FINDING COORDINATES OF TANGENT WHEN SLOPE IS GIVEN

Problem 1 :

Find the coordinates of the point (s) on 

f(x) = x² + 3x + 5

where the tangent is horizontal

Solution :

f(x) = x² + 3x + 5

Since the tangent is horizontal, its slope will be 0

f'(x) = 2x + 3(1) + 0

f'(x) = 2x + 3

f'(x) = 0

2x + 3 = 0

2x = -3

x = -3/2

Applying the value of x in the given function, we get

f(-3/2) = (-3/2)² + 3(-3/2) + 5

= (9/4) - (9/2) + 5

= (9 - 18 + 20)/4

= 11/4

So, the coordinates of the tangent is (-3/2, 11/4).

Problem 2 :

Find the coordinates of the point (s) on 

f(x) = √x

where the tangent has gradient 1/2

Solution :

f(x) = √x

f'(x) = 1/2√x

Given f'(x) = 1/2

1/2√x = 1/2

2 = 2√x

√x = 1

x = 1

When x = 1, y = √1

So, the required point is (1, 1).

Problem 3 :

Find the coordinates of the point (s) on 

f(x) = x³ + x² - 1

where the tangent has gradient 1

Solution :

f(x) = x³ + x² - 1

f'(x) = 3x² + 2x - 0

f'(x) = 3x² + 2x

Given that f'(x) = 1

3x² + 2x = 1

3x² + 2x - 1 = 0

(3x - 1) (x + 1 ) = 0

x = 1/3 and x = -1

So, the required points are (1/3, -23/27) and (-1, -1).

Problem 4 :

Find the coordinates of the point (s) on 

f(x) = x³ - 3x + 1

where the tangent has gradient 9

Solution :

f(x) = x³ - 3x + 1

f'(x) = 3x² - 3(1) + 0

f'(x) = 3x² - 3

Here f'(x) = 9

3x² - 3 = 9

3x² = 12

x² = 4

x = 2 and -2

So, the required points are (2, 3) and (-2, -1).

Problem 5 :

Find a if,

f(x) = ax² + 6x - 3

has a tangent with gradient with gradient 2 at the point where x = -1

Solution :

f(x) = ax² + 6x - 3

f'(x) = 2ax + 6 (1)

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

2a(-1) + 6 = 2

-2a = 2 - 4

-2a = -2

a = 1

Problem 6 :

Find a if,

f(x) = a/x

has a tangent with gradient with gradient 3 at the point where x = 2

Solution :

f(x) = a/x

f'(x) = a/x²

f'(x) = 3, at x = 2

3 = a/2²

a = 3(4)

a = 12