Find an equation for the tangent line to the graph of y=(x^3-25x)^8 at the point (5,0)?

Thomas2015-02-19T04:55:40Z

dy/dx=8(x^3-25x)^7(3x^2-25)

dy/dx(5)=0

so y=mx+b and we know m=0

0=5(0)+b and b=0 so in this case...

y=0 The tangent line is the x axis :)

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