The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$  for $(1,1)$  and
$5=(3-h)^2+k$  for $(3,5)$.
With a system of equations, I obtain that $h = 1$ and $k = 1$, so the new formula of the shifted parabola would be: $y=(x-1)^2+1$
What I don't understand is the answer of my exercise sheet, which says that the new equation would be $y= -x^2+6x-4$
Can someone help me understand this?
Thanks

"Shifted" here means translated.
In particular, after being shifted by $p$ units in the $y$ direction and $q$ units in the $x$ direction, the shifted parabola satisfies the equation
$$
(y-p)
= (x - q)^2 + 4(x - q) + 8.
$$
Now, the problem is to solve the above for $p$ and $q$ such that $(x, y)  =(1, 1)$ and $(3, 5)$ satisfy the equation.
Substituting these points, we find that
\begin{align*}
1 - p
&= -(1-q)^2 - 4(1 - q) + 8, \\
5 - p
&= -(3-q)^2 - 4(3 - q) + 8.
\end{align*}
Expanding these yeilds
\begin{align*}
p &= q^2 + 2q - 10, \\
p &= q^2 - 2q - 6,
\end{align*}
which we can solve to find $p = -7, q = 1$, which corresponds to the equation
\begin{align*}
(y + 7)
&= (x - 1)^2 + 4(x - 1) + 8,\\
y
&= -x^2 + 6x - 4.
\end{align*}
