How do you find the equation of the tangent line to the curve #y=x^2+2e^x# at (0,2)?

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Answer 1 · 2017-02-08T08:43:55

#y=2x+2#

#"we use the formula "y-y_1=m(x-x_1)" where:" #

#(x_1,y_1) " is a known co-ordinate on the line"#

#"& "m=" the gradient of the tangent"#

#y=x^2+2e^x " at "(0,2)#

#m=[(dy)/(dx)]_(x=0)#

#y=x^2+2e^x=>(dy)/(dx)=2x+2e^x #

#m=[(dy)/(dx)]_(x=0)=2xx0+2e^0#

#m=2#

#:."eqn of tgt. "y-y_1=m(x-x_1)#

#y-y_1=m(x-x_1)#

#y-2=2(x-0)#

#y-2=2x=>y=2x+2#