I used to think that the Gradient Method is for plotty functions in 2 variables. However, this answer may give you some pointers.
You could start by examining the function domain. In your case, all $x$ values are valid candidates. next, set $x=0$ then $y=0$ to get the intercepts. Setting $x=0$, yields $y=2$, so the point $(0,2)$ is on your graph. Now setting $y=0$ means that you need to solve the following for $x$:
Solving such equations is sometimes obvious at least for the first root, but in this case, it is not. You either follow a numeric method or use a calculator such as Cubic Eqn Solver or use the formula in Wolfarm-Cubic equation formula or Wiki-Cubic Function.
To get a plot of the function we'll just use the real root value found be either the above methods so the point $(-1.35320,0)$ is on the function graph as well. Note that the other 2 roots are complex, hence the function intersects the x-axis only once.
No we can move to find the critical points so that you can determine concavity of the function. Using the derivative method for testing, you can determine the local minimum and maximum of the function. The subject is a bit lengthy to include in detail here. I suggest you read about it in a book such as (page 191 and above) of: Google Books - Calculus of single variable.
In your case, the first derivative has no real roots. The second derivative
has a root at $x=-0.333$, this indicates that $(-0.333,1.741)$ is a point of inflection.
To further study the shape, take 2 points immediately before and after the point of inflection to determine the shape of the curve around this point. Use the obtained information so far together with few other points to determine the approximate curve shape.
There is a free web based graph plotter at: Desmos-Graph Plotter - Calculator, that may be useful for you also.
Here is a sample showing the function and its first derivative: