# Thread: Constructing the equation of an Irreducible Quadratic from it's graph.

1. ## Constructing the equation of an Irreducible Quadratic from it's graph.

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?

2. In turning point form you have $\displaystyle y = a(x-k)^2+h$ where $\displaystyle (k,h)$ is the turning point.

You should be able to find $\displaystyle a$ easily enough after substituting in these values.

3. did you get $(x-1)^2=(4)\left$$\frac{1}{8 }\right$$ (y-3)$

or $y=2x^2-4x+5$