Solutions — Mixed Non-Linear Graphs

1. Match graphs to equations. Fluency

   • Graph A → equation (i): Graph A is a U-shaped parabola with vertex below the x-axis and two x-intercepts at x = ±2. Matches (i) y = x² − 4 (vertex (0, −4), zeros at ±2).
   • Graph B → equation (ii): Graph B has two branches in Q1 and Q3 with the axes as asymptotes. This is a hyperbola; matches (ii) y = 6/x (k = 6 > 0).
   • Graph C → equation (iii): Graph C is a complete circle centred at the origin. Matches (iii) x² + y² = 9 (r = 3).
   • Graph D → equation (iv): Graph D is a decreasing curve that falls steeply on the left and flattens toward y = 0 on the right — exponential decay. Matches (iv) y = 2 × (½)^x.

2. Type and key feature. Fluency

   • (a) y = 5/x: Hyperbola. Asymptotes: x = 0 and y = 0.
   • (b) y = 3^x−2: Exponential. As x→−∞, 3^x→0, so y→−2. Asymptote: y = −2.
   • (c) x²+y²=9: Circle. Centre (0, 0), radius 3.
   • (d) y=(x−2)²+1: Parabola in vertex form. Vertex: (2, 1).

3. Domain and range. Fluency

   • (a) y = 4^x: Domain: all real numbers. Range: (0, ∞) — always positive.
   • (b) x²+y²=25: r=5. Domain: [−5, 5]. Range: [−5, 5].
   • (c) y = 3/x: Domain: x ≠ 0 (all reals except 0). Range: y ≠ 0.
   • (d) y=(x+1)²−4: Parabola opening upward, vertex at (−1,−4). Domain: all real numbers. Range: y ≥ −4, i.e. [−4, ∞).

4. Does it have a y-intercept? Fluency

   • (a) y = 2 × 3^x: Yes. At x=0: y=2×1=2. y-intercept (0, 2).
   • (b) y = 5/x: No. x=0 gives 5/0, which is undefined. The y-axis (x=0) is an asymptote.
   • (c) x²+y²=16: Yes — two y-intercepts. At x=0: y²=16, so y=±4. Intercepts: (0, 4) and (0, −4).
   • (d) y=x²+3x−1: Yes. At x=0: y=0+0−1=−1. y-intercept (0, −1).

5. Find the equation from a description. Understanding

   • (a) Hyperbola through (2, 6): y=k/x: 6=k/2 ⇒ k=12. y = 12/x
   • (b) Exponential through (0,3) and (1,6): At x=0: y=b×1=b=3. At x=1: y=3a=6 ⇒ a=2. y = 3 × 2^x
   • (c) Circle at origin through (5,12): r²=5²+12²=25+144=169. x²+y²=169
   • (d) Parabola, vertex (2,−3) through (4,1): y=a(x−2)²−3. At (4,1): 1=a(4)−3 ⇒ 1=4a−3 ⇒ a=1. y=(x−2)²−3

6. One point — four graphs. Understanding

   • (a) y = x² at (2,4): y=2²=4 ✓. The point lies on the parabola.
   • (b) y = 8/x at (2,4): y=8/2=4 ✓. The point lies on the hyperbola.
   • (c) y = 2^x at (2,4): y=2²=4 ✓. The point lies on the exponential.
   • (d) x²+y²=20 at (2,4): 2²+4²=4+16=20 ✓. The point lies on the circle. All four curves pass through (2, 4).

7. Choose the correct model. Understanding

   • (a) Soccer ball path: Parabola. Gravity creates a quadratic relationship between height and time (h = −5t² + vt + h₀).
   • (b) Medicine concentration decreasing 20% per hour: Exponential decay. A constant percentage decrease each period gives C = C₀ × (0.8)^t.
   • (c) Towns 50 km from a city: Circle. All points at a fixed distance from a centre define a circle: x²+y²=2500.
   • (d) Drive time as speed increases: Hyperbola. Time × speed = distance (constant), so T = d/v — an inverse proportion relationship.

8. Which graph type has this property? Understanding

   • (a) Fails vertical line test: Circle. For x-values near the centre, two y-values exist (upper and lower arc).
   • (b) Always passes through (0, 1): Exponential. a⁰ = 1 for any base a, so y = a^x always has y-intercept (0, 1).
   • (c) Two asymptotes along the coordinate axes: Hyperbola. y = k/x has x = 0 (y-axis) and y = 0 (x-axis) as asymptotes.
   • (d) Single vertex as max or min: Parabola. The vertex is the turning point — minimum if a > 0, maximum if a < 0.

9. Gas pressure and volume. Problem Solving

   • (a) PV products: 120×1=120, 60×2=120, 30×4=120, 15×8=120. PV is constant (= 120). This is an inverse proportion (hyperbola) relationship: P ∝ 1/V.
   • (b) Equation: PV = 120 ⇒ P = 120/V. (This is Boyle’s Law.)
   • (c) Find V when P = 40: 40 = 120/V ⇒ V = 120/40 = 3 litres.
   • (d) Second gas P = 120 × (0.5)^V at V = 3: P = 120 × (0.5)³ = 120 × 0.125 = 15 kPa. This is an exponential function (V is the exponent). Key difference: in P=120/V (hyperbola), V is in the denominator; in P=120×0.5^V (exponential), V is in the power.

10. Satellite dish design. Problem Solving

    • (a) Find the equation: y = ax². Point (4, 2): 2 = 16a ⇒ a = 1/8. y = x²/8
    • (b) Depth at x = 3: y = 3²/8 = 9/8 = 1.125 m
    • (c) Width at depth y = 1.125: 1.125 = x²/8 ⇒ x² = 9 ⇒ x = ±3. Width = 2 × 3 = 6 m.
    • (d) Larger dish, rim at x = 8: y = 8²/8 = 64/8 = 8 m deep. Doubling the width quadruples the depth (because y varies as x²).