Solutions — Reciprocal Functions

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1. Q1 — Asymptotes, domain and range

   (a) y=1/(x−3): VA x=3, HA y=0, Domain x≠3, Range y≠0   (b) y=1/x+2: VA x=0, HA y=2, Domain x≠0, Range y≠2   (c) y=4/(x+1)−3: VA x=−1, HA y=−3, Domain x≠−1, Range y≠−3   (d) y=−2/(x−5)+1: VA x=5, HA y=1, Domain x≠5, Range y≠1

2. Q2 — Intercepts

   (a) y=1/x+2: y-int (x=0): undefined — no y-intercept. x-int (y=0): 1/x=−2 → x=−1/2 → (−1/2, 0).   (b) y=3/(x−2)+1: y-int (x=0): 3/(−2)+1=−3/2+1=−1/2 → (0, −1/2). x-int (y=0): 3/(x−2)=−1 → x−2=−3 → x=−1 → (−1, 0).   (c) y=−2/(x+3): y-int (x=0): −2/3 → (0, −2/3). x-int (y=0): −2/(x+3)=0 has no solution — no x-intercept (HA is y=0).   (d) y=5/(x−1)−5: y-int (x=0): 5/(−1)−5=−10 → (0, −10). x-int (y=0): 5/(x−1)=5 → x−1=1 → x=2 → (2, 0).

3. Q3 — Evaluating

   (a) f(−4)=2/(−4)=−1/2.   (b) g(3)=1/(3−1)+3=1/2+3=3.5.   (c) h(2)=−4/(2+2)+1=−4/4+1=−1+1=0.

4. Q4 — Transformations from y = 1/x

   (a) y=1/(x−4): Shift right 4 units (VA moves to x=4).   (b) y=1/x−3: Shift down 3 units (HA moves to y=−3).   (c) y=−1/x+2: Reflect in x-axis then shift up 2. Branches now in Q2 and Q4.   (d) y=3/(x+2)+1: Shift left 2 (VA: x=−2), shift up 1 (HA: y=1), vertical stretch by factor 3.

5. Q5 — Sketching hyperbolas

   (a) y=2/(x−1)+3: VA x=1, HA y=3. y-int (x=0): 2/(−1)+3=1 → (0,1). x-int (y=0): 2/(x−1)=−3 → x−1=−2/3 → x=1/3 → (1/3, 0). Draw dashed lines at x=1 and y=3; upper-right branch passes through (0,1); lower-left branch passes through (1/3,0). a=2>0 so branches in Q1 and Q3 relative to (1,3).   (b) y=−1/(x+2)−1: VA x=−2, HA y=−1. y-int (x=0): −1/2−1=−3/2 → (0,−3/2). x-int (y=0): −1/(x+2)=1 → x+2=−1 → x=−3 → (−3, 0). a=−1<0 so branches in Q2 and Q4 relative to (−2,−1).

6. Q6 — Finding the equation

   (a) VA x=2, HA y=−1 → y=a/(x−2)−1. Sub (3,1): 1=a/1−1 → a=2. Equation: y=2/(x−2)−1.   (b) VA x=−3, HA y=4 → y=a/(x+3)+4. Sub (−2,1): 1=a/1+4 → a=−3. Equation: y=−3/(x+3)+4.   (c) VA x=0, HA y=0 → y=a/x. Sub (2,−3): −3=a/2 → a=−6. Equation: y=−6/x.

7. Q7 — Solving equations

   (a) 3/(x−1)=6 → x−1=3/6=1/2 → x=3/2.   (b) 2/(x+3)+1=5 → 2/(x+3)=4 → x+3=1/2 → x=−5/2.   (c) 4/(x−2)=2/(x+1). Cross-multiply: 4(x+1)=2(x−2) → 4x+4=2x−4 → 2x=−8 → x=−4.

8. Q8 — Rewriting in standard form

   (a) y=(3x+7)/(x+2). Divide: 3x+7 = 3(x+2)+1, so y=3+1/(x+2). Standard form: y=1/(x+2)+3. VA: x=−2, HA: y=3.   (b) y=(2x−3)/(x−1). Divide: 2x−3 = 2(x−1)−1, so y=2−1/(x−1). Standard form: y=−1/(x−1)+2. VA: x=1, HA: y=2.

9. Q9 — Workers modelling

   (a) T(4)=12/4=3 hours. T(8)=12/8=1.5 hours.   (b) Key points to plot: (1,12), (2,6), (3,4), (4,3), (6,2), (12,1). The curve is a decreasing hyperbola — steeply decreasing for small n, flattening for large n.   (c) 12/n=1.5 → n=12/1.5=8 workers. This is a whole number, so exactly 8 workers are needed.

10. Q10 — Intersections with lines

    (a) 2/(x−1)+3=5 → 2/(x−1)=2 → x−1=1 → x=2. y=5 → Point: (2, 5).   (b) 6/x=x+1 → 6=x(x+1)=x²+x → x²+x−6=0 → (x+3)(x−2)=0 → x=−3 or x=2. Points: (−3, −2) and (2, 3).   (c) 4/x=2x+3 → 4=x(2x+3)=2x²+3x → 2x²+3x−4=0. Discriminant: 9+32=41>0. Two real solutions: x=(−3±√41)/4. x⊂1;=(−3+√41)/4≈0.85, x⊂2;=(−3−√41)/4≈−2.35. The line does intersect y=4/x at two points: approximately (0.85, 5.00) and (−2.35, −1.70).