Given \( f(x)=\frac{-6}{x+2}-3 \). (1) Sketch the graph of \( f \). (2) Write down the equation of the vertical asymptote of \( f \). (3) Write down the equation of the horizontal asymptote of \( f \). (4) Write down the domain of \( f \). (5) Write down the range of \( f \). (6) Determine the equations of the axes of symmetry of \( f \).

We are given   f(x) = (–6/(x + 2)) – 3. Let’s answer each part one by one. ────────────────────────────── (1) Sketch the graph of f Step 1. Find basic features and transformations. • Write f(x) = –6/(x + 2) – 3. • Notice that the function is a transformed reciprocal function. Begin with y = 1/x, then multiply by –6 (which reflects it in the x‑axis and stretches it), then replace x by x + 2 (shifting the graph 2 units left), and finally subtract 3 (shifting the graph 3 units down). Step 2. Identify the asymptotes. • The vertical asymptote occurs when the denominator is zero, i.e. x + 2 = 0 ⟹ x = –2. • The horizontal asymptote is found by considering x large; since –6/(x + 2) → 0 as x → ±∞, we have y → –3. Thus y = –3. Step 3. Find intercepts. • y‑intercept: Set x = 0.   f(0) = –6/(0 + 2) – 3 = –6/2 – 3 = –3 – 3 = –6.   So, the point is (0, –6). • x‑intercept: Set f(x) = 0.   0 = –6/(x + 2) – 3  ⟹  –6/(x + 2) = 3  ⟹  x + 2 = –6/3 = –2  ⟹  x = –4.   So, the point is (–4, 0). Step 4. Draw the graph. • Sketch the vertical line x = –2 and the horizontal line y = –3. • Plot the intercepts (0, –6) and (–4, 0). • Draw the two branches of the hyperbola:   – One branch in the region x < –2 (to the left of the vertical asymptote), which approaches the asymptotes as x → –2⁻ and as x → –∞.   – The other branch in the region x > –2, approaching the asymptotes as x → –2⁺ and as x → +∞. • Also note that the graph is a reflection (with a vertical stretch) and translation of the basic reciprocal graph y = 1/x. A rough sketch is as follows:           y          |    Branch: coming from below, rising toward the asymptotes          | (approaches y = –3 from below as x → +∞)         -- – – – – – – – – – – – x = –2 (vertical asymptote)          |    (–4,0) •————• (0,–6)          |          +————————→ x The branch in x > –2 falls toward –∞ as x → –2⁺ and eventually levels off approaching y = –3 from below, while the branch in x < –2 comes from above (y → –3 from above as x → –∞) and falls to –∞ as x → –2⁻. ────────────────────────────── (2) Equation of the vertical asymptote Since the denominator vanishes when x + 2 = 0, the vertical asymptote is   x = –2. ────────────────────────────── (3) Equation of the horizontal asymptote As x → ±∞, the term –6/(x + 2) → 0. Hence, f(x) → –3. So the horizontal asymptote is   y = –3. ────────────────────────────── (4) Domain of f The function is undefined where the denominator is zero. So,   x + 2 ≠ 0 ⟹ x ≠ –2. Thus, the domain is all real numbers except –2, i.e.,   { x ∈ ℝ : x ≠ –2 } or (–∞, –2) ∪ (–2, ∞). ────────────────────────────── (5) Range of f Since f(x) is a transformed reciprocal function, its horizontal asymptote y = –3 is never reached. Therefore, the range is all real numbers except –3:   { y ∈ ℝ : y ≠ –3 } or (–∞, –3) ∪ (–3, ∞). ────────────────────────────── (6) Axes of symmetry of f For a function to have an axis of symmetry, it must be mirror-reflective about a line (such as vertical or horizontal). In our case, let’s check: Write x in terms relative to the vertical asymptote by letting t = x + 2 so that   f(x) = –6/t – 3. Now consider f(–2 + t) and f(–2 – t):   f(–2 + t) = –6/t – 3,   f(–2 – t) = –6/(–t) – 3 = 6/t – 3. These two are generally not equal (unless 6/t = –6/t, which happens only if 6/t = 0, an impossibility). Thus there is no vertical or horizontal line which is an axis of symmetry. Note, however, that the graph has 180° rotational symmetry about the point where the asymptotes meet. That is, if you rotate the graph 180° around (–2, –3) you will get the same graph. However, this is point symmetry rather than symmetry with respect to an axis (i.e. a line of reflection). Thus, the function has no axis of symmetry. ────────────────────────────── Summary of Answers 1. The graph of f(x) = –6/(x + 2) – 3 is a hyperbola with a vertical asymptote at x = –2 and a horizontal asymptote at y = –3. It intercepts the y‑axis at (0, –6) and the x‑axis at (–4, 0), with one branch for x > –2 and the other for x < –2. 2. Vertical asymptote: x = –2. 3. Horizontal asymptote: y = –3. 4. Domain: { x ∈ ℝ : x ≠ –2 } or (–∞, –2) ∪ (–2, ∞). 5. Range: { y ∈ ℝ : y ≠ –3 } or (–∞, –3) ∪ (–3, ∞). 6. f has no axis of symmetry (though it is symmetric under a 180° rotation about the point (–2, –3)). Any answer equivalent to the above is correct.