Solving an equation for 3 unknowns

Question

So I have the function $y=-a(x+b)^2+c$ And I also know that this equation goes through these three points $(0,0),\ (6,12),\ (2,8)$ I tried to make a system with $3$ equations from this info, but I was unable to solve it.

Answer 1

Hint: The system is given by $$0=-a(0+b)^2+c$$ $$12=-a(6+b)^2+c$$ $$8=-a(2+b)^2+c$$ From the first equation we get $$c=ab^2$$ plugging this in equation (2) and (3) and solving for $$a$$ we get $$\frac{8}{b^2-(2+b)^2}=a=\frac{12}{b^2-(6+b)^2}$$ Can you finish now? Simplifying we get $$2(12b+36)=3(4b+4)$$ This equation must be solved for $b$, it is linear in $$b$$

Answer 2

Hint:

Find quadratic polynomials $p(x)$, $q(x)$, $r(x)$ such that \begin{align}\begin{cases} p(6)=p(2)=0 ,\\ p(0)=1, \end{cases}\qquad\begin{cases} q(0)=q(6)=0 ,\\ q(2)=1, \end{cases} \qquad\begin{cases} r(2)=r(0)=0 ,\\ r(6)=1, \end{cases} \end{align} and take $$y(x)=0\cdot p(x)+ 8\cdot q(x)+12 \cdot r(x).$$

Then $-b\:$ is the abscissa of the axis if symmetry of the resulting parabola.