Let g(x) = x+c. The graph of f(x) = (a)/(x+p) + q has a vertical asymptote x=-p and a horizontal asymptote y=q. If the graph of g and the vertical asymptote of f intersect at the x-axis, and the graph

6.1
The equation of graph $g$ is $g(x) = x+c$.
To find the x-intercept of $g$, we set $g(x) = 0$:

$$x+c = 0 \implies x = -c$$

So the x-intercept of $g$ is $(-c; 0)$.

The equation of graph $f$ is $f(x) = \frac{a}{x+p} + q$.
The vertical asymptote of $f$ occurs when the denominator is zero:

$$x+p = 0 \implies x = -p$$

The problem states that graph $g$ and the vertical asymptote of $f$ intersect at the x-axis. This means their x-intercepts are the same.
Therefore, $-c = -p$, which implies $c=p$.
Substituting $c=p$ into the x-intercept of $g$:
The coordinates of the x-intercept of $g$ are $(-p; 0)$.

$$\boxed{(-p; 0)}$$

6.2
We need to determine the equation of $f(x) = \frac{a}{x+p} + q$. We already know $c=p$.

Step 1: Use the information that $g$ intersects the horizontal asymptote of $f$ at $x=1$.
The horizontal asymptote of $f(x)$ is $y=q$.
Since $g(x)$ intersects $y=q$ at $x=1$, we have $g(1) = q$.
Substitute $x=1$ into $g(x) = x+c$:

$$g(1) = 1+c$$

So, $q = 1+c$. Since $c=p$, we have $q = 1+p$.

Step 2: Use the information that $f$ and $g$ intersect on the y-axis.
This means $f(0) = g(0)$.
Substitute $x=0$ into $g(x) = x+c$:

$$g(0) = 0+c = c$$

Substitute $x=0$ into $f(x) = \frac{a}{x+p} + q$:

$$f(0) = \frac{a}{0+p} + q = \frac{a}{p} + q$$

Equating $f(0)$ and $g(0)$:

$$c = \frac{a}{p} + q$$

Substitute $q=1+c$:

$$c = \frac{a}{p} + (1+c)$$ $$0 = \frac{a}{p} + 1$$ $$\frac{a}{p} = -1 \implies a = -p$$

Step 3: Use the information that $f$ and $g$ intersect at $x=3$.
This means $f(3) = g(3)$.
Substitute $x=3$ into $g(x) = x+c$:

$$g(3) = 3+c$$

Substitute $x=3$ into $f(x) = \frac{a}{x+p} + q$:

$$f(3) = \frac{a}{3+p} + q$$

Equating $f(3)$ and $g(3)$:

$$3+c = \frac{a}{3+p} + q$$

Substitute $a=-p$ and $q=1+c$:

$$3+c = \frac{-p}{3+p} + (1+c)$$

Subtract $(1+c)$ from both sides:

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