Tangent equation

Quote:

Originally Posted by totalnewbie

My aim is to compose a tangent equation for the line y=(x-1)*(x+1)*(x+2) provided that tangent points are intersections with x-axis.

Hello,

you know already the intersections between x-axis and curve:
(-2,0), (-1,0), (1,0).

Now you need the slope of the tangent lines:

Your function is $y=f(x)=(x-1)(x+1)(x+2)=x^3+2x^2-x-2$

The gradient of the function gives the slope of the tangents:

$f'(x)=3x^2+4x-1$

First slope: $m_1=f'(-2)=3 \cdot 4 +4 \cdot(-2) - 1=3$
2nd slope: $m_2=f'(-1)=3 \cdot 1 +4 \cdot(-1) - 1=-2$
3rd slope: $m_3=f'(1)=3 \cdot 1 +4 \cdot(1) - 1=6$

Now use the point-slope-formula and you'll get:
$t_1:y=3 \cdot (x+2)$
$t_2:y=-2 \cdot (x+1)$
$t_3:y=6 \cdot (x-1)$

Greetings

EB