# What is the equation of the line tangent to # f(x)=2x^3-x^2+3x # at # x=3 #?

The equation is

We figure out the initial derivative.

The tangent's equation is

The tangent is

graph{(y-51x+99)=0 [-37.5, 35.6, -4.3, 32.23]}=(y-2x^3-x^2+2x)

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Finding the gradient is always the first step in determining the tangent or normal to a line.

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The equation of the line tangent to f(x)=2x^3-x^2+3x at x=3 is y=32x-33.

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