Finding the equation of the parable and build it
Axis parallel to y = 0 and goes through A(-2,4) B(-3,2) C(-11,-2)
Answer
$\displaystyle y^2-8y+4x+24=0$
All right. We have an equation now. That's a step in the right direction.
A parabola is defined by three points. Since we have three points here, we can define it.
In our equation $\displaystyle (y-k)^2=2p(x-h)$, we can use $\displaystyle (x,y)=(-2,4)$ to get an equation in three unknowns: $\displaystyle h,k,p$. Doing the same with the other two points will give us three equations in three unknowns, and from there, we can solve for $\displaystyle h,k,p$.
So
$\displaystyle (4 - k)^2 = 2p(-2 - h)$
$\displaystyle (2 - k)^2 = 2p(-3 - h)$
$\displaystyle (-2 - k)^2 = 2p(-11 - h)$
So we get
$\displaystyle k^2 - 8k + 16 = -4p - 2ph$ (1)
$\displaystyle k^2 - 4k + 4 = -6p - 2ph$ (2)
$\displaystyle k^2 + 4k + 4 = -22p - 2ph$ (3)
Subtract equation (2) from equation (1) and then subtract equation (3) from equation (1)
$\displaystyle -4k + 12 = 2p$
$\displaystyle -12k + 12 = 18p$
You can find values of k and p from these and then use one of the original equations to find h.
Another (perhaps simpler in this case) way to approach this is to use the "standard" form for this kind of parabola:
$\displaystyle ay^2 + by + c = x$
and plug the points into that. This gives you a linear system in a, b, and c.
-Dan
$\displaystyle (-2,4), (-3,2), (-11,-2)$
$\displaystyle ay^2+by+c=x$
Substituting each ordered pair into the standard equation, we get:
$\displaystyle 1.\ \ a(4)^2+b(4)+c=-2$
$\displaystyle 16a+4b+c=-2$
$\displaystyle 2. \ \ a(2)^2+b(2)+c=-3$
$\displaystyle 4a+2b+c=-3$
$\displaystyle 3. \ \ a(-2)^2+b(-2)+c=-11$
$\displaystyle 4a^2-2b+c=-11$
Solve the system using matrix equation (or any method you like):
$\displaystyle \left[\begin{array}{cccc}16 & 4 & 1 & \\ 4 & 2 & 1 & \\ 4 & -2 & 1 \\ \end{array}\right] \cdot \left[\begin{array}{c}a \\ b \\ c \end{array}\right] = \left[\begin{array}{c}-2 \\ -3 \\ -11 \end{array}\right]$
$\displaystyle \left[\begin{array}{cccc}16 & 4 & 1 & \\ 4 & 2 & 1 & \\ 4 & -2 & 1 \\ \end{array}\right]^{-1} \cdot \left[\begin{array}{c}-2 \\ -3 \\ -11 \end{array}\right] = \left[\begin{array}{c}a \\ b \\ c \end{array}\right]$
$\displaystyle \left[\begin{array}{c}a \\ b \\ c \end{array}\right]= \left[\begin{array}{c}-\frac{1}{4} \\ 2 \\ -6 \end{array}\right]$
Substituting back into your original equaton:
$\displaystyle ay^2+by+c=x$
$\displaystyle -\frac{1}{4}y^2+2y-6=x$
$\displaystyle \boxed{y^2-8y+4x+24=0}$