EQUATION OF TANGENT

Example 1 :

Find the equation of tangent to the curve y = x² – x – 2 at x = 1.

Solution :

Substitute x = 1 into the equation of curve.

y = 1² – 1 – 2

= 1 - 1 - 2

= -2

Point of tangency is (1, -2).

y = x² – x – 2

dy/dx = 2x - 1

Substitute x = 1.

= 2(1) - 1.

= 2 - 1

= 1

Slope of the tangent is 1.

Equation of the tangent :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (1, -2) and m = 1.

y - (-2) = 1(x - 1)

y + 2 = x - 1

y = x - 3

Example 2 :

Find the equation of tangent to the curve y = x² – 4x – 5 at x = -2.

Solution :

Substitute x = -2 into the equation of curve.

y = (-2)² – 4(-2) – 5

= 4 + 8 - 5

= 7

Point of tangency is (-2, 7).

y = x² – 4x – 5

dy/dx = 2x - 4

Substitute x = -2.

= 2(-2) - 4

= -4 - 4

= -8

Slope of the tangent is -8.

Equation of the tangent :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (-2, 7) and m = -8.

y - 7 = -8[x - (-2)]

y - 7 = -8(x + 2)

y -7 = -8x - 16

y = -8x - 9

Example 3 :

Find the equation of tangent to the curve y = x⁴ + 1 and it is parallel to the line 32x - y = 15.

Solution :

y = x⁴ + 1 ----(1)

Slope of the tangent :

dy/dx = 4x³ ----(1)

Write the equation 32x - y = 15 in slope intercept form.

32x - y = 15

y = 32x - 15

Slope of the line 32x - y = 15 :

m = 32

Since tangent is parallel to the line 32x - y = 15, the slopes are equal.

4x³ = 32

x³ = 8

x³ = 2³

x = 2

Substitute x = 2 in (1).

y = 2⁴ + 1

= 16 + 1

= 17

Point of tangency is (2, 17).

Substitute x = 2 in (2).

dy/dx = 4(2³)

= 4(8)

= 32

Slope of the tangent is 32.

Equation of the tangent :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (2, 17) and m = 32.

y - 17 = 32(x - 2)

y - 17 = 32x - 64

y = 32x - 47

Example 4 :

Find the equations of tangent lines to the curve 

y = x³ - 3x² + 3x - 3

and are parallel to the line 3x - y = 15.

Solution :

y = x³ - 3x² + 3x - 3 ----(1)

Slope of the tangent :

dy/dx = 3x² - 6x + 3 ----(2)

Write the equation 3x - y = 15 in slope intercept form.

3x - y = 15

y = 3x - 15

Slope of the line 3x - y = 15 :

m = 3

Since tangent is parallel to the line 3x - y = 15, the slopes are equal.

3x² - 6x + 3 = 3

3x² - 6x = 0

Divide each side by 3.

x² - 2x = 0

x(x - 2) = 0

x = 0 or x = 2

Substitute x = 0 and 2 in (1).

Points of tangency are (0, -3) and (2, -1).

Substitute x = 0 and 2 in (2).

Slope of the tangent is 3.

Equation of the tangent at (0, -3) with slope 3.

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (0, -3) and m = 3.

y - (-3) = 3(x - 0)

y + 3 = 3x

y = 3x - 3

Equation of the tangent at (2, -1) with slope 3.

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (2, -1) and m = 3.

y - (-1) = 3(x - 2)

y + 1 = 3x - 6

y = 3x - 7

Example 5 :

Find a cubic function in the form y = ax³ + bx² + cx + d  whose graph has horizontal tangents at the points (-2, 6) and (2, 0).

Solution :

Slope of the tangent :

dy/dx = 3ax² + 2bx + c

Slope at (-2, 6) :

dy/dx = 3a(-2)² + 2b(-2) + c

dy/dx = 12a - 4b + c

Slope at (2, 0) :

dy/dx = 3a(2)² + 2b(2) + c

dy/dx = 12a + 4b + c

Since the tangents are horizontal, slope = 0

The curve passes through the point (-2, 6).

a(-2)³ + b(-2)² + c(-2) + d = 6

-8a + 4b - 2c + d = 6 ----(3)

The curve passes through the point (2, 0).

a(2)³ + b(2)² + c(2) + d = 0

8a + 4b + 2c + d = 0 ----(4)

(1) - (2) :

(12a - 4b + c) - (12a + 4b + c) = 0

12a - 4b + c - 12a - 4b - c = 0

-8b = 0

b = 0

(3) + (4) :

(-8a + 4b - 2c + d) + (8a + 4b + 2c + d) = 6

8b + 2d = 6

Substitute b = 0.

2d = 6

d = 3

Substitute b = 0 in (2).

12a + 4(0) + c = 0

12a + c = 0 ----(5)

Substitute b = 0 and d = 3 in (4).

8a + 4(0) + 2c + 3 = 0

8a + 2c + 3 = 0 ----(6)

(6) - 2(5) :

(8a + 2c + 3) - 2(12a + c) = 0

8a + 2c + 3 - 24a - 2c = 0

-16a + 3 = 0

a = 3/16

Substitute a = 3/16 in (5).

12(3/16) + c = 0

9/16 + c = 0

c = -9/16

Substitute the values of a, b c in y = ax³ + bx² + cx + d. 

y = (3/16)x³ + 0x² - (9x/16) + 3

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