When I sketch the graph for a general second degree polynomial $y = ax^2 + bx + c$ it is easy to "see" its roots by looking at the points where $y=0$. This is true also for any $n$-degree polynomial.
But that's assuming the roots are real. For $y = x^2 + 10$, the solutions are complex and I (of course) won't find the zeros when $y=0$. My question is:
Is there some way to approximately find out what the complex roots of a polynomial $p(x)$ are by just looking at the graph (as we can for the real roots)?
I ask this mainly for second degree polynomials, but if there is a nice generalisation for higher degrees that would also be very much appreciated. I know about Argand diagrams
Suppose the root is $$a\pm bi$$ Then the polynomial will have a factor of the form $$(x-a-bi)(x-a+bi)=(x-a)^2-(bi)^2=(x-a)^2+b^2$$ So we will have for some polynomial $g$, $$f(x)=(x-a)^2g(x)+b^2g(x)$$ Thus $$f'(x)=2(x-a)g(x)+\big((x-a)^2+b^2\big)\,g'(x)$$ and $$f'(a)=b^2\,g'(a)$$ If f is quadratic, then g will be constant, so we will get $$f'(a)=0$$ In other words, for quadratics with complex roots, the roots correspond to the vertex of the parabola. If the solutions are $a\pm bi$, $a$ will be the x-coordinate of the vertex, and $b$ will be the function squared at that point (times a constant perhaps) because $f(a)=b^2g(a)=cb^2$ ($c=f''(x)$) from above.
If f is cubic, g will be linear, and we will get $$f'(a)=b^2(c)=c_1b^2$$ Since $$f(a)=b^2g(a)=b^2(ma+n)$$ we get $$f'(a)=\frac{c_1}{ma+n}f(a)$$ In other words, after finding $c_1,m,n$, you would need to look for points on the graph where the slope is roughly "inversely proportional" to the function value.
For higher order polynomials I'm not aware of there being any simple way to visually find roots.