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Given the curve $y = \sqrt{x}+ 2$. Find a point on the curve where the tangent line is parallel to $y = \frac{1}{3}x − 10$.

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closed as off-topic by Namaste, user21820, Jack, kingW3, Did Oct 2 '17 at 15:47

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    $\begingroup$ Welcome to Math.SE! You're much more likely to get help if you let people know what you've tried and where you're getting stuck; that way, we can address the ACTUAL question you have, instead of just solving your problem for you. :-) $\endgroup$ – kingW3 Feb 8 '17 at 18:06
  • $\begingroup$ Iv'e edited your post. Please ensure it is written as supposed to be. $\endgroup$ – Galc127 Feb 8 '17 at 18:07
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Hint:

The gradient of the other line is clearly $m=\frac{1}{3}$.

Evaluate the derivative with respect to $x$ of the curve $y=\sqrt{x}+2$ and let $\frac{dy}{dx}=\frac{1}{3}$.

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The slope of tangent of the curve $y = \sqrt{x} + 2$ is given by the derivative of $y$ with respect to $x$, viz.

$$ \frac{dy}{dx} = \frac{1}{2\sqrt{x}}. $$

The other curve, $y = \frac13 x - 10$, is a straight line given in the form $y=mx+c$, where $m$ is the slope of the line. For the tangent of the first curve to be parallel to the straight line,

$$ m = \frac{dy}{dx} \\ \implies x = \frac94. $$

Hence, the corresponding point on the curve is $(9/4, 5)$.

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