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anonymous
 3 years ago
Determine the equation of the tangent to the curve
y=
anonymous
 3 years ago
Determine the equation of the tangent to the curve y=

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1371802792014:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0our 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:$$ye=2e(x1)\\y=2ex2e+e=2exe$$

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

anonymous
 3 years 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.

anonymous
 3 years 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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