★ Topic Review — Functions and Relations — Solutions

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This review covers all four lessons in this topic: Functions, Relations and Function Notation; Graphs of Relations — Circles and Horizontal Parabolas; Reciprocal Functions; and Square Root Functions. Questions are mixed across difficulty levels. Click each answer button to reveal the solution.

1. Fluency — Function or relation?

   1. (a) Determine whether each of the following is a function or a relation only. State a reason.
         • y = 3x − 2
         • x² + y² = 9
         • y = x² + 1
         • x = y² − 4
   2. (b) Explain what the vertical line test is and how it is used to determine whether a graph represents a function.
   3. (c) A graph passes through the points (1, 2), (2, 5), (3, 2) and (1, −1). Is it a function? Explain.
   (a) y=3x−2: Function (each x gives exactly one y).   x²+y²=9: Relation (circle — vertical lines cross it twice e.g. x=0 gives y=±3).   y=x²+1: Function (parabola opening up, passes VLT).   x=y²−4: Relation (horizontal parabola — two y-values for each x in its range).    (b) A vertical line is drawn at any x-value. If the line crosses the graph at most once, it is a function. If it crosses more than once, it is a relation only.    (c) Not a function. The x-value 1 is paired with both y=2 and y=−1, so the rule assigns two outputs to the same input.

2. Fluency — Domain and range

   1. (a) State the domain and range of each function.
         • f(x) = 2x + 3, x ∈ ℝ
         • g(x) = x² − 4
         • h(x) = √(x + 5)
         • p(x) = 3/(x − 2)
   2. (b) Find f(3) and f(−1) for f(x) = x² − 2x + 4.
   3. (c) If g(x) = 2x − 1, find x when g(x) = 7.
   (a) f(x)=2x+3: Domain ℝ, Range ℝ.   g(x)=x²−4: Domain ℝ, Range y≥−4.   h(x)=√(x+5): Domain x≥−5, Range y≥0.   p(x)=3/(x−2): Domain x≠2, Range y≠0.    (b) f(3)=9−6+4=7; f(−1)=1+2+4=7.    (c) 2x−1=7 → x=4.

3. Fluency — Piecewise functions

   Let f(x) =
      x + 3,   x < 0
      x²,       0 ≤ x ≤ 3
      2x − 1, x > 3

   1. (a) Evaluate f(−2), f(0), f(2), f(4).
   2. (b) Is the piecewise function continuous at x = 0? Justify by checking left and right values.
   3. (c) Is it continuous at x = 3? Justify.
   4. (d) State the range of each piece and hence the overall range.
   (a) f(−2)=−2+3=1; f(0)=0²=0; f(2)=4; f(4)=2(4)−1=7.    (b) At x=0: left limit = 0+3=3; right value = 0²=0. Not continuous (jump from 3 to 0).    (c) At x=3: piece 2 gives 3²=9; piece 3 gives 2(3)−1=5. Not continuous (jump from 9 to 5).    (d) Piece 1 (x<0): range y<3. Piece 2 (0≤x≤3): range 0≤y≤9. Piece 3 (x>3): range y>5. Overall range: y<3 combined with 0≤y≤9 and y>5, giving y<3 or 0≤y≤9 or y>5. In simpler terms, all real values except the gap between the pieces (carefully: y=3 not achieved in piece 1, but y=3 is covered in piece 2 at x=√3). Range is effectively all real y.

4. Fluency — Circles

   1. (a) State the centre and radius of each circle.
         • x² + y² = 25
         • (x − 3)² + (y + 1)² = 16
         • (x + 2)² + y² = 7
   2. (b) Write the equation of a circle with centre (4, −3) and radius 5.
   3. (c) Does the point (3, 4) lie on the circle x² + y² = 25? Show working.
   (a) x²+y²=25: centre (0,0), radius 5.   (x−3)²+(y+1)²=16: centre (3,−1), radius 4.   (x+2)²+y²=7: centre (−2,0), radius √7.    (b) (x−4)²+(y+3)²=25.    (c) 3²+4²=9+16=25 ✓. Yes, the point lies on the circle.

5. Fluency — Horizontal parabolas

   1. (a) For the relation x = (y − 2)² + 1, state the vertex, axis of symmetry, and the direction it opens.
   2. (b) Write the equation of the horizontal parabola with vertex (−3, 1) opening to the left.
   3. (c) Find the x-intercept and y-intercepts (if any) of x = y² − 4.
   4. (d) Explain why horizontal parabolas are not functions, and describe how to write them as two functions.
   (a) Vertex (1, 2); axis of symmetry y=2; opens right (coefficient of (y−2)² is positive).    (b) x=−a(y−1)²−3 with a>0, e.g. x=−(y−1)²−3.    (c) x-int: set y=0: x=−4. x-int: (−4, 0). y-ints: set x=0: y²=4, y=±2. y-ints: (0, 2) and (0, −2).    (d) For a given x-value in the range, there are two y-values (top and bottom halves), so VLT fails. Top half: y=+√(x+4) (a function); bottom half: y=−√(x+4) (a function).

6. Fluency — Reciprocal functions

   1. (a) For y = 1/(x − 3) + 2, state the equations of the asymptotes, domain, and range.
   2. (b) For y = −4/(x + 1) − 1, state the asymptotes and describe the transformation from y = 1/x.
   3. (c) Find the x- and y-intercepts of y = 2/(x + 1) − 4.
   (a) Vertical asymptote: x=3; horizontal asymptote: y=2. Domain: x≠3; Range: y≠2.    (b) Vertical asymptote: x=−1; horizontal asymptote: y=−1. Transformations from y=1/x: shift left 1, vertical dilation by factor 4 with reflection in x-axis, shift down 1.    (c) y-int (x=0): y=2/1−4=−2. y-int: (0, −2). x-int (y=0): 0=2/(x+1)−4 → 4=2/(x+1) → x+1=0.5 → x=−0.5. x-int: (−0.5, 0).

7. Fluency — Square root functions

   1. (a) State the endpoint, domain and range of y = √(x + 4) − 3.
   2. (b) State the endpoint, domain and range of y = −2√(x − 1) + 5.
   3. (c) A square root function has endpoint (2, −1) and passes through (6, 3). Find its equation.
   (a) Endpoint: (−4, −3); Domain: x≥−4; Range: y≥−3.    (b) Endpoint: (1, 5); Domain: x≥1; Range: y≤5 (reflected).    (c) y=a√(x−2)−1. Sub (6,3): 3=a√4−1=2a−1 → 2a=4 → a=2. Equation: y=2√(x−2)−1.

8. Understanding — Connecting function types

   For each question, think about which function type applies and what the key features are.
   1. (a) The top half of the circle x² + y² = 36 can be written as a function. State this function and its domain and range.
   2. (b) A reciprocal function y = a/(x − h) + k has a vertical asymptote x = 2, passes through (0, 0), and has horizontal asymptote y = 1. Find a, h, and k.
   3. (c) The function f(x) = √x and the relation g: y² = x are connected. Explain the relationship and state the domain and range of each.
   (a) Top half: y = √(36 − x²). Domain: −6≤x≤6, Range: 0≤y≤6.    (b) h=2, k=1. Sub (0,0): 0=a/(0−2)+1 → −1=a/(−2) → a=2. So y=2/(x−2)+1.    (c) f(x)=√x is the top half of the relation y²=x (a horizontal parabola). f(x): domain x≥0, range y≥0. Relation y²=x: domain x≥0, range y∈ℝ. f(x) is a function; y²=x is not.

9. Understanding — Transformations and asymptotes

   1. (a) Describe fully the transformations from y = 1/x to y = 3/(x + 2) − 4. State asymptotes and sketch key features.
   2. (b) A hyperbola has asymptotes x = −1 and y = 3, and passes through (1, 4). Find its equation in the form y = a/(x − h) + k.
   3. (c) For y = √(x − a) + b, what is the minimum value of y, and for what x does it occur?
   (a) From y=1/x: shift left 2 (x-asymptote moves to x=−2); vertical stretch by factor 3; shift down 4 (y-asymptote moves to y=−4). Asymptotes: x=−2, y=−4. Passes through (0, −4+1.5)=(0,−2.5).    (b) h=−1, k=3. Sub (1,4): 4=a/(1−(−1))+3=a/2+3 → a=2. Equation: y=2/(x+1)+3.    (c) Minimum value of y is b, occurring at x = a (the endpoint of the square root function).

10. Problem Solving — Modelling with functions

    Multi-step challenge. A drone's altitude A (metres) as a function of horizontal distance d (metres) from its launch point is modelled by A(d) = −√(d − 100) + 20, for d ≥ 100.
    1. (a) Find the altitude at d = 100 m and d = 500 m.
    2. (b) At what horizontal distance does the drone reach ground level (A = 0)?
    3. (c) State the domain and range in context, explaining what each represents.
    4. (d) Is A(d) an increasing or decreasing function? What does this tell us about the drone's flight path?
    (a) A(100)=−√0+20=20 m. A(500)=−√400+20=−20+20=0 m.    (b) Set A=0: −√(d−100)+20=0 → √(d−100)=20 → d−100=400 → d=500 m.    (c) Domain: 100≤d≤500 (the drone is in flight). Range: 0≤A≤20 (altitude from ground up to max 20 m).    (d) Decreasing function (as d increases, A decreases due to the negative in front of √). The drone descends gradually from 20 m at d=100 to ground level at d=500. The square root shape means it descends steeply at first then more slowly.

11. Problem Solving — Mixed function identification

    Multi-step.
    1. (a) A curve has the shape of a hyperbola with asymptotes x = 3 and y = 0, passes through (5, 4) and (1, −4). Verify these points are consistent with y = a/(x − 3) and find a.
    2. (b) A relation is described as all points inside or on a circle centred at (−2, 1) with radius 3. Write an inequality to describe this region.
    3. (c) Consider f(x) = 2/x and g(x) = √(x − 1). Find f(g(5)) and g(f(2)). State any domain restrictions that apply.
    (a) Sub (5,4): 4=a/(5−3)=a/2 → a=8. Check (1,−4): 8/(1−3)=8/(−2)=−4 ✓. Equation: y=8/(x−3).    (b) (x+2)² + (y−1)² ≤ 9.    (c) g(5)=√(5−1)=√4=2. f(g(5))=f(2)=2/2=1.   f(2)=2/2=1. g(f(2))=g(1)=√(1−1)=√0=0. Domain: for f(g(x)), need x≥1 (for g) and g(x)≠0, i.e. x≠1; so domain x>1. For g(f(x)), need f(x)≥1, i.e. 2/x≥1, so 0<x≤2.

12. Understanding — Reciprocal and square root combined

    1. (a) For f(x) = 4/(x − 1) + 2 and g(x) = √(x − 2) + 1:
          • State all asymptotes and the endpoint.
          • Find any x- and y-intercepts.
          • State domain and range of each.
    2. (b) At what x-value(s) do f(x) and g(x) intersect? (Use technology or substitution — note: this may not have a neat algebraic solution.)
    (a) f(x)=4/(x−1)+2: VA x=1, HA y=2. x-int: 0=4/(x−1)+2 → x−1=−2 → x=−1; x-int (−1,0). y-int (x=0): 4/(−1)+2=−2; y-int (0,−2). Domain x≠1, Range y≠2.   g(x)=√(x−2)+1: endpoint (2,1). y-int: x=0 gives √(−2) undefined — no y-intercept. x-int: √(x−2)=−1 impossible (sqrt ≥0) — no x-intercept. Domain x≥2, Range y≥1.    (b) Set 4/(x−1)+2=√(x−2)+1. For x≥2: at x=2, f(2)=4/1+2=6, g(2)=1. At x=5, f(5)=4/4+2=3, g(5)=√3+1≈2.73. At x=6, f(6)=4/5+2=2.8, g(6)=√4+1=3. Intersection is between x=5 and x=6, approximately x≈5.5 (use technology for exact value).