We know that $y=2e^{0.5x}$, and that the tangent line is a line that passes through the origin.

So $y'=e^{0.5x}$, therefore we know that $l(x)=e^{0.5a}(x-a)+2e^{0.5a}$ by the equation for calculating a tangent line.

But how do I find $a$, that is to say the value of $x$ that the tangent line shears through $y=2e^{0.5x}$? And how do I find the point the tangent line shears through? I know the tangent line to be $ex$.

This question is part of a problem that was given in a Mathematics 4 course on a Swedish Gymnasium.

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    $\begingroup$ You are looking for the point on the curve for which the tangent line passes through the origin? Well, in that case what is $l(0)$? $\endgroup$ – John Brevik Dec 5 '17 at 21:28
  • $\begingroup$ @JohnBrevik Ah, yes of course. Thank you! $\endgroup$ – ಠ ಠ Dec 5 '17 at 21:47

$(y - y(a)) = y'(a) (x-a)$ is a generic equation for the tangent line.

It goes through zero if $(0 - y(a)) = y'(a) (0-a)\\ y(a) = ay'(a)$

And for this curve.

$2e^\frac a2 = a e^\frac a2\\ a = 2$


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