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. A good way to understand and sketch f(x) is to view it as a transformation of the basic reciprocal function g(u) = 1/u. Step 1. Write f in transformation form. Notice that   f(x) = –6/(x + 2) – 3 = [–6/(x + 2)] – 3. If we let u = x + 2, then g(u) = 1/u is multiplied by –6 and shifted downward by 3. Thus, compared to y = 1/x, the graph:  • is horizontally shifted left by 2 (since x has changed to x + 2),  • is vertically stretched (by a factor of 6) and reflected about the horizontal axis (because of the –6 in the numerator),  • and then is shifted downward by 3. Now we answer each part. ──────────────────────── (1) Sketch the graph of f To sketch f(x) = –6/(x + 2) – 3: • Draw the vertical asymptote (see (2) below) at x = –2. • Draw the horizontal asymptote (see (3) below) at y = –3. • Notice that the graph will have two separate branches. • When x is very large (or very small), the –6/(x + 2) term becomes very small, so the graph approaches y = –3. • Find a couple of points. For example:  – When x = 0, u = 0 + 2 = 2 so f(0) = –6/2 – 3 = –3 – 3 = –6.  – To find an x-intercept, set f(x) = 0:    –6/(x + 2) – 3 = 0 ⟹ –6/(x + 2) = 3 ⟹ x + 2 = –6/3 = –2 ⟹ x = –4. So the graph passes through (–4, 0) and (0, –6). • Finally, note that because of the –6/(x+2) factor the branch for x > –2 will be one “type” (falling toward the horizontal asymptote as x → ∞ and shooting off to ±∞ as x→ –2⁺), while the branch for x < –2 will be of the opposite curve. Sketch these two branches along with the asymptotes to complete the graph. ──────────────────────── (2) Vertical Asymptote The vertical asymptote occurs when the denominator is zero. Since   x + 2 = 0 ⟹ x = –2, the vertical asymptote is   x = –2. ──────────────────────── (3) Horizontal Asymptote For large |x| the term –6/(x + 2) tends to 0, so   f(x) ≈ –3. Thus the horizontal asymptote is   y = –3. ──────────────────────── (4) Domain We cannot allow a denominator of zero. Since x + 2 = 0 when x = –2, the domain is all real numbers except –2. In set‐notation,   Domain = { x ∈ ℝ : x ≠ –2 }. ──────────────────────── (5) Range Since f(x) = –6/(x + 2) – 3 can never equal –3 (because –6/(x + 2) is never 0), the horizontal asymptote is actually a value that f does not attain. Hence the range is all real numbers except –3:   Range = { y ∈ ℝ : y ≠ –3 }. ──────────────────────── (6) Axes of Symmetry Many transformed reciprocal functions do not have any lines (axes) of symmetry. Instead, they have point symmetry. If we rewrite f in the form   f(x) + 3 = –6/(x + 2), and let u = x + 2, then we have g(u) = –6/u, which is an odd function because   g(–u) = –6/(–u) = 6/u = –[–6/u] = –g(u). The graph of an odd function is symmetric with respect to the origin. In our case, because we made the substitution u = x + 2, the “origin” in u corresponds to the point where x + 2 = 0 and f(x) + 3 = 0; that is, the point   (x, y) = (–2, –3). Thus the graph of f is symmetric with respect to the point (–2, –3) (a central symmetry), but it is not symmetric about any line (vertical or horizontal). ──────────────────────── Summary of Answers (1) The graph of f(x) = –6/(x + 2) – 3 is that of a transformed reciprocal function with:  • a vertical asymptote x = –2,  • a horizontal asymptote y = –3,  • the point (–4, 0) (x-intercept) and (0, –6) (y-intercept), and two branches which are reflections of each other about the point (–2, –3). (2) Vertical asymptote: x = –2. (3) Horizontal asymptote: y = –3. (4) Domain: {x ∈ ℝ : x ≠ –2}. (5) Range: {y ∈ ℝ : y ≠ –3}. (6) Axes of symmetry: There are no lines of symmetry; however, the graph is symmetric with respect to the point (–2, –3). Any sketch showing these features is correct.