Posted by **rita** on Friday, July 22, 2011 at 9:34am.

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....

- math -
**Reiny**, Friday, July 22, 2011 at 10:45am
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)

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