Find the equation of the line tangent to the graph of $f$ at $(1, -3)$, where $f$ is given by $f(x) = 6x^3 − 13x^2 + 4$. Use $y$ as the dependent variable when you write your equation.
My answer which was false:
The $f'(x)= 18x^2-26x$
$f'(1) = 18-26 = -8$
the equation $(y+3)=-8(x-1)$
$y= -8x+5 $
but it was false, so could any one help please :)
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