Q:

What are the coordinates of red point?

The function  $f (x)=-x^2+4$ "in red" is moving along the line $y=x+4$  " in black "  from green point  to  black point  and becomes in the place of blue graph  as shown in the following  graph

What are the coordinates  of red point?
I got the green point $(0,4)$ by using the equation  $y=x+4$ ,  and got the intersection  point of the two  parabolas $(2,0)$  and I stopped here.

A:

We can solve this problem by remembering function transformations.
The equation of the original parabola is $y = -x^2 + 4$. You are shifting the parabola along a line of slope $1$, which means that the parabola is shifting up the same ammount as it is shifting to the right. So the equation of the blue parabola will be $y = -(x-h)^2 + 4+h$, where $h$ is a positive constant. 
Since the blue parabola and the red parabola share an $x$ intercept, we know that $(2, 0)$ satisfies $y = -(x-h)^2 + 4 + h$. If we plug in $(2, 0)$, we can solve for $h$, and then we are done. Can you do the rest?

A:

If you slide the green point some distance $d$ to the right along the line, then both the $x$ and $y$ coordinates of the point increase by $d$. This means that the shifted parabola has the equation $(y-d)=4-(x-d)^2$. You know that this shifted parabola passes through the intersection point $(2,0)$, so plug that in and solve for $d$, then find the other $x$-intercept of the resulting equation (you can use symmetry to make that last step really easy).

A:

The other answers are good but here's another way to do it. Given a parabola with it's vertex on the line $y=x+4$ and $a=-1$, and a zero at $x=2$, the height of the parabola will be $x+4$, and $x$-distance from the vertex to the zero will be $|x-2|$. Thus, $(x+4)-|x-2|^2 = 0$, or $x^2-5x = 0$, which means that $x=0$ (the red parabola), or $x=5$ (the blue parabola). Since the distance from the vertex to each of it's roots are the same, we calculate $|x-2|$, and we can add and subtract that to $x$, the $x$-coordinate of the vertex. This gives us the two zeros of the blue parabola.

