FINDING EQUATION OF TANGENT LINE WITH DERIVATIVES

Example 1 :

Find the equation of the tangent line at x = 5 to the curve f(x) = √(2x - 1).

Solution :

f(x) = √(2x - 1)

Substitute x = 5.

f(5) = √9

= 3

y = 3

The point on the curve at where the tangent line is drawn is (5, 3).

Find the slope of the tangent line at ay x = 3 :

f(x) = √(2x - 1)

f'(x) = {(1/[2√(2x-1)]} ⋅ 2

f'(x) = 1/√(2x - 1)

Substitute x = 5.

f'(5) = 1/√9

slope (m) = 1/3

Equation of the tangent line :

y - y₁ = m(x - x₁)

Substitute m = 1/3 and (x₁, y₁) = (5, 3).

y - 3 = (1/3)(x - 5)

3y - 9 = x - 5

x - 3y - 5 + 9 = 0

x - 3y + 4 = 0

Example 2 :

If the line y = 3x + 9 is tangent to the graph of the function f at (2, 15), what is f'(2) ?

Solution :

Equation of the tangent line y = 3x + 9 is in slope-intercept form. 

Comparing y = mx + b and y = 3x + 9, we get

slope (m) = 3

The first derivative of a function always represents the slope.

Since the tangent line is drawn at (2, 15),

slope at (2, 15) = 3

f'(2) = 3

Example 3 :

What is the x-coordinate of the point where the tangent line to the curve y = x² + 12x + 11 is parallel to x-axis.

Solution :

y = x² + 12x + 11

Find the first derivative to get the slope of the tangent line.

dy/dx = 2x + 12

Slope of any line which is parallel to x-axis is equal to zero.

Since the tangent line is parallel to x-axis, its slope is equal to zero. 

dy/dx = 0

2x + 12 = 0

2x = -12

x = -6

Example 4 :

Find the equation of the tangent line which goes through the point (2, -1) and is parallel to the line given by the equation 2x - y = 1.

Solution :

2x - y = 1

Write the above equation in slope-intercept form :

-y = -2x + 1

y = 2x - 1

Comparing y = mx + b and y = 2x - 1, we get

slope (m) = 2

Since the tangent line is parallel to the line 2x - y = 1, the slopes are equal.

slope of the tangent line = 2

The tangent line is passing through the point (2, -1) with the slope 2.

Equation of the tangent line :

y - y₁ = m(x - x₁)

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

y + 1 = 2(x - 2)

y + 1 = 2x - 4

2x - y - 5 = 0

Example 5 :

For a particular g, we know g'(5) = 2 and g(5) = 3. Write the equation of a tangent to g at x = 5.

Solution :

g'(5) = 2  and  g(5) = 3

Slope of the tangent to g at x = 5 is 2.

We draw the tangent line at the point (5, 3).

Equation of the tangent line :

y - y₁ = m(x - x₁)

Substitute m = 2 and (x₁, y₁) = (5, 3).

y - 3 = 2(x - 5)

y - 3 = 2x - 10

2x - y - 7 = 0

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