FINDING THE EQUATION OF A TANGENT LINE

To find slope of the tangent line drawn at a particular point, we have to follow the procedure.

Find the slope of the tangent line at a point (x1, y1by finding  the derivative of the curve.

Apply the given point in the derivative to get the slope exactly a the point (x1, y1)

Using the formula 

(y - y1) = m (x - x1

We can find equation of the tangent line.

For each problem, find the equation of the tangent line to the function at the given point.

Problem 1 :

f(x) = x2 + 1; (1, 2)

A)  y = 2x

C)  y = (1/2)x + 3/2

B)  y = -8x + 10

D)  y = -6x + 8

Solution :

Given, f(x) = x2 + 1

f'(x) at (1, 2)

x = 1 and y = 2

f'(x) = m = 2x 

m = 2(1)

Slope m = 2

Equation of a tangent line is y - y1 = m(x - x1)

x1 = 1 and y1 = 2

y - 2 = 2(x - 1)

y - 2 = 2x - 2

y = 2x - 2 + 2

y = 2x 

Equation of a tangent line is y = 2x.

So, option A) is correct.

Problem 2 :

f(x) = 2x2 + 2x + 2; (-1, 2)

A)  y = -8x – 6

C)  y= (1/2)x + 5/2

B)  y = -2x

D)  y = 6x + 8

Solution :

Given, f(x) = 2x2 + 2x + 2

f'(x) at (-1, 2)

x = -1 and y = 2

f'(x) = m = 4x + 2 

m = 4(-1) + 2

m = -4 + 2

Slope m = -2

Equation of a tangent line is y - y1 = m(x - x1)

x1 = -1 and y1 = 2

y - 2 = -2(x + 1)

y - 2 = -2x - 2

y = -2x - 2 + 2

y = -2x 

Equation of a tangent line is y = -2x.

So, option B) is correct.

Problem 3 :

y = x2 + x - 2; (1, 0)

A) y = 34x - 34

            C) y = 3x - 3

B) y = -32x + 32

       D)  y = -12x + 12

Solution :

Given, y = x2 + x - 2

y' at (1, 0)

x = 1 and y = 0

y' = m = 2x + 1 

m = 2(1) + 1

m = 2 + 1

Slope m = 3

Equation of a tangent line is y - y1 = m(x - x1)

x1 = 1 and y1 = 0

y - 0 = 3(x - 1)

y - 0 = 3x - 3

y = 3x - 3

Equation of a tangent line is y = 3x - 3.

So, option C) is correct.

Problem 4 :

y = 2x2 + 1; (-1, 3)

A)  y = 2x + 5        B)  y = -4x – 1     C)  y = 12x + 15

Solution :

Given, y = 2x2 + 1

y' at (-1, 3)

x = -1 and y = 3

y' = m = 4x 

m = 4(-1)

Slope m = -4

Equation of a tangent line is y - y1 = m(x - x1)

x1 = -1 and y1 = 3

y - 3 = -4(x + 1)

y - 3 = -4x - 4

y = -4x - 4 + 3

y = -4x - 1

Equation of a tangent line is y = -4x - 1.

So, option B) is correct.

Problem 5 :

f(x)= 1x + 1 ; (0, 1)

A) y = (1/4)x + 1

C) y = -x + 1

B)  -2x + 1

D)  y = 1

Solution :

Given, f(x)= 1x + 1 f'(x) at (0, 1)x = 0 and y = 1f'(x) = m = -1(x + 1)2 m = -1(0 + 1)2 m = -1

Equation of a tangent line is y - y1 = m(x - x1)

x1 = 0 and y1 = 1

y - 1 = -1(x - 0)

y - 1 = -1x

y = -x + 1

Equation of a tangent line is y = -x + 1.

So, option C) is correct.

Problem 6 :

f(x)= -1 x ;2, -12
A ) y = - 34x + 1
C) y = 34x - 2
B) y = 14x - 1
D) y = x - 52

Solution :

Given, f(x)= -1x f(x) = -x-1f'(x) at 2, -12x = 2 and y = -12f'(x) = m = -x-1= -(-x)-1 - 1m = x-2m = 1x2m = 122m = 14Equation of a tangent line is y - y1 = m(x - x1)y - -12 = 14(x - 2)y + 12 = 14 - 24y = 14x - 24 - 12y = 14x - 24 - 24y = 14x - 44y = 14x - 1So, Equation of a tangent line is y = 14x - 1.

Hence, option B) is correct.

Problem 7 :

f(x)= -1x + 3 ; (-2, -1)

A)  y = -1 

C)  y = -(1/2)x - 2

B)  y = x + 1

D)  y = 4x + 7

Solution :

Given, f(x)= -1x + 3 f'(x) at (-2, -1)x = -2 and y = -1f'(x) = m = -1(x + 3)2 m = -1(-2 + 3)2 m = 11m = 1

Equation of a tangent line is y - y1 = m(x - x1)

x1 = -2 and y1 = -1

y  + 1 = 1(x + 2)

y = x + 2 - 1

y = x + 1

Equation of a tangent line is y = x + 1.

So, option B) is correct.

Problem 8 :

f(x)= 1x - 1 ; -1, -12
A) y = -14x - 34
B) y = -112x - 712
C) y = 112x - 512

Solution :

Given, f(x)= 1x - 1 f'(x) at -1, -12x = -1 and y = -12f'(x) = m = -1(x - 1)2 m = -1(-1 - 1)2 m = -1(-2)2m = -14Equation of a tangent line is y - y1 = m(x - x1)y + 12 = -14(x + 1)y + 12 = -14x - 14y = -14x - 14 - 12y = -14x - 14 - 24y = -14x - 34So, equation of a tangent line is y = -14x - 34.

So, option A) is correct.

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