The given function cannot pass through (-2,0) for the simple reason that it isn't defined there at all.

It is the line whose equation we have to find which passes through the given point.

suppose the line is of the form y=mx+c

m is its slope and c is its y-intercept

As (-2,0) satisfies this line,

0=m(-2)+c

hence c=2m

so the equation is y=mx+2m

In order to find m, we use the fact that the line y=mx-2 is tangent to the curve y=sqrt(x) at some point which is not known to us.

So, for some x,

the slope m= derivative of sqrt(x)

as you've calculated,

m=0.5/sqrt(x) for some particular x.

so 1/sqrt(x) +0.5sqrt(x)=sqrt(x)

0.5sqrt(x)=1/sqrt(x)

so x=2

So m=0.5/sqrt(2)

m=1/2sqrt(2)

hence the equation is y=1/2sqrt(2) + 1/sqrt(2)

x-2sqrt(2)y+2=0

EDITED!