How would the graph of $$y=1-\sqrt[3]{x+3}$$ look?

Question

mani_jha · Answer

A curve of course. Put x=-3, and you get 0.
So, the curve cuts the x-axis at (-3,0).
Similarly, put y=0, and you get the point at which the curve cuts the y-axis.
Another way of analysis would be to differentiate the function with respect to x. That would give you the slope at any point of the curve.
See in what region the slope is positive and in what region the slope is negative. Also, see where the slope becomes zero(That gives the peak point of the curve either above or below the x-axis).
First differentiate it, and post the result.

campbell_st · Answer

a vertical asymptote exists at x = -3