# Seeing complex roots on the graph of a polynomial

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.