Posted by **nic** on Tuesday, January 25, 2011 at 10:30pm.

please help! Find an equation of the tangent line to the curve at the point (-1, 1).

y = 5x3 - 6x.

y=? ...............i keep getting y=x+2 and its wrong. can someone please help?

- calculus -
**MathMate**, Tuesday, January 25, 2011 at 11:34pm
Let f(x)=5x³-6x, then

f(-1)=5(-1)³-6(-1)=1

Slope at (x, f(x)) is given by dy/dx=f'(x)

f'(x) = dy/dx = 15x³-6

f'(-1)= 15(-1)²-6=9

Therefore you need a line with a slope of 9 to pass through the point (-1,1).

The standard equation for this is:

(y-y1)=m(x-x1)

where m=9, y1=-1, x1=1

(y-(-1)) = 9(x-1)

y=9x-9-1

y=9x-10 (slope of 9 and passes through (-1,1))

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