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 $\boxed{no x-intercepts}$.
  • 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 $\boxed{discontinuous}$.

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

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

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