Compute the equation of the line which is tangent to the curve given by?

Steps:

1. Find the derivative of the function -- f'(x) --

2. Plug in the x value in the derivative function to find the gradient of the tangent -- dy/dx --

3. use y-y1 = m(x-x1) or y = mx + c to find the equation of the tangent

It is important to know that:

1. Constants are zero after you have found the derivative, ie the - 1 will become zero o

2. The derivative is the same as the gradient; and is found using the form nx^(n-1) where n is the power, in this case 2.

3. Anything to the power of o is 1

f(x) = -3x^2 + 2x - 1

f'(x) derivative function

f'(x) = (-3)(2)(x^(2-1)) + (2)(1)(x^(1-1)) - (1)

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

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

f'(x) = - 6x + 2

Using the value of x = 1 the gradient of the tangent (dy/dx) at the given point is:

dy/dx = - 6x + 2

dy/dx = - 6(1) + 2

dy/dx = - 6 + 2

dy/dx = - 4

To find the y value at the point where the gradient is -4 and the x value is 1;

y = - 3x^2 + 2x - 1

y = - 3(1)^1 + 2(1) - 1

y = - 2

Using:

gradient = - 4

x coordinate = 1

y coordinate = - 2

Subsititue into y = mx + c to find the y intercept

- 2 = (- 4)(1) + c

- 2 = - 4 + c

- 2 + 4 = c

2 = c

Therefore the equation of the line is y = mx + c

y = - 4x + 2

f'(x) = -6x + 2.

f'(1) = -6(1) + 2 = -4. So this is m = slope of the tangent line.

f(1) = -3 + 2 - 1 = -2. So your (x, y) point is at (1, -2).

Plug into y = mx + b.

-2 = (-4)(1) + b.

-2 = -4 + b.

b = 6.

Equation: y = -4x + 6.

First you find the derivative of the function:

f´(x) = -6x + 2... now you substitute the point f`(1) = -4 ... this is the angular coefficient of the line.

Now you may use the formula: y - y1 = m(x - x1) , where (x1,y1) are the coordinates of the point and m is the angular coefficient.

y-(-2) = -4(x-1)... and you can write in the form: y = -4x +2