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Let $y=(x^3-25x)^8$ at the point $(-5,0)$. I know that you have to find the first derivative, but I dont know where to go from there.

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Equation of the tangent line of $f(x)=(x^3-25x)^8$ at $(-5,0)$ is $$y−y_0=f′(x_0).(x−x_0)$$ where $y_0=0,x_0=-5$. $$f'(x)=8(x^3-25x)^7(3x^2-25)\Rightarrow f'(-5)=0$$ So equation of the tangent line is $$y-0=0(x+5)\Rightarrow y=0$$

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Once a value for the slope $m$ is known from the derivative at the given point, use the parameterized definition of a line to get values for x and y in terms of the parameter t and the tangent point (-5,0):

$x = 1 * t - 5$
$y = m * t + 0$

If desired, eliminate t from this system of equations to obtain a standard equation fo the line just in terms of $x$ and $y$.

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