Question

Find the derivative of y=2x^2-3x+1 and the slope of the curve at the point where x=3

Answer 1

The slope of the tangent line to y(x) at any given x is equal to the derivative dy/dx at that point. Differentiate to get

`(\mathrm dy)/(\mathrm dx)=\boxed{4x-3}`

and evaluate it at x = 3 to get the slope, 4•3 - 3 = 9.