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.
1
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)
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
(–1, –4)
2
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
f ' (x) = 2x3 + 1
1–1–
0
Slope of tangent at x = –2→ f '(–2) = 2(–2)3 + 1 = –15
–4 = –15(–2) + b
b = –26
y = – 15x – 26
(– 2 , 4)
3
Example 3: Find the points where the graph of f(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
𝟕
x=−
𝟑
𝟕
𝟕
𝒇 −
= −
𝟑
𝟑
𝟑
𝟕
+𝟐 −
𝟑
𝟐
−𝟓 −
𝟕
+𝟏
𝟑
𝟕
𝟐𝟗𝟑
𝒇 −
=
𝟑
𝟐𝟕
or x = 1
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
𝟕 𝟐𝟗𝟕
)
𝟑 𝟐𝟕
(− ,
and (1, – 1)
4
Check by graphing
𝟕 𝟐𝟗𝟑
(− ,
)
𝟑 𝟐𝟕
(1, -1)
5
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
3x2 = 27
x2 = 9
or
3x2 – 27 = 0
3(x2 – 9) = 0
3(x – 3)(x + 3) = 0
x = 3 or – 3
Points are (3, – 9) and (– 3, 9)
6
y = 15x + 54
m = 15
(-3, 9)
y = 15x – 54
m = 15
(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
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
f (x) = – 2x 2
x=1
→ f (1) = – 2(1)2 = – 2
m = – 4 and P(1, – 2)
– 2 = – 4(1) + b
b=2
(1, -2)
8
y=–4x+2
y = – 4x – 5
y = 2x2
Example 6
Find the equation of a tangent to the graph of f (x) = 3x2 – 4
𝟏
that is perpendicular to 𝒚 = − 𝒙 + 𝟏
𝟔
Solution: The derivative of any function determines the slope of the tangent line
f ' (x ) = 6x and the slope of the given line is
so the slope of the perpendicular line is 6
1
−
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=6x–7
9
Example 6 Continued
Graph and check.
𝟏
𝒚=− 𝒙 + 𝟏
𝟔
y = 6x – 7
y = 3x2 – 4
10
Example 7
Find the equation of the tangents to the curves
1
9
𝑓 (𝑥) = 𝑥 2 and 𝑔(𝑥) =
at their point of intersection
3
𝑥
Solution: Graphs intersect when
𝒙𝟐 𝟗
=
𝟑
𝒙
x3 = 27
x=3
𝟗
𝒚= =𝟑
𝟑
Point of intersection is (3, 3)
11
Example 7 Continued
1 2
f x   x
3
9
g x    9 x 1
x
𝟐
𝒇′ 𝒙 = 𝒙
𝟑
𝟐
𝒇 ′(𝟑) = (𝟑) = 𝟐
𝟑
𝒈′
𝟗
𝒙 =− 𝟐
𝒙
𝒈′
𝟗
𝟑 = − 𝟐 = −𝟏
𝟑
Equations of Tangents
3 = 2(3) + b
b=–3
3 = – 1(3) + b
b=6
y = 2x – 3
y=–x+6
12
Graph and Check
𝟏 𝟐
𝒇(𝒙) = 𝒙
𝟑
(3, 3)
y = -x + 6
9
𝑔(𝑥) =
𝑥
y = 2x – 3
13
Lesson 3
Tangent Problems Assignment
Complete Questions 1-14
Hand-in for marks
14