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

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, user21820, Jack, kingW3, Did
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. :-) – kingW3 Feb 8 '17 at 18:06
• Iv'e edited your post. Please ensure it is written as supposed to be. – Galc127 Feb 8 '17 at 18:07

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 $\frac{dy}{dx}=\frac{1}{3}$.

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 $(9/4, 5)$.