How do you find the equation of a line tangent to the function #y=(x-2)(x^2+1)# at x=-1?

2 Answers
Yahia M.
Apr 14, 2018

#y=8x+2#

Explanation:

Tangent Line to a Curve
simplify
#y=x^3-2x^2+x-2#
differentiate
#y'=3x^2-4x+1#
substitute for #x=-1#
#y'=8#
the value of #y'# is the slope of the tangent line at #x=-1#
and to get the point which lies on the tangent we substitute with #x=-1# in the function
#y=-6#
so the point on the line is#(-1,-6)# and its slope is 8
substitute in the following formula to get the equation

#y-y_1=m(x-x_1)#

#y+6=8(x+1)#

#y=8x+2#

I hope this was helpful.

Alex
Apr 14, 2018

#y = 8x + 2#

Explanation:

At #x = -1#:

#y = (-1-2)((-1)^2 + 1)#

#y = -3 xx 2#

#y = -6#

The tangent passes through #(-1,-6)#

The gradient:

#y = (x-2)(x^2 + 1)#

#y = x^3 - 2x^2 + x - 2#

#y' = 3x^2 - 4x + 1#

At #x = -1#:

#m = 3(-1)^2 - 4(-1) + 1#

#m = 3 + 4 + 1#

#m = 8#

Since we have the gradient of the tangent and a point through which it passes we can use the point-gradient formula:

#y - y_1 = m(x - x_1)#

#y +6 = 8(x +1 )#

#y = 8x + 8 - 6#

#y = 8x + 2#

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