# FIND THE EQUATION OF THE TANGENT LINE TO THE CURVE AT THE GIVEN POINT

Tangent is also a line which touches the curve. To find the equation of tangent, we have to follow the given below.

i) Find the slope of the tangent drawn at the point (x1, y1) from the given equation of curve.

Use the formula,

y - y1 = m(x - x1)

ii)  Here m is the slope of the tangent line at the point of contact. (x1, y1) is the point of contact.

For each problem, find the equation of the line tangent to the function at the given point. Your answer should be in slope – intercept form.

Problem 1 :

y = x3 – 3x2 + 2 at (3, 2)

Solution :

y = x3 – 3x2 + 2

Differentiating with respect to x, we get

dy/dx = 3x2 - 6x

Slope at (3, 2)

dy/dx = 3(3)2 - 6(3)

dy/dx = 27 - 18

dy/dx = 9

Slope of tangent = 9

Equation of tangent :

(y - y1) = m(x - x1)

(y - 2) = 9(x - 3)

y - 2 = 9x - 27

y = 9x - 27 + 2

y = 9x - 25

Problem 2 :

Solution :

y = -5(x2 + 1)-1

Differentiating with respect to x, we get

dy/dx = -5(-1) (x2 + 1)-2 (2x)

dy/dx = 10x (x2 + 1)-2

Equation of tangent :

(y - y1) = m(x - x1)

Problem 3 :

y = x3 – 2x2 + 2 at (2, 2)

Solution :

y = x3 – 2x2 + 2

Differentiating with respect to x, we get

dy/dx = 3x2 - 4x

Slope at (2, 2)

dy/dx = 3(2)2 - 4(2)

dy/dx = 12 - 8

dy/dx = 4

Slope of tangent = 4

Equation of tangent :

(y - y1) = m(x - x1)

(y - 2) = 4(x - 2)

y - 2 = 4x - 8

y = 4x - 8 + 2

y = 4x - 6

Problem 4 :

Solution :

y = -3(x2 - 25)-1

Differentiating with respect to x, we get

dy/dx = -3(-1) (x2 - 25)-2 (2x)

dy/dx = 6x (x2 - 25)-2

Equation of tangent :

(y - y1) = m(x - x1)

Problem 5 :

Solution :

y = -3(x2 - 4)-1

Differentiating with respect to x, we get

dy/dx = -3(-1) (x2 - 4)-2 (2x)

dy/dx = 6x (x2 - 4)-2

Equation of tangent :

(y - y1) = m(x - x1)

Problem 6 :

Solution :

Differentiating with respect to x, we get

Equation of tangent :

(y - y1) = m(x - x1)

Problem 7 :

y = In (-x) at (-2, In 2)

Solution :

y = In (-x)

Differentiating with respect to x, we get

dy/dx = -1/-x

dy/dx = 1/x

Slope at (-2, In 2)

dy/dx = 1/-2

dy/dx = -1/2

Slope of tangent = -1/2

Equation of tangent :

(y - y1) = m(x - x1)

(y - In 2) = -1/2(x + 2)

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

y - In 2 =  -1/2 x - 1

y = -1/2 x + In 2 - 1

Problem 8 :

y = -2tan (x) at (-π, 0)

Solution :

y = -2tan (x)

Differentiating with respect to x, we get

dy/dx = -2 ⋅ sec2(x)

Slope at (-π, 0)

dy/dx = -2 ⋅ sec2(-π)

= -2 ⋅ 1

= -2

Slope of tangent = -2

Equation of tangent :

(y - y1) = m(x - x1)

y  - 0 = -2(x + π )

y - 0 = -2x - 2π

y = -2x - 2π

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