calculus
posted by nic .
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?

Let f(x)=5x³6x, then
f(1)=5(1)³6(1)=1
Slope at (x, f(x)) is given by dy/dx=f'(x)
f'(x) = dy/dx = 15x³6
f'(1)= 15(1)²6=9
Therefore you need a line with a slope of 9 to pass through the point (1,1).
The standard equation for this is:
(yy1)=m(xx1)
where m=9, y1=1, x1=1
(y(1)) = 9(x1)
y=9x91
y=9x10 (slope of 9 and passes through (1,1))
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