙𝒉)𝒉𝒙 (𝒙+𝒉)
continued→
Practice Question 4
Find the slope of the tangent to at the point where x = 1.
23
𝒇 (𝒙 )=𝟏−𝟐 𝒙𝒙
𝒎=𝒍𝒊𝒎𝒉→𝟎
( 𝒙−𝟐 𝒙𝟐−𝟐 𝒙𝒉 ) −(𝒙−𝟐 𝒙𝟐+𝒉−𝟐 𝒙𝒉)𝒉𝒙 (𝒙+𝒉)
continued→
Practice Question 4 con`t
𝒎=𝒍𝒊𝒎𝒉→𝟎
−𝒉𝒉𝒙 (𝒙+𝒉)
𝒎=𝒍𝒊𝒎𝒉→𝟎
−𝟏𝒙 (𝒙+𝒉)
=−𝟏
𝒙 (𝒙+𝟎)=
−𝟏𝒙𝟐
24
𝒎=−𝟏𝟏𝟐 =−𝟏