How do you solve #y > sqrt(x + 3)#?

1 Answer
Apr 16, 2016

Draw the parabola #y^2=x+3#, with vertex #V(-3, 0)#. Draw the tangent at the vertex, #x =-3#. Shade the region in the first two quadrants, between x=-3 and the parabola, for solution {(x. y)}...

Explanation:

#y=+sqrt(x+3)# is the equation of the half of the parabola #y^2=x+3#, above the its axis, the x-axis. The other half is represented by #y=-sqrt(x+3)#,

Here, x>-3.
For points above the parabola and the right side of the tangent at the vertex #x = -3#, #y>+sqrt(x+3)#.