y=sqrt(x^2-5x+4)
find:
- local min/max
- inflection points
- horizontal/vertical asymptote
- inc/dec
y = √(x^2-5x+4) = √((x-1)(x-4))
y' = (2x-5)/(2√(x^2-5x+4))
so vertical tangents at x = 1,4
only defined for x<=1, x>=4
No asymptotes, min or max
increasing where y' > 0 ... x > 4
decreasing where y' < 0 ... x < -1
Note that this is just the top half of a hyperbola: (x - 5/2)^2 - y^2 = 1/4