Find the equation of the tangent line to y=xy = \sqrt{x}y=x at the point (4,2)(4, 2)(4,2).
y=14x+1y = \frac{1}{4}x + 1y=41x+1
y=12xy = \frac{1}{2}xy=21x
y=14x+2y = \frac{1}{4}x + 2y=41x+2
y=2x−6y = 2x - 6y=2x−6