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burhan101
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge y=e^{2x1} \] at x=1

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0dw:1371802792014:dw

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.3our 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:$$ye=2e(x1)\\y=2ex2e+e=2exe$$

burhan101
 one year ago
Best ResponseYou've already chosen the best response.0i think y= \[\huge 2e(x1)+e\]

Azteck
 one year ago
Best 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.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Then 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
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