# Thread: equation of a tangent

1. ## equation of a tangent

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

2. 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
$\displaystyle \displaystyle \frac{1}{4} = \lim_{h\to 0 }\frac{\sqrt{x_{1}+h}-\sqrt{x_{1}}}{h} \cdot \frac{\sqrt{x_{1}+h}+\sqrt{x_{1}}}{\sqrt{x_{1}+h}+ \sqrt{x_{1}}}$

ring any bells?

3. So I would then simplify from there?

4. Originally Posted by nightrider456
So I would then simplify from there?
I gave you the first step in finding the limit on the right hand side of the equation ... finish it by finding that limit, set it equal to 1/4 and solve for $\displaystyle x_1$.

5. Are you required to use the definition of limit? If you know the "power rule", that the derivative of $\displaystyle x^r$, where r is a constant, is $\displaystyle rx^{r-1}$, then you can use that with $\displaystyle r= \frac{1}{2}$.