UNIT 1B LESSON 7

USING LIMITS TOFIND TANGENTS

1

Slopes of Secant Lines

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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

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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

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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.

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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