Question

question 2 (7 points)

consider the reciprocal function \\(y = f(x) = \frac{1}{g(x)}\\) of the quadratic function \\(g(x) = 8 - 2x - x^2\\).

#1) state the equation(s) of any vertical asymptote(s) of \\(f(x)\\).

#2) state the equation(s) of any horizontal asymptote(s) of \\(f(x)\\).

#3) write the interval(s) of negative of \\(f(x)\\).

#4) write the interval(s) of increase of \\(g(x)\\).

#5) identify any symmetry of \\(f(x)\\).

#6) determine the general end behaviour of \\(g(x)\\).

#7) which feature(s), written for #1 to #6 above, would be exactly the same for both functions?

Explanation:

🆕 New Concept Discovered: Reciprocal Functions
How the features of a function relate to its reciprocal

Step 1: Vertical Asymptotes of \( f(x) \)

Vertical asymptotes of a reciprocal function \( f(x) = \frac{1}{g(x)} \) occur where the denominator is zero, \( g(x) = 0 \).

\[
8 - 2x - x^2 = 0
\]

Multiply by \(-1\):

\[
x^2 + 2x - 8 = 0
\]

Factor the quadratic equation:

\[
(x + 4)(x - 2) = 0
\]

Thus, the vertical asymptotes are:

\[
x = -4 \quad \text{and} \quad x = 2
\]

Step 2: Horizontal Asymptote of \( f(x) \)

For the reciprocal function \( f(x) = \frac{1}{8 - 2x - x^2} \), as \( x \to \pm\infty \), the denominator \( g(x) \to -\infty \).

Therefore, the function values approach \( 0 \):

\[
y = 0
\]

Step 3: Intervals of Negativity of \( f(x) \)

The sign of a reciprocal function \( f(x) = \frac{1}{g(x)} \) is identical to the sign of its base function \( g(x) \). We find where \( g(x) < 0 \):

\[
-x^2 - 2x + 8 < 0 \implies (x + 4)(x - 2) > 0
\]

This inequality holds true outside the roots:

\[
x \in (-\infty, -4) \cup (2, \infty)
\]

Step 4: Interval of Increase of \( g(x) \)

The quadratic function \( g(x) = -x^2 - 2x + 8 \) is a parabola opening downward. The \( x \)-coordinate of its vertex is:

\[
x = -\frac{b}{2a} = -\frac{-2}{2(-1)} = -1
\]

Since the parabola opens downward, it increases on the interval to the left of the vertex:

\[
x \in (-\infty, -1)
\]

Step 5: Symmetry of \( f(x) \)

To check for even symmetry (\( f(-x) = f(x) \)) or odd symmetry (\( f(-x) = -f(x) \)):

\[
f(-x) = \frac{1}{8 - 2(-x) - (-x)^2} = \frac{1}{8 + 2x - x^2}
\]

Since \( f(-x)
eq f(x) \) and \( f(-x)
eq -f(x) \), the function has no even or odd symmetry (neither).

However, it is symmetric about the vertical line passing through the vertex of the quadratic denominator:

\[
x = -1
\]

Step 6: End Behaviour of \( g(x) \)

Since \( g(x) = -x^2 - 2x + 8 \) is a polynomial with a negative leading coefficient and an even degree:

\[
\text{As } x \to \infty, \, g(x) \to -\infty
\]
\[
\text{As } x \to -\infty, \, g(x) \to -\infty
\]

Step 7: Shared Features Between \( f(x) \) and \( g(x) \)

Let's compare the features of \( g(x) \) and \( f(x) \):

• Vertical Asymptotes: Only \( f(x) \) has them.
• Horizontal Asymptotes: Only \( f(x) \) has them.
• Intervals of Negativity: Since \( f(x) = \frac{1}{g(x)} \), both functions share the exact same sign for any \( x \). Thus, the intervals of negativity are identical.
• Intervals of Increase/Decrease: Reciprocal functions have opposite behaviors; where \( g(x) \) increases, \( f(x) \) decreases.
• Symmetry: Both functions share the same axis of symmetry at \( x = -1 \).
• End Behaviour: \( g(x) \to -\infty \) while \( f(x) \to 0 \).

The features that are exactly the same are the intervals of negativity (and the axis of symmetry).

Answer:

1. Vertical Asymptotes of \( f(x) \): \( x = -4 \) and \( x = 2 \)
2. Horizontal Asymptote of \( f(x) \): \( y = 0 \)
3. Intervals of Negativity of \( f(x) \): \( (-\infty, -4) \cup (2, \infty) \)
4. Interval of Increase of \( g(x) \): \( (-\infty, -1) \)
5. Symmetry of \( f(x) \): Symmetric about the line \( x = -1 \) (neither even nor odd)
6. End Behaviour of \( g(x) \): As \( x \to \pm\infty \), \( g(x) \to -\infty \)
7. Identical Features: The intervals of negativity (and the axis of symmetry \( x = -1 \))