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.

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Tangent Line Problems

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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 tangent at x = –1→ f ' (–1) = 2(–1)+ 2 = 0

– 4 = 0(–1) + b b = – 4

y = – 4

Point of tangency is f (–1) = (–1)2 + 2(–1) – 3= – 4 → (– 1, – 4)

(–1, –4)

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Example 2: Find the equation of the tangent to f (x) = ½ x 4 + x – 2 at x = – 2

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

(– 2 , 4)

Slope of tangent at x = –2→ f '(–2) = 2(–2)3 + 1 = –15

–4 = –15(–2) + b b = –26

y = – 15x – 26

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Example 3: Find the points where the graph of f(x) = x3 + 2x2 – 5x + 1 has tangent lines with a slope of 2.Solution f '(x) = 3x2 + 4x – 5

3x2 + 4x – 5 = 23x2 + 4x – 7 = 0(3x + 7)(x – 1) = 0x =

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 and (1, – 1)

Derivative of the function gives the slope of the tangent lines

𝒇 (−𝟕𝟑 )=(−𝟕

𝟑 )𝟑+𝟐 (−𝟕

𝟑 )𝟐

−𝟓(−𝟕𝟑 )+𝟏

𝒇 (−𝟕𝟑 )=𝟐𝟗𝟑

𝟐𝟕

or x = 1

Check by graphing

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(−𝟕𝟑 ,𝟐𝟗𝟑𝟐𝟕 )

(1, -1)

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Example 4Given 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

3x2 = 27 3x2 – 27 = 0

x2 = 9 3(x2 – 9) = 0

3(x – 3)(x + 3) = 0

or

x = 3 or – 3

Points are (3, – 9) and (– 3, 9)

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y = 15x – 54 m = 15

y = 15x + 54m = 15

(3, -9)

(-3, 9)

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Example 5: Find the equation of a tangent to the graph of f (x) = – 2x 2 that is parallel to y = – 4x – 5

f ' (x ) = – 4x and the slope of the given line is – 4

Solution: The derivative of any function determines the slope of the tangent line

– 4 x = – 4x = 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

(1, -2)

y = 2x2

y = – 4x – 5

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Example 6Find the equation of a tangent to the graph of f (x) = 3x2 – 4 that is perpendicular to

f ' (x ) = 6x and the slope of the given line is so the slope of the perpendicular line is 6

Solution: The derivative of any function determines the slope of the tangent line

6x = 6x =1

f (x) = 3x 2 – 4 → f (1) = 3(1)2 – 4 = – 1m = 6 and P(1, – 1)– 1 = 6(1) + b b = – 7

y = 6 x – 7

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Graph and check.

𝒚=−𝟏𝟔 𝒙+𝟏

y = 6x – 7

y = 3x2 – 4

Example 6 Continued

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Example 7Find 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)

𝒚=𝟗𝟑=𝟑

𝒙𝟐

𝟑 =𝟗𝒙

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𝒇 ′ (𝒙 )=𝟐𝟑 𝒙

𝒇 ′ (𝟑)=𝟐𝟑 (𝟑)=𝟐

𝒈 ′ (𝒙 )=− 𝟗𝒙𝟐

𝒈 ′ (𝟑 )=− 𝟗𝟑𝟐=−𝟏

3 = 2(3) + b 3 = – 1(3) + b b = – 3 b = 6

Example 7 Continued

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31 xxf 199 x

xxg

Equations of Tangents

y = 2x – 3 y = – x + 6

Graph and Check

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(3, 3)𝒇 (𝒙)=

𝟏𝟑 𝒙𝟐

𝑔 (𝑥)=9𝑥 y = 2x – 3

y = -x + 6

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Lesson 3Tangent Problems Assignment

Complete Questions 1-14

Hand-in for marks