what (in slope form) is the equation for a tangent line where y=2X^2 with the point (1,2)

Answer 1

The slope of the function y=2x^2 is given by the derivative dy/dx = 4x. If we evaluate dy/dx at the point (1,2) we get dy/dx = 4 because x = 1 at (1,2). We now know the slope of this line and a point it goes through so we can put it into point slope form, so y-2 = 4(x-1). Putting this into slope-intercept form we get y=4x-2.

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