UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS

1 / 24

# UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS - PowerPoint PPT Presentation

UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS. Slopes of Secant Lines. The slope of secant PQ is given by. 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. .

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS' - lora

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Slopes of Secant Lines

The slope of secant PQ is given by

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.

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 – 1using thegeneral slope formula and having h (the change in x) approach 0.

m = lim [2(x + h)2 + 4(x + h) – 1] – [2x2 + 4x – 1]

h→0 (x + h) – x

m = lim [2(x2 + 2xh + h2)+ 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[ 4xh+ 2h2+ 4h]

h→0 h

Lesson 7 EXAMPLE 1 (continued) Page 1

m = lim[ 4xh+ 2h2+ 4h]

h→0 h

m = limh(4x+ 2h+ 4)

h→0 h

m = lim [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

Lesson 7 Page 1 con’t

The equation for the slope of any tangent linem = 4x + 4

This equation for the slope of any tangent linecan be used for any x value

slope of tangent line at 4(2) + 4 = 12

Point of tangency at (2,15)

Equation of tangent line

15 = 12(2) + b

• b = – 9
• y = 12x – 9
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 linecan be used for any x value

slope of tangent line at 4(–1) + 4 = 0

Point of tangency at

• Equation of tangent line

– 3 = 0(– 1) + b

• b = – 3
• y = – 3
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 linecan be used for any x value

slope of tangent line at 4(–3) + 4 = –8

Point of tangency at

Equation of tangent line

• 5 = –8(–3) + b
• b = –19
• y = – 8x– 19

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 = lim2x + 2h – x2 – 2xh – h2– 2x + x2

h→0h

m = lim 2h – 2xh – h2

h→0 h

m = lim2 – 2x – h = 2 – 2x – 0 = 2 – 2x

h→0

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

Point of tangency (2, 0)

0 = – 2(2) + b

b = 4

y = –2 x + 4

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

Point of tangency (-3, -15)

–15 = 8(–3) + b

b = 9

y = 8x + 9

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

Point of tangency (0, 0)

0= – 2(0) + b

b = 0

y = 2x

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 = limx2+ 2xh + h2 + 4x + 4h – 1 – x2 – 4x + 1

h→0 h

m = lim2xh + h2+ 4h

h→0 h

m = lim2x + h + 4

h→0

Slope = m =2x + 0 + 4 = 2x + 4

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

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

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

Lesson 7 Page 4

Consider this:

Lesson 7 Page 4 Example 2

Find the slope of the tangent to

at the point where x = 3.

continued→

Lesson 7 Page 4 Example 2 con’t

so at x = 3 the slope of the tangent is

PRACTICE QUESTION 3

Find the slope of the tangent to

at the point where x = 5.

=

continued→

Practice question 3 con’t

=

so at x = 5 the slope of the tangent is

Practice Question 4

Find the slope of the tangent to

at the point where x = 1.

continued→