#### Answer

$y=\frac{1}{4}x+1$

#### Work Step by Step

To find the equation of a tangent to the function $f(x)=\sqrt x$, it might be useful to first find the derivative function.
$f(x)=\sqrt x=x^\frac{1}{2}$
$f'(x)=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^\frac{-1}{2}=\frac{1}{2\sqrt x}$
To find the x value at which the function has a slope of $\frac{1}{4}$, we equate the derivative function to $\frac{1}{4}$ and solve for $x$.
$\frac{1}{4}=\frac{1}{2\sqrt x}$
Taking the reciprocal of both sides.
$4=2\sqrt x$
$2=\sqrt x$
$x=4$
As such, it is the coordinate $(4,f(4))=(4,2)$ that has a slope of $\frac{1}{4}$.
Solving for the y intercept of the tangent line;
$\frac{2-c}{4-0}=\frac{1}{4}$
$c=1$
As such, the equation of the tangent is $y=\frac{1}{4}x+1$