# How do you find the equation of the line tangent to #y=x^3 - 2x# at the point (2,4)?

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To find the equation of the line tangent to the curve y=x^3 - 2x at the point (2,4), we need to find the slope of the tangent line at that point.

To find the slope, we take the derivative of the function y=x^3 - 2x with respect to x.

The derivative of y=x^3 - 2x is dy/dx = 3x^2 - 2.

Substituting x=2 into the derivative, we get dy/dx = 3(2)^2 - 2 = 12 - 2 = 10.

So, the slope of the tangent line at the point (2,4) is 10.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Substituting (2,4) and m=10 into the equation, we get y - 4 = 10(x - 2).

Simplifying the equation, we have y - 4 = 10x - 20.

Rearranging the equation, we get y = 10x - 16.

Therefore, the equation of the line tangent to y=x^3 - 2x at the point (2,4) is y = 10x - 16.

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