UNIT 1B LESSON 7
USING LIMITS TO
FIND TANGENTS
1
Slopes of Secant Lines
The slope of secant PQ is given by
𝒇 𝒙 + 𝒉 − 𝒇(𝒙)
𝒎=
𝒙+𝒉 −𝒙
𝑸 𝒙𝟐, 𝒚𝟐 𝒐𝒓 𝑸(𝒙 + 𝒉, 𝒇 𝒙 + 𝒉 )
𝑷 𝒙𝟏, 𝒚𝟏 𝒐𝒓 𝑷(𝒙, 𝒇 𝒙 )
2
Slopes of Tangent Lines
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.
𝒇 𝒙 + 𝒉 − 𝒇(𝒙)
𝒎 = 𝐥𝐢𝐦
𝒉→𝟎
𝒙+𝒉 −𝒙
3
Lesson 7
EXAMPLE 1
Page 1
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.
m = lim
h→0
[2(x + h)2 + 4(x + h) – 1] – [2x2 + 4x – 1]
(x + h) – x
[2(x2 + 2xh + h2) + 4(x + h) – 1] – [2x2 + 4x – 1]
h→0
(x + h) – x
m = lim
m = lim
h→0
[2x2 + 4xh + 2h2 + 4x + 4h – 1 – 2x2 – 4x + 1]
h
m = lim [ 4xh + 2h2 + 4h]
h→0
h
4
Lesson 7
EXAMPLE 1 (continued)
m = lim
h→0
m = lim
h→0
m = lim
Page 1
[ 4xh + 2h2 + 4h]
h
h(4x + 2h + 4)
h
[4x + 2h + 4]
h→0
m =[4x + 2(0) + 4]
m = 4x + 4
The equation for the slope of any tangent line = m = 4x + 4
5
Lesson 7 Page 1 con’t
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 𝒙 = 𝟐
4(2) + 4 = 12
Point of tangency at 𝒙 = 𝟐 𝟐 𝟐
𝟐
+ 𝟒 𝟐 − 𝟏 = 𝟏𝟓
(2,15)
Equation of tangent line
15 = 12(2) + b
b= –9
y = 12x – 9
6
Lesson 7 Page 1 con’t
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 𝒙 = −𝟏 4(–1) + 4 = 0
Point of tangency at 𝒙 = −𝟏 𝟐 −𝟏
𝟐
+ 𝟒 −𝟏 − 𝟏 = −𝟑
(−1,−3)
Equation of tangent line
– 3 = 0(– 1) + b
b=–3
y=–3
7
Lesson 7 Page 1 con’t
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 𝒙 = −𝟑
Point of tangency at 𝒙 = −𝟑 𝟐 −𝟑
𝟐
4(–3) + 4 = –8
+ 𝟒 −𝟑 − 𝟏 = 𝟓
(−3, 5)
Equation of tangent line
5 = –8(–3) + b
b = –19
y = – 8x – 19
8
Practice Question #1
Find the equation for the slope of the tangent line to the parabola y = 2x – x2
[2(x + h) – (x + h)2] – [2x – x2]
(x + h) – x
m = lim
h→0
m = lim
h→0
2x + 2h – x2 – 2xh – h2 – 2x + x2
h
𝒉
m = lim
h→0
2h – 2xh – h2
h
m = lim 2 – 2x – h = 2 – 2x – 0 = 2 – 2x
h→0
9
Practice Question #1a
Find the equation the tangent line to the parabola y = 2x – x2 when x = 2
Slope = m = 2 – 2x = 2 – 2(2) = – 2
𝒚 = 𝟐 𝟐 − 𝟐
𝟐
=𝟎
Point of tangency (2, 0)
0 = – 2(2) + b
b=4
(𝟐, 𝟎)
y = –2 x + 4
10
Practice Question #1b
Find the equation the tangent line to the parabola y = 2x – x2 when x = –3
Slope = m = 2 – 2x = 2 – 2(–3 ) = 8
𝒚 = 𝟐(–3) − (–3)𝟐
=
–15
Point of tangency (-3, -15)
–15 = 8(–3) + b
b=9
y = 8x + 9
(−𝟑 , −𝟏𝟓 )
11
Practice Question #1c
Find the equation the tangent line to the parabola y = 2x – x2 when x = 0
Slope = m = 2 – 2x = 2 – 2(0) = 2
𝒚 = 𝟐 0 − 0
𝟐 =
0= – 2(0) + b
𝟎
Point of tangency (0, 0)
(𝟐, 𝟎)
b=0
y = 2x
12
Practice Question #2a
Find the equation for the slope of the tangent line to the parabola y = x2 + 4x – 1
m = lim
h→0
m = lim
h→0
[(x + h)2 + 4(x + h) – 1 ] – [x2 + 4x – 1]
(x + h) – x
x2+ 2xh + h2 + 4x + 4h – 1 – x2 – 4x + 1
h
𝒉
m = lim
h→0
2xh + h2 + 4h
h
m = lim 2x + h + 4
h→0
Slope = m =2x + 0 + 4 = 2x + 4
13
Practice Question #2 a
Find the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = –3
Slope = m = 2x + 4 = 2(–3) + 4 = –2
y = (–3)2 + 4(–3) – 1 = – 4
Point of tangency (-3, –4)
– 4 = –2(–3) + b
b = – 10
y = –2x – 10
(−𝟑, −𝟒)
14
Practice Question #2 b
Find the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = –2
Slope = m = 2x + 4 = 2(–2) + 4 = 0
y = (–2)2 + 4(–2) – 1 = – 5
Point of tangency (-2, –5)
– 5 = 0(–2) + b
b = –5
y = –5
(−𝟐, −𝟓)
15
Practice Question #2 c
Find the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = 0
Slope = m = 2x + 4 = 2(0) + 4 = 4
y = (0)2 + 4(0) – 1 = – 1
Point of tangency (0, – 1 )
– 1 = 4(0) + b
b= –1
y = 4x – 1
16
Lesson 7 Page 4
Consider this:
7  2  7  2  3  7  2  1  7  2
5  1 5  1
5  1 3 35  1
37  2 7  2

7  2 1
72
5  1  5  1  3  5  1  3  35  1
3 7  2  7  2  3  7  2  1  7  2
5  1 75  21 7  25 1 37  235 11 7  2

3 
 
5  1 5  1
3
5  1 3 35  1
3
17
Working with Compound fractions
3
𝟑
𝟑 𝟏
𝟑
 
 4   𝟒 ÷ 𝟕 =𝟒 × 𝟕 = 𝟐𝟖
7
 5x  7 


 4 
3
𝟓𝒙 + 𝟕
=
÷𝟑
𝟒
OR
OR
𝟑
𝟑
=
(𝟒)(𝟕) 𝟐𝟖
𝟓𝒙 + 𝟕 𝟏 𝟓𝒙 + 𝟕
=
× =
𝟒
𝟑
𝟏𝟐
𝟓𝒙 + 𝟕 𝟓𝒙 + 𝟕
=
(𝟒)(𝟑)
𝟏𝟐
18
Lesson 7 Page 4 Example 2
Find the slope of the tangent to
𝒙−𝟓
𝒇 𝒙 =
𝒙
at the point where x = 3.
𝒙+𝒉−𝟓 𝒙−𝟓
−
𝒙
𝒙
+
𝒉
𝐥𝐢𝐦
𝒉→𝟎
𝒙+𝒉 −𝒙
(𝒙 + 𝒉) 𝒙 − 𝟓
𝒙 𝒙+𝒉−𝟓
−
𝒙(𝒙 + 𝒉)
𝒙 𝒙+𝒉
𝐥𝐢𝐦
𝒉→𝟎
𝒉
𝒙 𝒙 + 𝒉 − 𝟓 − (𝒙 + 𝒉)(𝒙 − 𝟓)
𝒉→𝟎
𝒉𝒙 𝒙 + 𝒉
𝐥𝐢𝐦
continued→
19
Lesson 7 Page 4 Example 2 con’t
𝒙𝟐 + 𝒙𝒉 − 𝟓𝒙 − (𝒙𝟐 − 𝟓𝒙 + 𝒙𝒉 − 𝟓𝒉)
𝐥𝐢𝐦
𝒉→𝟎
𝒉𝒙 𝒙 + 𝒉
𝒙𝟐 + 𝒙𝒉 − 𝟓𝒙 − 𝒙𝟐 + 𝟓𝒙 − 𝒙𝒉 + 𝟓𝒉
𝐥𝐢𝐦
𝒉→𝟎
𝒉𝒙 𝒙 + 𝒉
𝟓𝒉
𝐥𝐢𝐦
𝒉→𝟎 𝒉𝒙 𝒙 + 𝒉
𝟓
𝐥𝐢𝐦
𝒉→𝟎 𝒙 𝒙 + 𝒉
𝟓
𝟓
= 𝟐
𝒙(𝒙 + 𝟎) 𝒙
so at x = 3 the slope of the tangent is
𝟓
𝟓
=
𝟐
𝟑
𝟗
20
PRACTICE QUESTION 3
𝒙+𝟏
𝒇
𝒙
=
Find the slope of the tangent to
𝒙
at the point where x = 5.
𝒙+𝒉+𝟏 𝒙+𝟏
−
𝒙
𝒙
+
𝒉
𝐥𝐢𝐦
𝒉→𝟎
𝒙+𝒉 −𝒙
(𝒙 + 𝒉) 𝒙 + 𝟏
𝒙 𝒙+𝒉+𝟏
−
𝒙(𝒙 + 𝒉)
𝒙 𝒙+𝒉
𝐥𝐢𝐦
𝒉→𝟎
𝒉
𝒙 𝒙 + 𝒉 + 𝟏 − (𝒙 + 𝒉)(𝒙 + 𝟏)
𝐥𝐢𝐦
𝒉→𝟎
𝒉𝒙 𝒙 + 𝒉
continued→
21
Practice question 3 con’t
𝒙𝟐 + 𝒙𝒉 + 𝒙 − (𝒙𝟐 + 𝒙 + 𝒙𝒉 + 𝒉)
𝐥𝐢𝐦
𝒉→𝟎
𝒉𝒙 𝒙 + 𝒉
𝒙𝟐 + 𝒙𝒉 + 𝒙 − 𝒙𝟐 − 𝒙 − 𝒙𝒉 − 𝒉
𝐥𝐢𝐦
𝒉→𝟎
𝒉𝒙 𝒙 + 𝒉
−𝟏
−𝒉
𝒉→𝟎 𝒉𝒙 𝒙 + 𝒉
𝐥𝐢𝐦
−𝟏
−𝟏
=
𝒙(𝒙 + 𝟎) 𝒙𝟐
so at x = 5 the slope of the tangent is
−𝟏 −𝟏
=
𝟐
𝟓
𝟐𝟓
22
𝒙+𝟏
𝒇 𝒙 =
𝒙
𝟓+𝟏
𝒇 𝟓 =
𝟓
𝟔
𝒇 𝟓 =
𝟓
so at x = 5
the slope of
the tangent is
Slope =
−𝟏
𝒙𝟐
−𝟏 −𝟏
=
𝟓𝟐
𝟐𝟓
23
Practice Question 4
Find the slope of the tangent to 𝒇 𝒙 =
𝟏 − 𝟐𝒙
𝒙
at the point where x = -2.
𝟏 − 𝟐(𝒙 + 𝒉) 𝟏 − 𝟐𝒙
− 𝒙
𝒙
+
𝒉
𝒎 = 𝐥𝐢𝐦
𝒉→𝟎
𝒉
𝟏 − 𝟐𝒙 − 𝟐𝒉 𝟏 − 𝟐𝒙
− 𝒙
𝒙
+
𝒉
𝒎 = 𝐥𝐢𝐦
𝒉→𝟎
𝒉
𝒙(𝟏 − 𝟐𝒙 − 𝟐𝒉)
𝒙 + 𝒉 (𝟏 − 𝟐𝒙)
−
𝒙(𝒙 + 𝒉)
𝒙(𝒙 + 𝒉)
𝒎 = 𝐥𝐢𝐦
𝒉→𝟎
𝒉
𝒙 − 𝟐𝒙𝟐 − 𝟐𝒙𝒉 − (𝒙 − 𝟐𝒙𝟐 + 𝒉 − 𝟐𝒙𝒉)
𝒎 = 𝒍𝒊𝒎
𝒉→𝟎
𝒉𝒙(𝒙 + 𝒉)
continued→
24
Practice Question 4 con`t
𝒙 − 𝟐𝒙𝟐 − 𝟐𝒙𝒉 − (𝒙 − 𝟐𝒙𝟐 + 𝒉 − 𝟐𝒙𝒉)
𝒎 = 𝒍𝒊𝒎
𝒉→𝟎
𝒉𝒙(𝒙 + 𝒉)
−𝒉
𝒎 = 𝒍𝒊𝒎
𝒉→𝟎 𝒉𝒙(𝒙 + 𝒉)
−𝟏
−𝟏
−𝟏
𝒎 = 𝒍𝒊𝒎
=
=
𝒉→𝟎 𝒙(𝒙 + 𝒉)
𝒙(𝒙 + 𝟎) 𝒙𝟐
−𝟏
𝟏
𝒎=
=−
−𝟐 𝟐
𝟒
continued→
25
𝟏 − 𝟐𝒙
𝒇 𝒙 =
𝒙
𝟏 − 𝟐(−𝟐)
𝟓
𝒇 −𝟐 =
=−
(−𝟐)
𝟐
−𝟏
𝒎= 𝟐
𝒙
−𝟏
𝟏
𝒎=
=−
𝟐
(−𝟐)
𝟒
26