Question
An equation of the tangent line to the curve y=2x^3-x^2+2 at the point (1,3) is? So my approach is first calculate the derivative of y=2x^3-x^2+2
An equation of the tangent line to the curve y=2x^3-x^2+2 at the point (1,3) is?
So my approach is first calculate the derivative of y=2x^3-x^2+2 which is y'= 6x^2-2x, and then plug in (1,3) to get (1,4)
I know the correct answer using the y-yo=a(x-xo) and a should indicate the slope? But why is a 4 here? Can you explain that to me please?
Thank you very much for your help
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Bayesian Data Analysis
Authors: Andrew Gelman, John B Carlin, Hal S Stern, David B Dunson, Aki Vehtari, Donald B Rubin
3rd Edition
