[ATTACH=full]3439[/ATTACH] Question 26 y = sqrt{2x - 1} Let x = 0 y = sqrt{2(0) - 1} y = sqrt{0 - 1} y = sqrt{-1} The sqrt{-1} = I. I will say no y-intercept over the real numbers. Let y = 0 0 = sqrt2x - 1 } (0)^2 = [sqrt{2x - 1}]^2 0 = 2x - 1 1 = 2x 1/2 = x The x-intercept is 1/2.