f(x) = x^4 + 18x^3 + 81x^2
You can sketch a polynomial function if you know its zeros.
Factor out the Greatest Common Factor:
f(x) = x^2(x^2 + 18x + 81)
The trinomial factor looks like a perfect square trinomial; let's try factoring it:
f(x) = x^2(x + 9)(x + 9)
f(x) = x^2(x + 9)^2
0 = x^2(x + 9)^2
Use Zero Product Property:
x = 0 with multiplicity 2, and
x = –9 with multiplicity 2
Graph the two zeros as points on the x-axis. I like to put a little 2 above each to remind me that the multiplicity is 2.
Now determine the right side end behavior.
As x → +∞, f(x) → +1*(+∞)^4 → +∞.
So end behavior is up to the right, and since the degree is even (4), the end behavior to the left is also up.
[Use lines and parabolas to remember the relationship between left and right end behaviors of odd (line) and even (parabola) polynomial functions.]
Both zeros have even multiplicity so the graph will "bounce" off the x-axis at each zero, the graph will not pass through the x-axis.
[A multiplicity greater than 1 causes graph to flatten near the zero, for both even and odd multiplicities. Odds go through the x-axis, evens "bounce" off x-axis.]
So we can sketch something like this:
\ ^ /
\ | /
\ /\ | /
\ / \ | /
\ / \ |/
I can't round the turning points with this sketch, but you should do so on graph paper.
If you want to "drive some nails" to define the function better, you can find some points; e.g., (1,100), (-1,64).
Use graphing calculator or GeoGebra: