Question

What is the equation of the line tangent to `f(x)=2x^3 - x^2-3x` at `x=-1`?

Answer 1

y = 5x + 5

Explanation:

The equation of the tangent is in the form y-b=m(x-a)

where m is gradient and (a,b) a point on the line. These are
required to be found.

The derivative f'(x) is the gradient of the tangent to the curve

and f'(-1) will give it's value. the x-coord a , is given x=-1 and

the y-coord, b , can be found by evaluating f(-1).

f'(x) `= 6x^2 -2x - 3`

and f'(-1) `= 6(-1)^2 - 2 (-1) -3 = 6 + 2 - 3 = 5 =` m

`f(-1)=2(-1)^3-(-1)^2-3(-1)=-2-1+3=0`

equation: y-0 = 5 (x+1) `rArr y = 5x + 5`