Question

find the coordinates of the point where the tangent to the curve y=x^3 +x +2 at the point (1,4) meets the curve again. [ans:-2,-8]
pls help me i don't understand the question....

Answer 1

dy/dx = 3x^2 + 1
at (1,4) , dy/dx = 4
tangent equation
4 = 4(1) + b
b = 0
the tangent equation at (1,4) is y = 4x
where does this cross or touch y = x^3 + x + 2 ?
x^3 + x + 2 = 4x
x^3 - 3x + 2 = 0
let f(x) = x^3 - 3x + 2
f(1) = 1 - 3 + 2 = 0 , so x-1) is a factor

by synthetic division ...
x^3 - 3x + 2 = (x-1)(x^2 + x - 2)
= (x-1)(x-1)(x+2)

then the roots of x^3 - 3x + 2 = 0 are
x = 1 and x = -1 and x = -2
we knew x = 1 from the tangent part, so the new x = -2 is the other intersection point value
if x = -2
y = -8 - 2 + 2 = -8

the other point is (-2,-8)