Question

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?

Answer 1

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))