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
 A: If $f$ is a function, its graph has equation $$y=f(x).$$
Shifting this graph horizontally by $a\in \Bbb R$ and vertically by $b\in \Bbb R$ yields to the graph of equation
$$
y = f(x-a)+b.
$$
Here, you have the parabola of equation $y=f(x)$ with $f(x)=-x^2 + 4x + 8$.
Its shift by $(a,b)$ has then equation
$$
y = -(x-a)^2 + 4(x-a) + 8 + b.
$$
You can now solve this exercise by substituting $(x,y) = (1,1)$ and $(x,y) = (3,5)$ and solve for $(a,b)$.
A: The original parabola is
$ y = - x^2 + 4 x + 8 $
Shifting by $(h, k)$ gives the parabola,
$ y = - (x - h)^2 + 4 (x - h) + 8 + k $
Points $(1,1)$ and $(3,5)$ are on the new parabola, so
$ 1 = - (1 - h)^2 + 4 (1 - h) + 8 + k $
$ 5 = - (3 - h)^2 + 4 (3 - h) + 8 + k $
Substracting,
$ - 4 = - (1^2 - 3^2 - 2 h (1 - 3 )) + 4 (1 - 3) $
$ - 4 = - (-8 + 4 h ) - 8 $
Therefore,
$ -8 + 4 h = - 4 $
From which
$ 4 h = 4 $
Hence, $ h = 1$.
Now substitute this into one of the new parabola equations to find $k$.
$ 1 = - (1 - 1)^2 + 4 (1 - 1) + 8 + k $
Therefore, $ k = -7 $
Now the equation of the new parabola is
$ y = - (x - 1)^2 + 4 (x - 1) + 8 + (-7) $
Expanding, this becomes
$ y = - x^2 +6 x - 4 $
A: Here's a slightly different approach.
Assuming that a "shift" strictly means a translation, then the vector between points $A = (1,  1)$ and $B = (3, 5)$, which is $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ is preserved. Thus the following system of equations holds:
$$\begin{cases} b = -a^2 + 4a + 8 \\ b + 4 = -(a + 2)^2 + 4(a + 2) + 8 \end{cases} \tag{2}$$
and this gives $4 = -(a + 2)^2 - (-a^2) + 4(a + 2) - 4a$ or $4 = -4a - 4 + 8 \implies a = 0$, and $b = 8$.
Hence the point $A$ corresponds to $(0, 8)$ on the original quadratic, and so the new quadratic is a translation of $1$ unit to the right and $7$ units down.
Therefore the equation of the new parabola is:
$$y = -(x - 1)^2 + 4(x - 1) + 8 - 7 = -(x^2 - 2x + 1) + 4x - 4 + 1$$
$$ = \boxed{-x^2+ 6x - 4}.$$
A: "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*}
