1 / 14

Tangent Line Problems

Tangent Line Problems. Find the equations of tangents at given points Find the points on the curve if tangent slope is known Find equations of tangents parallel to given lines through a given point. Find equations of tangents perpendicular to given lines through a given point.

padma
Download Presentation

Tangent Line Problems

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Tangent Line Problems • Find the equations of tangents at given points • Find the points on the curve if tangent slope is known • Find equations of tangents parallel to given lines through a given point. • Find equations of tangents perpendicular to given lines through a given point. • Find equations of tangents to curves at their point of intersection.

  2. Point of tangency is f (–1) = (–1)2+ 2(–1) – 3= – 4 → (– 1, – 4) (–1, –4) Tangent Line Problems Example 1: Find the equation of the tangent to f (x) = x 2 + 2x – 3 at x = –1 Solution Slope of tangent = f ' (x) = 2x + 2 Slope of tangentat x = –1→ f ' (–1) = 2(–1)+ 2 = 0 – 4 = 0(–1) + b b = – 4 y= – 4

  3. Example 2: Find the equation of the tangent to f (x) = ½ x 4 + x – 2 at x = – 2 (– 2 , 4) Solution Point of tangency is f (–2) = 4 → (–2, 4) Slope of tangentf ' (x) = 4( ½) x4 – 1 + 1(1)x 1 – 1 – 0 f ' (x) = 2x3 + 1 Slope of tangentat x = –2→ f '(–2) = 2(–2)3+ 1 = –15 –4 = –15(–2) + b b = –26 y=–15x –26

  4. Example 3: Find the points where the graph off(x) = x3 + 2x2 – 5x + 1 has tangent lines with a slope of 2. f'(x) = 3x2 + 4x – 5 Solution Derivative of the function gives the slope of the tangent lines 3x2 + 4x – 5 = 2 3x2 + 4x – 7 = 0 (3x + 7)(x – 1) = 0 x = • or x = 1 The points on the graph off (x)where the slope willbe 2 areand (1, – 1)

  5. (1, -1) Check by graphing

  6. Example 4 Given the curve g(x) = x3 – 12 x, at what points is the slope of the tangent line equal to 15? g(x) = x3 – 12 x g' (x) = 3x2 – 12 3x2 – 12 = 15 or 3x2 – 27 = 0 3x2 = 27 3(x2 – 9) = 0 x2 = 9 3(x – 3)(x + 3) = 0 x = 3 or – 3 Points are (3, –9) and (–3, 9)

  7. y = 15x + 54 m = 15 y = 15x – 54 m = 15 (-3, 9) (3, -9)

  8. Example 5: Find the equation of a tangent to the graph of f (x) = – 2x 2that is paralleltoy = – 4x – 5 (1, -2) Solution:The derivative of any function determines the slope of the tangent line f' (x ) = – 4x and the slope of the given line is – 4 – 4 x =– 4 x = 1 f (x) = – 2x 2→ f (1) = – 2(1)2 = – 2 m = – 4 and P(1, – 2) – 2 = – 4(1) + b b = 2 y = – 4 x + 2 y = 2x2 y = – 4x – 5

  9. Example 6 Find the equation of a tangent to the graph of f (x) = 3x2 – 4 that is perpendicularto Solution:The derivative of any function determines the slope of the tangent line f' (x ) = 6xand the slope of the given line is so the slope of the perpendicular line is6 6x = 6 x =1 f (x) = 3x 2– 4→ f (1) = 3(1)2 – 4 = – 1 m = 6 and P(1, – 1) – 1 = 6(1) + b b = – 7 y = 6 x – 7

  10. Example 6 Continued Graph and check. y = 6x – 7 y = 3x2 – 4

  11. Example 7 Find the equation of the tangents to the curves and at their point of intersection Solution: Graphs intersect when x3 = 27 x = 3 Point of intersection is (3, 3)

  12. Example 7 Continued • Equations of Tangents • 3 = 2(3) + b 3 = – 1(3) + b • b = – 3 b = 6 • y = 2x – 3y = – x + 6

  13. (3, 3) Graph and Check y = -x + 6 y = 2x – 3

  14. Lesson 3 Tangent Problems Assignment Complete Questions 1-14 Hand-in for marks

More Related