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3) Find an equation of the tangent line to the graph of y = 3x2 – ex + 2 at (0,1)
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\[y = 3x ^{2} – e ^{x} + 2\]
ghass, thanks soo mch for replying to the translating probelm! you are a genius!
Do you recall the geometrical meaning of the derivative?
yes
first differentiate the equation,f y = 3x^2 – e^x + 2 you'll get the slope of the tangent y'=6x-e^x y']at (x=0)=-1 so the slope of the line =-1 then the equation is => y-1=-1(x-0) => y-1 =-x
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and so , x+y-1=0 is the tangent
thanks!!
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