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. One very useful way to view this function is to write it in the “shifted reciprocal” form. Notice that if we define   u = x + 2  and  v = y + 3 then   v = –6⁄u  or equivalently  u·v = –6. This shows that the graph of f is just a translation of the basic reciprocal function, with: • a horizontal shift 2 units to the left (x + 2 in the denominator), and • a vertical shift 3 units downward (–3 added at the end). Below are the answers to the numbered parts. ──────────────────────────── (1) Sketch the graph of f The graph is a hyperbola made up of two branches. Key features include: • Vertical asymptote: x = –2. • Horizontal asymptote: y = –3. • Intercepts:   – To find the x–intercept, set f(x) = 0.     0 = –6⁄(x + 2) – 3 ⟹ –6⁄(x + 2) = 3 ⟹ x + 2 = –2 ⟹ x = –4.    So the x–intercept is (–4, 0).   – To find the y–intercept, set x = 0.     f(0) = –6⁄(0 + 2) – 3 = –3 – 3 = –6.    So the y–intercept is (0, –6). • End Behavior:   – As x → ∞ or x → –∞, –6⁄(x + 2) → 0 so f(x) → –3.   – As x → –2⁺ (from the right), (x + 2) is positive and very small, so –6⁄(x + 2) → –∞ giving f(x) → –∞.   – As x → –2⁻ (from the left), (x + 2) is negative and very small, so –6⁄(x + 2) → +∞ giving f(x) → +∞. Hint for your sketch: Draw the dashed vertical line x = –2 and dashed horizontal line y = –3. Plot the intercepts (–4, 0) and (0, –6), and sketch two branches of a hyperbola approaching these asymptotes in the appropriate way. (One branch lies in the region where x < –2 and y > –3; the other branch lies where x > –2 and y < –3.) ──────────────────────────── (2) Vertical asymptote The vertical asymptote occurs where the denominator is zero. Since   x + 2 = 0 ⟹ x = –2, the vertical asymptote is   x = –2. ──────────────────────────── (3) Horizontal asymptote As x → ±∞, the term –6⁄(x+2) tends to 0 so that   f(x) → –3. Thus, the horizontal asymptote is   y = –3. ──────────────────────────── (4) Domain of f The only restriction is that the denominator cannot be zero. Thus, we must have   x + 2 ≠ 0 ⟹ x ≠ –2. So the domain is   { x ∈ ℝ : x ≠ –2 }. ──────────────────────────── (5) Range of f Because f(x) = –6⁄(x+2) – 3 can take any real value except that the horizontal asymptote y = –3 is never reached, the range is   { y ∈ ℝ : y ≠ –3 }. ──────────────────────────── (6) Axes of symmetry of f It is sometimes useful to express the function in “shifted” coordinates. Recall that we may write   u = x + 2  and  v = y + 3, so that the equation becomes   v = –6⁄u  or  u·v = –6. This equation, like the basic reciprocal function y = 1⁄x (except for the constant –6), has 180° rotational symmetry about the origin (for the (u, v)–plane). In other words, the graph is symmetric with respect to its center at (u, v) = (0, 0). Returning to the original variables, the center is the point where   x + 2 = 0 ⟹ x = –2  and  y + 3 = 0 ⟹ y = –3. Thus, the center of the hyperbola is (–2, –3), and the whole graph has point (rotational) symmetry about (–2, –3). Moreover, for a hyperbola of the form (x + 2)(y + 3) = –6 (which is equivalent to our function’s equation), the two lines that bisect the angles between the asymptotes are also symmetry axes. In the (u, v)–plane these bisectors are the lines u = v and u = –v. Converting back to x and y: • u = v means  x + 2 = y + 3 ⟹ y = x – 1. • u = –v means  x + 2 = –(y + 3) ⟹ x + 2 = –y – 3 ⟹ y = –x – 5. Thus, the two symmetry axes (i.e. the lines through the center along which if you reflect the complete graph you get the same graph) are   y = x – 1  and  y = –x – 5. Note: Although the function f (being defined only as “y = …”) is a branch of its overall hyperbola, the two branches taken together form a graph that has these two symmetry axes. ──────────────────────────── Summary of Answers (1) The graph of f(x) = –6⁄(x+2) – 3 is a hyperbola with   • vertical asymptote x = –2,   • horizontal asymptote y = –3,   • x–intercept (–4, 0),   • y–intercept (0, –6),   and with branches that approach these asymptotes (one branch with x < –2, the other with x > –2). (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 (of the full hyperbola):  y = x –