Solutions — Square Root Functions

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1. Q1 — Domain, range and endpoint

   (a) y=√(x−4): Endpoint (4,0), Domain x≥4, Range y≥0   (b) y=√(x+3)+2: Endpoint (−3,2), Domain x≥−3, Range y≥2   (c) y=−√x+5: Endpoint (0,5), Domain x≥0, Range y≤5 (reflected, decreasing)   (d) y=4√(x−1)−3: Endpoint (1,−3), Domain x≥1, Range y≥−3

2. Q2 — Evaluating

   (a) f(4)=√(8+1)=√9=3   (b) g(11)=3√(9)+1=3(3)+1=10   (c) h(9)=−2(3)+6=−6+6=0

3. Q3 — Transformations

   (a) y=√(x−5)+3: Shift right 5, up 3. Domain x≥5, Range y≥3.   (b) y=2√(x+1): Shift left 1, vertical stretch factor 2. Domain x≥−1, Range y≥0.   (c) y=−√(x−2)+4: Shift right 2, reflect in x-axis, shift up 4. Domain x≥2, Range y≤4.

4. Q4 — x-intercepts

   (a) √x−3=0: x=9. x-intercept: (9,0)   (b) 2√(x+4)−6=0 → √(x+4)=3 → x+4=9 → x=5. x-intercept: (5,0)   (c) −√(x−1)+2=0 → √(x−1)=2 → x=5. x-intercept: (5,0)   (d) y=√(x−9)+3: Minimum value of y is 3 (at endpoint x=9), so y≥3 always. y never equals 0. No x-intercept.

5. Q5 — Finding the equation

   (a) y=a√x; sub (4,6): 6=2a → a=3. Equation: y=3√x   (b) y=a√(x−2)+1; sub (6,5): 5=2a+1 → a=2. Equation: y=2√(x−2)+1   (c) y=a√(x+1)+3; sub (3,−1): −1=2a+3 → a=−2. Equation: y=−2√(x+1)+3

6. Q6 — Solving equations

   (a) √(2x+3)=5 → 2x+3=25 → x=11. Check: √25=5 ✓   (b) 3√(x−1)+2=8 → √(x−1)=2 → x−1=4 → x=5. Check: 3(2)+2=8 ✓   (c) √(x+5)=x−1. Square: x+5=(x−1)²=x²−2x+1 → x²−3x−4=0 → (x−4)(x+1)=0. x=4 or x=−1. Check x=4: √9=3=4−1 ✓. Check x=−1: √4=2, −1−1=−2≠2 ✗ (extraneous). Solution: x=4.

7. Q7 — Connecting to parabolas

   (a) y=√(x−2)+1 → y−1=√(x−2) → (y−1)²=x−2. Full parabola: (y−1)²=x−2 (horizontal parabola, vertex (2,1)).   (b) Bottom half: y=−√(x−2)+1   (c) f(9)=a√9=3a=6 → a=2. f(25)=2√25=2(5)=10.

8. Q8 — Range with restricted domain

   (a) f(x)=√(x+4) on −4≤x≤12. f(−4)=0, f(12)=√16=4. Range: 0≤y≤4.   (b) g(x)=−2√x+8 on 0≤x≤16. g(0)=8, g(16)=−8+8=0. Range: 0≤y≤8.   (c) √(x+4)=−2√x+8. Let u=√x; then √(x+4)=√(u²+4). This requires numerical approach. Alternatively: square both sides after isolating: √(x+4)+2√x=8. Let u=√x: √(u²+4)=8−2u. Square: u²+4=64−32u+4u² → 3u²−32u+60=0 → u=(32±√(1024−720))/6=(32±√304)/6. √304=4√19. u=(32−4√19)/6=(16−2√19)/3. x=u²≈x≈3.08 (check required).

9. Q9 — Pendulum period

   (a) T=2π√(2.45/9.8)=2π√(0.25)=2π(0.5)=π≈3.14 s.   (b) 2=2π√(L/9.8) → 1/π=√(L/9.8) → L/9.8=1/π² → L=9.8/π² m ≈0.993 m.   (c) As L increases, T increases but at a decreasing rate (concave down). The relationship is not linear — the square root function grows quickly at first then more slowly. Doubling L does not double T; it multiplies T by √2.

10. Q10 — Skateboarder modelling

    (a) v=√(20×5)=√100=10 m/s.   (b) Endpoint (0,0). At h=5, v=10. At h=1.25, v=5. Plot curve from (0,0) through (1.25,5) and (5,10).   (c) 15=√(20h) → 225=20h → h=11.25 m. To reach 15 m/s, the ramp must be 11.25 m high.