There seems to be some technicalities with this problem, so I need your help.
Is graph of $f(x)= $$x^2 - 4 \sqrt{x}$ symmetric about the y-axis or is it technically not symmetric because it's only partly a parabola?
Does it have any asymptotes? I believe it has an x intercept at x=0, but when I graph it on a Ti-84 Plus C the graph does not intercept 0 and looks to have a vertical asymptote of 0.
And dependent upon whether or not it has an aymptote, would the interval of increase be 0 to infinity? Would the interval of concavity be the same as the interval increase?
Is the point of inflection the vertex in this case?
Thank you in advance, your help is always appreciated!
The graph is not symmetric about the y-axis because the function is not real for x < 0. The graph has no asymptotes. It has an x-intercept at x=0, but the Ti-84's resolution is not high enough to show it.
Differentiate the function to find the interval of increase:
$$\frac{dy}{dx}=2x-\frac{2}{\sqrt x}$$
Since the first derivative is greater than zero when x > 1, the function is increasing on the interval x > 1. For the interval of concavity, differentiate once more:
$$\frac{d^2y}{dx^2}=2+\frac{1}{x^\frac{3}{2}}$$
Since the second derivative is positive for all in the domain, the function is concave up on x > 0. Points of inflection occur when the second derivative changes sign, so this function has no points of inflection.