104. The graph of $y = -x^2 + 2x + 5$ is shifted 3 units to the right along the positive $x$-axis, then 2 units to the negative $y$-axis. What is the new vertex of the parabola?
(1) $(3,4)$ (2) $(2,5)$ (3) $(3,5)$ (4) $(2,6)$

Let $a$ and $b$ be constants where $a > 0$. Translate the graph of the quadratic function
$$y = 4x^2 + 2ax + b$$
by $a$ in the $x$-direction and by $1 - 7a$ in the $y$-direction. If this graph passes through the point $(0, 4)$, we have
$$b = \mathbf{AB}\, a^2 + \mathbf{C}\, a + \mathbf{D},$$
and the quadratic function representing the graph resulting from these translations is
$$y = \mathbf{E}\, x^2 - \mathbf{F}\, ax + \mathbf{G}.$$
When the graph of quadratic function (1) is tangent to the $x$-axis, we have $a = \frac{\mathbf{H}}{\mathbf{I}}$, and the $x$-coordinate of the point of tangency is $x = \mathbf{J}$.

Let $a$ and $b$ be constants where $a > 0$. Translate the graph of the quadratic function
$$y = 4x^2 + 2ax + b$$
by $a$ in the $x$-direction and by $1 - 7a$ in the $y$-direction. If this graph passes through the point $(0, 4)$, we have
$$b = \mathbf{AB}\, a^2 + \mathbf{C}\, a + \mathbf{D}$$
and the quadratic function representing the graph resulting from these translations is
$$y = \mathbf{E}\, x^2 - \mathbf{F}\, ax + \mathbf{G}.$$
When the graph of quadratic function (1) is tangent to the $x$-axis, we have $a = \frac{\mathbf{H}}{\mathbf{I}}$, and the $x$-coordinate of the point of tangency is $x = \mathbf{J}$.

Consider the quadratic function
$$y = 3 x ^ { 2 } - 6 .$$
(1) Suppose that the graph obtained by a parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ passes through the two points $( 1,5 )$ and $( 4,14 )$. The quadratic function of this graph is
$$y = \mathbf { A } x ^ { 2 } - \mathbf { B C } x + \mathbf { D E } .$$
This graph is the parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ by $\mathbf{F}$ in the $x$-direction and by $\mathbf { G }$ in the $y$-direction.
(2) The quadratic function having the graph which is symmetric to the graph of $y = 3 x ^ { 2 } - 6$ with respect to the straight line $y = c$ is
$$y = - \mathbf { H } x ^ { 2 } + \mathbf { I } c + \mathbf { J } .$$
When the graphs of the two quadratic functions (1) and (2) have just one common point, it follows that $c = \mathbf { K }$, and the coordinates of the common point are ( $\mathbf { L } , \mathbf { M }$ ).

Consider the quadratic function
$$y = 3 x ^ { 2 } - 6 .$$
(1) Suppose that the graph obtained by a parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ passes through the two points $( 1,5 )$ and $( 4,14 )$. The quadratic function of this graph is
$$y = \mathbf { A } x ^ { 2 } - \mathbf { BC } x + \mathbf { D E } .$$
This graph is the parallel translation of the graph of $y = 3 x ^ { 2 } - 6$ by $\mathbf{F}$ in the $x$-direction and by $\mathbf { G }$ in the $y$-direction.
(2) The quadratic function having the graph which is symmetric to the graph of $y = 3 x ^ { 2 } - 6$ with respect to the straight line $y = c$ is
$$y = - \mathbf { H } x ^ { 2 } + \mathbf { I } c + \mathbf { J } .$$
When the graphs of the two quadratic functions (1) and (2) have just one common point, it follows that $c = \mathbf { K }$, and the coordinates of the common point are ( $\mathbf { L } , \mathbf { M }$ ).

Let $a$ and $b$ be positive real numbers. In the Cartesian coordinate plane, using the parabola
$$p ( x ) = ( x - a ) ^ { 2 } - b$$
that passes through the origin, three parabolas defined as
$$\begin{aligned} & p ( x + a ) + b \\ & p ( x + a ) - b \\ & p ( x - a ) - b \end{aligned}$$
have their vertices at the vertices of a triangle with an area of 16 square units.
Accordingly, what is the sum $a + b$?
A) 6 B) 9 C) 12 D) 15 E) 18