Find an equation of the parabola with zeros $0$ and $6$ and a minimum value of $-9$ The zeros of a quadratic relation are $0$ and $6$. The relation has a minimum value of $-9$. Find an equation of the parabola.
Hello, I've tried to come up with a solution to the exercise but it's complicated. I don't really know if it's right or not.
The quadratic equation is: $ax^2+bx+c=0$
Then we have $3$ points: $(0,0),(0,6),(x,-9)$
\begin{align*}   
c & = 0\\
36a+6b+c & = 0\\
-b^2+4ac+36a & = 0
\end{align*}
The $y$-coordinate of the vertex is $-\Delta/4a$.
Then we will use $c$ and the substitution method to solve $a,b,c$. Is that right?
Thank you for your help.

If by a "quadratic relation" you mean $y=ax^2+bx+c,$ then you have $y = a(x-0)(x-6),$ so that $y=0$ if $x$ is either $0$ or $6.$ Then you need to figure out which value of $a$ will make the minimum $y$-value $-9.$ Symmetry shows that if the $x$-intercepts are $0$ and $6$ then the minimum $y$-value occurs when $x$ is halfway between $0$ and $6$. So $y = a(3-0)(3-6) = -9,$ so what is $a$?
