Unit 2 Lesson #3 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.

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

Unit 2 Lesson #3 Tangent Line Problems Example 2: Find the equation of the tangent to f (x) = ½ x 4 + x – 2 at x = – 2 Unit 2 Lesson #3 Tangent Line Problems Solution Point of tangency is f (–2) = 4 → (–2, 4) Slope of tangent f ' (x) = 4( ½) x 4 – 1 + 1(1)x 1 – 1 – 0 f ' (x) = 2x3 + 1 Slope of tangent at x = –2→ f '(–2) = 2(–2)3 + 1 = –15 –4 = –15(–2) + b b = –26 y = – 15x – 26 (– 2 , 4)

Unit 2 Lesson #3 Tangent Line Problems Example 3: Find the points where the graph of f(x) = x3 + 2x2 – 5x + 1 has tangent lines with a slope of 2. Unit 2 Lesson #3 Tangent Line Problems 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 = − 𝟕 𝟑 𝒇 − 𝟕 𝟑 = − 𝟕 𝟑 𝟑 +𝟐 − 𝟕 𝟑 𝟐 −𝟓 − 𝟕 𝟑 +𝟏 or x = 1 𝒇 − 𝟕 𝟑 = 𝟐𝟗𝟑 𝟐𝟕 𝒇(𝟏) = (𝟏)𝟑 + 𝟐(𝟏)𝟐 – 𝟓(𝟏) + 𝟏 = −𝟏 The points on the graph of f (x) where the slope will be 2 are (− 𝟕 𝟑 , 𝟐𝟗𝟕 𝟐𝟕 ) and (1, – 1)

Unit 2 Lesson #3 Tangent Line Problems Check by graphing (− 𝟕 𝟑 , 𝟐𝟗𝟑 𝟐𝟕 ) (1, -1)

Unit 2 Lesson #3 Tangent Line Problems 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 3x2 – 27 = 0 3(x2 – 9) = 0 x2 = 9 3(x – 3)(x + 3) = 0 x = 3 or – 3 Points are (3, – 9) and (– 3, 9)

Unit 2 Lesson #3 Tangent Line Problems y = 15x + 54 m = 15 y = 15x – 54 m = 15 (-3, 9) (3, -9)

Unit 2 Lesson #3 Tangent Line Problems Example 5: Find the equation of a tangent to the graph of f (x) = – 2x 2 that is parallel to y = – 4x – 5 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 (1, -2) y = – 4 x + 2 y = 2x2 y = – 4x – 5

Unit 2 Lesson #3 Tangent Line Problems Example 6 Find the equation of a tangent to the graph of f (x) = 3x2 – 4 that is perpendicular to 𝒚 =− 𝟏 𝟔 𝒙 +𝟏 Unit 2 Lesson #3 Tangent Line Problems Solution: The derivative of any function determines the slope of the tangent line f ' (x ) = 6x and the slope of the given line is − 1 6 so the slope of the perpendicular line is 6 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

Unit 2 Lesson #3 Tangent Line Problems Example 6 Continued Unit 2 Lesson #3 Tangent Line Problems Graph and check. y = 6x – 7 𝒚=− 𝟏 𝟔 𝒙 + 𝟏 y = 3x2 – 4

Unit 2 Lesson #3 Tangent Line Problems Example 7 Find the equation of the tangents to the curves 𝑓 (𝑥) = 1 3 𝑥 2 and 𝑔(𝑥) = 9 𝑥 at their point of intersection Solution: Graphs intersect when 𝒙 𝟐 𝟑 = 𝟗 𝒙 x3 = 27 x = 3 𝒚= 𝟗 𝟑 =𝟑 Point of intersection is (3, 3)

Unit 2 Lesson #3 Tangent Line Problems Example 7 Continued 𝒈 ′ 𝒙 =− 𝟗 𝒙 𝟐 𝒇 ′ 𝒙 = 𝟐 𝟑 𝒙 𝒇 ′(𝟑) = 𝟐 𝟑 (𝟑) = 𝟐 𝒈 ′ 𝟑 =− 𝟗 𝟑 𝟐 = −𝟏 Equations of Tangents 3 = 2(3) + b 3 = – 1(3) + b b = – 3 b = 6 y = 2x – 3 y = – x + 6

Unit 2 Lesson #3 Tangent Line Problems Graph and Check 𝒇(𝒙) = 𝟏 𝟑 𝒙𝟐 (3, 3) y = -x + 6 𝑔(𝑥)= 9 𝑥 y = 2x – 3

Unit 2 Lesson #3 Tangent Line Problems Tangent Problems Assignment Unit 2 Lesson #3 Tangent Line Problems Complete Questions 1-14 Hand-in for marks