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burhan101Best ResponseYou've already chosen the best response.0
\[\huge y=e^{2x1} \] at x=1
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
dw:1371802792014:dw
 10 months ago

oldrin.batakuBest ResponseYou've already chosen the best response.3
our tangent passes through point \((1,e)\) with slope determined by:$$y'=e^{2x1}\cdot2=2e^{2x1}\\y'(1)=2e$$thus we just put it all together:$$ye=2e(x1)\\y=2ex2e+e=2exe$$
 10 months ago

burhan101Best ResponseYou've already chosen the best response.0
i think y= \[\huge 2e(x1)+e\]
 10 months ago

AzteckBest ResponseYou've already chosen the best response.0
@burhan101 When they give you the x value of the point, you should start finding the yvalue for the point by plugging the xvalue which is 1, into the curve's equation. Once you've done that and found the derivative (which you've done), you can find the gradient of the tangent by plugging the point into the derivative.
 10 months ago

AzteckBest ResponseYou've already chosen the best response.0
Then you can use this to find your equation of the tangent: \[xx_{1}=m(yy_{1})\] where: \[(x_{1}, y_{1})\] is the point given and m is the gradient of the tangent
 10 months ago
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