In turning point form you have where is the turning point.
You should be able to find easily enough after substituting in these values.
I have a Quadratic with Minima (1,3) and y-intercept of 5. When I did this with quadratics with rational zeros I just reconstructed the polynomials off of those Zeros and then added the terms necessary to get the correct maxima/minima. Since I can not, or do not know, how to do this with an Irreducible Quadratic I'm running into trouble.
I have a Quadratic that satisfies the (1,3) and the y-intercept 5 requirements but the facing is incorrect. (-2x^2)+5 = f(#9)
Also, some what related, I assume that I could find my complex zeros by inspection (as this is the procedure that I am using to find rational zeros) if the drawing was set up correctly disregarding the fact that I can not visually discern between values that are near each other. Another way of putting this is to say; I assume that the complex plane can be "drawn". I'm I thinking of this correctly?