Originally Posted by

**nightrider456** This is probably pretty easy but i am confused with how to continue

The problem: Find an equation of the straight line having a slope $\displaystyle \frac{1}{4}$ that is tangent to the curve $\displaystyle y=\sqrt{x}$

I pluged some values in to a formula my prof. gave me $\displaystyle \frac{f(x_{1}+h)-f(x_{1})}{h}$

Here is what I have so far:

$\displaystyle \frac{1}{4}=\lim_{h\rightarrow0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h}$

But i don't know where to go from here

Thanks for looking