Question

The tangent to y=x^3 + ax^2 - 4x + 3 at x=1 is parallel to the line y=3x. Find the value of a and the equation of the tangent at x=1. Where does the tangent cut the curve again?

Answer 1

`a=1`, tangent is `y=3x-1` and it cuts the curve again at `(-4,-13)`

Explanation:

As slope of tangent to a curve `y=f(x)` is given by `f'(x)` or `(dy)/(dx)`, the tangent to a curve `y=f(x)` at `x=x_0` is

`(y-f(x_0))=f'(x_0)(x-x_0)`

Here we have `y=x^3+ax^2-4x+3` and hence `f'(x)=3x^2+2ax-4`

and hence at `x=1`, `y=f(1)=1+a-4+3=a` and `f'(1)=3+2a-4=2a-1`

and equation of tangent is `y-a=(2a-1)(x-1)`

or `y=2ax-2a-x+1+a` or `y=(2a-1)x-a+1`

as it is parallel to `y=3x`, we have `2a-1=3` or `a=2`

and equation of tangent is `y=3x-1`

and equation of curve is `y=x^3+2x^2-4x+3`

to fond point of intersection let us put `y=3x-1` in it and we get

`3x-1=x^3+2x^2-4x+3` or `x^3+2x^2-7x+4=0`

i.e. `(x-1)(x^2+3x-4)=0` or `(x-1)(x+4)(x-1)=0`

Hence it will cut the curve again at `x=-4`,

where `y=(-4)^3+2(-4)^2-4(-4)+3=-13` i.e. at `(-4,-13)`

graph{(x^3+2x^2-4x+3-y)(y-3x+1)=0 [-39.83, 40.17, -25.12, 14.88]}