Find the slope of a straight line that passes through the point (-2,0) and is tangent to the curve y = √x

y' = 1/(2√x)

y = √x
√x => x≥0
m=y`=1/[2√x]

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


√x=m(x+2)
√x=[1/[2√x]](x+2)
2|x|=x+2
|x|=x , x≥0
2x=x+2 => x=2
m=y`(2)=1/[2√2] => y=m(x+2)=(x+2)/[2√2]

y=(x+2)/[2√2]
http://www.wolframalpha.com/input/?i=y+%…

Answer
m=1/[2√2]