The graph of the function f(x) = racax+p + q has a vertical asymptote at x = -1, a horizontal asymptote at y = 1, and passes through the point P(-2; 4). Determine the values of a, p, and q.

The general form of a hyperbola is $f(x) = \frac{a}{x+p} + q$.
The vertical asymptote is given by $x+p = 0$, so $x = -p$.
From the problem, the vertical asymptote is $x = -1$.
Therefore, $-p = -1 \implies p = 1$.

The horizontal asymptote is $y = q$.
From the problem, the horizontal asymptote is $y = 1$.
Therefore, $q = 1$.

Substitute $p=1$ and $q=1$ into the function:

The point P$(-2; 4)$ lies on the graph, so substitute $x=-2$ and $f(x)=4$:

The values are: $\boxed{a = -3, p = 1, q = 1}$

Using the function $f(x) = \frac{-3}{x+1} + 1$:

Multiply both sides by $(x+1)$:

The $x$-intercept is at $x=2$, and the $y$-coordinate is 0.

The coordinates of T are: $\boxed{(2; 0)}$

The asymptotes are $x = -1$ and $y = 1$. Their intersection point is $(-1; 1)$.
For a hyperbola, the lines of symmetry pass through the point of intersection of its asymptotes.

The given line of symmetry is $y = x+c$.
Since this line passes through $(-1; 1)$, substitute $x=-1$ and $y=1$ into the equation:

The value of $c$ is: $\boxed{2}$