Given the graphs of f(x) = ax² + bx + c and g(x) = m/x-p + q, answer the following questions: Give the value(s) for which f is increasing? Determine the equation of f. Is g continuous or discontinuous? Give the equation of the asymptote(s) of g. Hence, determine the equation of g.

Here are the solutions to Question 3:

3.1.1
To draw the graphs of $f(x) = x^2 - 2x - 8$ and $g(x) = 2^x + 2$, you would need to plot the following key features:

• For $f(x) = x^2 - 2x - 8$ (Parabola):

  • x-intercepts: Set $f(x)=0 \implies x^2 - 2x - 8 = 0 \implies (x-4)(x+2) = 0$. The x-intercepts are at $\boxed{x=4}$ and $\boxed{x=-2}$.
  • y-intercept: Set $x=0 \implies f(0) = 0^2 - 2(0) - 8 = -8$. The y-intercept is at $\boxed{y=-8}$.
  • Turning point (vertex): $x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$. $f(1) = 1^2 - 2(1) - 8 = 1 - 2 - 8 = -9$. The turning point is at $\boxed{(1, -9)}$.
  • The parabola opens upwards.

• For $g(x) = 2^x + 2$ (Exponential function):

  • x-intercepts: Set $g(x)=0 \implies 2^x + 2 = 0 \implies 2^x = -2$. There are $\boxed{\text{no x-intercepts}}$.
  • y-intercept: Set $x=0 \implies g(0) = 2^0 + 2 = 1 + 2 = 3$. The y-intercept is at $\boxed{y=3}$.
  • Asymptote: The horizontal asymptote is $y=k$ for $g(x) = a \cdot b^x + k$. Here, $k=2$, so the horizontal asymptote is $\boxed{y=2}$.

3.1.2
To write $f(x) = x^2 - 2x - 8$ in the form $f(x) = a(x-p)^2 + q$, we complete the square:

$$f(x) = x^2 - 2x - 8$$ $$f(x) = (x^2 - 2x + 1) - 1 - 8$$ $$f(x) = (x-1)^2 - 9$$

The equation in the desired form is $\boxed{f(x) = (x-1)^2 - 9}$.

3.2.1
From the graph of $f$, the parabola opens upwards and its turning point is at $x=1$. The function $f$ is increasing for all $x$-values to the right of the turning point.
Therefore, $f$ is increasing for $\boxed{x > 1}$.

3.2.2
From the graph, the x-intercepts of $f$ are at $x=-2$ and $x=4$.
We can write the equation in factored form as $f(x) = a(x - (-2))(x - 4) = a(x+2)(x-4)$.
The y-intercept is at $(0, -8)$. Substitute this point into the equation:

$$-8 = a(0+2)(0-4)$$ $$-8 = a(2)(-4)$$ $$-8 = -8a$$ $$a = 1$$

So, the equation of $f$ is $f(x) = 1(x+2)(x-4)$.

$$f(x) = x^2 - 4x + 2x - 8$$ $$f(x) = x^2 - 2x - 8$$

The equation of $f$ is $\boxed{f(x) = x^2 - 2x - 8}$.

3.2.3
The graph of $g$ is a hyperbola with a vertical asymptote at $x=1$. A function is discontinuous if there is a break or a jump in its graph. Since the graph of $g$ has a vertical asymptote, it is not continuous at that point.
Therefore, $g$ is $\boxed{\text{discontinuous}}$.

3.2.4
From the graph of $g$:
The vertical dashed line is at $x=1$. This is the $\boxed{\text{vertical asymptote } x=1}$.
The horizontal dashed line is at $y=-3$. This is the $\boxed{\text{horizontal asymptote } y=-3}$.

3.2.5
The general form of the equation for $g$ is $g(x) = \frac{m}{x-p} + q$.
From the asymptotes identified in 3.2.4:
The vertical asymptote is $x=p$, so $p=1$.
The horizontal asymptote is $y=q$, so $q=-3$.
Substitute these values into the general equation:

$$g(x) = \frac{m}{x-1} - 3$$

From the graph, the y-intercept of $g$ is at $(0, -2)$. Substitute this point into the equation to find $m$:

$$-2 = \frac{m}{0-1} - 3$$ $$-2 = \frac{m}{-1} - 3$$ $$-2 = -m - 3$$

Add 3 to both sides:

$$1 = -m$$ $$m = -1$$

Substitute $m=-1$ back into the equation:

$$g(x) = \frac{-1}{x-1} - 3$$

The equation of $g$ is $\boxed{g(x) = \frac{-1}{x-1} - 3}$.