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WACE Maths Methods study materials
WACE Maths Methods study resources

WACE Maths Methods practice questions

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Question 1Functions and graphs

The function f is defined by f(x) = 2x² − 8x + 3. What are the coordinates of the vertex of the graph of y = f(x)?

  1. (2, −5) Correct answer.
  2. (2, 3)
  3. (4, 3)
  4. (−2, 27)

Explanation

For a quadratic written as y = ax² + bx + c, the axis of symmetry is x = −b/(2a). Here a = 2 and b = −8, so x = 8/4 = 2. Substituting gives f(2) = 8 − 16 + 3 = −5, so the vertex is (2, −5).

Question 2Functions and graphs

Which statement correctly describes the domain and range of f(x) = √(x − 4)?

  1. Domain x ≥ 4, range y ≥ 4
  2. Domain x ≥ 0, range y ≥ 4
  3. Domain x ≥ 4, range y ≥ 0 Correct answer.
  4. Domain x ≤ 4, range y ≤ 0

Explanation

A square root requires a non-negative argument, so x − 4 ≥ 0, giving x ≥ 4. The principal square root is never negative, so the range is y ≥ 0, with the minimum value 0 occurring at x = 4.

Question 3Functions and graphs

The graph of y = f(x) is transformed to y = f(x − 3) + 2. Which description of the transformation is correct?

  1. Translation 3 units right and 2 units up Correct answer.
  2. Translation 2 units right and 3 units up
  3. Translation 3 units left and 2 units up
  4. Translation 3 units right and 2 units down

Explanation

Replacing x with x − 3 shifts the graph 3 units in the positive x-direction, because the original output now occurs 3 units later. Adding 2 outside the function moves every point 2 units up.

Question 4Functions and graphs

What is the inverse of the function f(x) = 3x − 6?

  1. f⁻¹(x) = (x + 6)/3 Correct answer.
  2. f⁻¹(x) = 1/(3x − 6)
  3. f⁻¹(x) = (x − 6)/3
  4. f⁻¹(x) = 3x + 6

Explanation

To find an inverse, write y = 3x − 6 and solve for x, giving x = (y + 6)/3, then interchange the variable names. The result is f⁻¹(x) = (x + 6)/3, which undoes the operations in reverse order.

Question 5Functions and graphs

The functions f(x) = x² and g(x) = x + 1 are given. What is f(g(x))?

  1. (x + 1)² Correct answer.
  2. x² + x + 1
  3. x² − 1
  4. x² + 1

Explanation

In the composite f(g(x)), the output of g becomes the input of f. Since f squares its input, substituting g(x) = x + 1 gives f(g(x)) = (x + 1)².

Question 6Functions and graphs

Consider the graph of y = 1/x. Which statement about its asymptotes is correct?

  1. It has asymptotes x = 0 and y = 0 Correct answer.
  2. It has no asymptotes
  3. It has asymptotes x = 1 and y = 1
  4. It has only the asymptote y = 0

Explanation

The expression 1/x is undefined at x = 0, and as x approaches 0 the magnitude of y grows without bound, giving a vertical asymptote x = 0. As x becomes large in magnitude, 1/x approaches 0, giving a horizontal asymptote y = 0.

Question 7Functions and graphs

The graph of y = f(x) passes through the point (2, 5). Which point must lie on the graph of y = −f(x)?

  1. (−2, 5)
  2. (2, −5) Correct answer.
  3. (−2, −5)
  4. (5, 2)

Explanation

The transformation y = −f(x) reflects the graph in the x-axis, so each y-coordinate changes sign while x stays the same. The point (2, 5) therefore maps to (2, −5).

Question 8Functions and graphs

For the exponential function f(x) = 2ˣ, which statement is true?

  1. The range is y ≥ 0 and the graph has a vertical asymptote x = 0
  2. The range is y > 2 and the graph has a horizontal asymptote y = 2
  3. The range is all real numbers and there are no asymptotes
  4. The range is y > 0 and the graph has a horizontal asymptote y = 0 Correct answer.

Explanation

A positive base raised to any real power is always positive, so 2ˣ > 0 for every x and the range is y > 0. As x becomes very negative, 2ˣ approaches 0, so the x-axis is a horizontal asymptote.

Question 9Functions and graphs

The graph of the quadratic function y = x² − 6x + 5 crosses the x-axis at which pair of points?

  1. (2, 0) and (3, 0)
  2. (1, 0) and (5, 0) Correct answer.
  3. (0, 0) and (6, 0)
  4. (−1, 0) and (−5, 0)

Explanation

Setting y = 0 gives x² − 6x + 5 = 0, which factorises as (x − 1)(x − 5) = 0. The solutions x = 1 and x = 5 are the x-intercepts, so the graph crosses at (1, 0) and (5, 0).

Question 10Functions and graphs

A parabola has x-intercepts at x = −2 and x = 4 and passes through the point (0, −8). What is its equation?

  1. y = 2(x + 2)(x − 4)
  2. y = −(x + 2)(x − 4)
  3. y = (x + 2)(x − 4) Correct answer.
  4. y = 0.5(x + 2)(x − 4)

Explanation

With intercepts at −2 and 4, the equation has the form y = a(x + 2)(x − 4). Substituting (0, −8) gives −8 = a(2)(−4) = −8a, so a = 1 and the equation is y = (x + 2)(x − 4).

Question 11Functions and graphs

The function f(x) = x² − 4x + 7 is written in the form f(x) = (x − h)² + k. What are the values of h and k?

  1. h = 2, k = 7
  2. h = −2, k = 3
  3. h = 2, k = 3 Correct answer.
  4. h = 4, k = 7

Explanation

Completing the square gives x² − 4x + 7 = (x − 2)² − 4 + 7 = (x − 2)² + 3. Comparing with (x − h)² + k gives h = 2 and k = 3.

Question 12Functions and graphs

Which of the following functions has an inverse that is also a function?

  1. f(x) = |x|
  2. f(x) = x³ + 1 Correct answer.
  3. f(x) = x² − 4
  4. f(x) = x² + 2x

Explanation

A function has an inverse function only if it is one-to-one, which the horizontal line test confirms. The cubic x³ + 1 is strictly increasing, so it passes the test, whereas the other options are many-to-one over the reals.

Question 13Functions and graphs

The graph of y = f(x) is reflected in the line y = x to give y = g(x). If f(3) = 7, which statement about g is correct?

  1. g(−3) = 7
  2. g(3) = 7
  3. g(7) = −3
  4. g(7) = 3 Correct answer.

Explanation

Reflecting in y = x swaps the coordinates of every point, so the point (3, 7) on f becomes (7, 3) on g. This means g(7) = 3, which is the defining property of an inverse relation.

Question 14Functions and graphs

The functions f(x) = 2x + 1 and g(x) = x² − 3 are given. What is g(f(x))?

  1. 4x² + 1
  2. 2x² − 5
  3. 4x² + 4x + 4
  4. 4x² + 4x − 2 Correct answer.

Explanation

Substituting f(x) into g gives g(f(x)) = (2x + 1)² − 3. Expanding, (2x + 1)² = 4x² + 4x + 1, so subtracting 3 yields 4x² + 4x − 2.

Question 15Functions and graphs

The graph of y = 1/(x − 2) + 3 has asymptotes at which lines?

  1. x = 3 and y = 2
  2. x = 2 and y = 3 Correct answer.
  3. x = −2 and y = 3
  4. x = 2 and y = −3

Explanation

The denominator x − 2 is zero at x = 2, where the function is undefined and grows without bound, so x = 2 is a vertical asymptote. The added constant 3 shifts the horizontal asymptote from y = 0 to y = 3.

Question 16Functions and graphs

The function f(x) = x² − 6x + 5 is restricted to the domain x ≥ 3. What is the range of this restricted function?

  1. y ≥ −4 Correct answer.
  2. y ≥ 0
  3. y ≤ −4
  4. y ≥ 5

Explanation

Completing the square gives f(x) = (x − 3)² − 4, so the vertex is at (3, −4). On the domain x ≥ 3 the function is increasing from that vertex, so the minimum value is −4 and the range is y ≥ −4.

Question 17Functions and graphs

The graph of y = f(x) has a local maximum at (1, 4). After the transformation y = f(2x), where is the corresponding local maximum?

  1. (2, 4)
  2. (1, 8)
  3. (1, 2)
  4. (0.5, 4) Correct answer.

Explanation

The transformation y = f(2x) compresses the graph horizontally by a factor of 2, so x-coordinates are halved while y-coordinates are unchanged. The point (1, 4) therefore becomes (0.5, 4).

Question 18Functions and graphs

For which values of k does the equation x² + kx + 9 = 0 have two distinct real solutions?

  1. −6 < k < 6
  2. k = ±6
  3. k < −6 or k > 6 Correct answer.
  4. k > 6 only

Explanation

A quadratic has two distinct real roots when its discriminant b² − 4ac is positive. Here the discriminant is k² − 36, which is positive when k² > 36, that is k < −6 or k > 6.

Question 19Functions and graphs

The graph of y = aˣ passes through the point (2, 9). What is the value of a, given a > 0?

  1. 81
  2. 9
  3. 4.5
  4. 3 Correct answer.

Explanation

Substituting the point gives a² = 9. Since a is required to be positive, the only valid solution is a = 3, after which the graph passes through (2, 9).

Question 20Functions and graphs

The graph of y = f(x) is shown to pass through (−1, 2) and (3, 2), and f is a quadratic with a line of symmetry midway between these points. What is the axis of symmetry?

  1. x = 3
  2. x = 2
  3. x = 1 Correct answer.
  4. x = −1

Explanation

For a parabola, points with equal y-values are symmetric about the axis of symmetry, which lies halfway between their x-coordinates. The midpoint of −1 and 3 is (−1 + 3)/2 = 1, so the axis is x = 1.

Question 21Functions and graphs

The function f(x) = (x + 1)/(x − 3) is defined for x ≠ 3. What is f⁻¹(x)?

  1. (3x − 1)/(x + 1)
  2. (3x + 1)/(x − 1) Correct answer.
  3. (x + 1)/(x − 3)
  4. (x − 3)/(x + 1)

Explanation

Write y = (x + 1)/(x − 3) and solve for x: y(x − 3) = x + 1, so yx − x = 3y + 1 and x(y − 1) = 3y + 1. Hence x = (3y + 1)/(y − 1), and interchanging variables gives f⁻¹(x) = (3x + 1)/(x − 1).

Question 22Functions and graphs

The graph of y = f(x) is transformed to y = 2f(−x) + 1. The point (3, −4) on y = f(x) maps to which point?

  1. (−3, −9)
  2. (−3, −7) Correct answer.
  3. (−3, 9)
  4. (3, −7)

Explanation

Replacing x with −x reflects in the y-axis, so the x-coordinate becomes −3. Multiplying by 2 gives 2(−4) = −8, and adding 1 gives −7, so the image point is (−3, −7).

Question 23Functions and graphs

The quadratic y = x² + bx + c has its vertex on the line y = x. If the vertex has x-coordinate 2, what is the value of c?

  1. 2
  2. 4
  3. 8
  4. 6 Correct answer.

Explanation

The vertex x-coordinate satisfies x = −b/2, so −b/2 = 2 gives b = −4. The vertex lies on y = x, so its y-coordinate is 2, and substituting (2, 2) into y = x² − 4x + c gives 2 = 4 − 8 + c, hence c = 6.

Question 24Functions and graphs

The function f(x) = x² − 2x + 3 is restricted to x ≤ 1. What is the rule for f⁻¹(x)?

  1. f⁻¹(x) = 2 − √(x − 3)
  2. f⁻¹(x) = 1 − √(x − 2) Correct answer.
  3. f⁻¹(x) = √(x − 2) + 3
  4. f⁻¹(x) = 1 + √(x − 2)

Explanation

Completing the square gives f(x) = (x − 1)² + 2, so y − 2 = (x − 1)² and x = 1 ± √(y − 2). Because the domain is x ≤ 1, the negative branch applies, giving f⁻¹(x) = 1 − √(x − 2).

Question 25Functions and graphs

The graph of y = f(x) has a vertical asymptote at x = 2 and a horizontal asymptote at y = −1. Which asymptotes does the graph of y = f(x + 3) − 2 have?

  1. x = 5 and y = −3
  2. x = 5 and y = 1
  3. x = −1 and y = 1
  4. x = −1 and y = −3 Correct answer.

Explanation

Replacing x with x + 3 shifts the graph 3 units left, so the vertical asymptote moves from x = 2 to x = −1. Subtracting 2 shifts the graph down, so the horizontal asymptote moves from y = −1 to y = −3.

Question 26Functions and graphs

The function f(x) = 3x − 6 is defined for all real x. What is f⁻¹(x)?

  1. f⁻¹(x) = (x − 6)/3
  2. f⁻¹(x) = (x + 6)/3 Correct answer.
  3. f⁻¹(x) = 3x + 6
  4. f⁻¹(x) = 1/(3x − 6)

Explanation

To invert, write y = 3x − 6 and solve for x: add 6 to both sides then divide by 3, giving x = (y + 6)/3. Swapping the variable names gives f⁻¹(x) = (x + 6)/3, and checking f(f⁻¹(x)) = x confirms it.

Question 27Functions and graphs

The function f(x) = x² + 4 is defined on the domain x ≥ 0. Which statement describes the range of f and the domain of f⁻¹?

  1. Range of f is y ≥ 4; domain of f⁻¹ is x ≥ 0
  2. Range of f is all real y; domain of f⁻¹ is all real x
  3. Range of f is y ≥ 0; domain of f⁻¹ is x ≥ 0
  4. Range of f is y ≥ 4; domain of f⁻¹ is x ≥ 4 Correct answer.

Explanation

On x ≥ 0 the smallest output is f(0) = 4 and f increases without bound, so the range is y ≥ 4. The domain of the inverse equals the range of the original function, hence f⁻¹ has domain x ≥ 4.

Question 28Functions and graphs

The graph of y = f(x) is transformed to y = f(x − 3) + 2. How is the graph of y = f(x) moved?

  1. 3 units right and 2 units up Correct answer.
  2. 3 units left and 2 units up
  3. 3 units left and 2 units down
  4. 3 units right and 2 units down

Explanation

Replacing x by x − 3 shifts the graph 3 units in the positive x-direction, because the feature now occurs when the input to f is unchanged, i.e. 3 larger. Adding 2 outside the function shifts every point 2 units up.

Question 29Functions and graphs

The graph of y = f(x) is transformed to y = f(2x). A point (4, 7) lies on y = f(x). Which point lies on y = f(2x)?

  1. (4, 14)
  2. (2, 7) Correct answer.
  3. (8, 7)
  4. (4, 3.5)

Explanation

For y = f(2x) to equal f(4), the input must satisfy 2x = 4, giving x = 2. The output is unchanged, so the point becomes (2, 7), a horizontal compression by factor ½ about the y-axis.

Question 30Functions and graphs

Consider the function f(x) = 1/(x + 2). Which statement about its graph is correct?

  1. It has a vertical asymptote at x = −2 and a horizontal asymptote at y = 1.
  2. It has a vertical asymptote at x = −2 and a horizontal asymptote at y = 0. Correct answer.
  3. It has a vertical asymptote at x = 2 and a horizontal asymptote at y = 0.
  4. It has no asymptotes because the function is continuous everywhere.

Explanation

The denominator is zero when x = −2, so the graph has a vertical asymptote there. As x grows large in magnitude the fraction approaches 0, giving the horizontal asymptote y = 0.

Question 31Functions and graphs

The function f(x) = x² − 8x + 11 is expressed in the form f(x) = (x − h)² + k. What is the minimum value of f(x)?

  1. −16
  2. 11
  3. 4
  4. −5 Correct answer.

Explanation

Completing the square gives x² − 8x + 11 = (x − 4)² − 16 + 11 = (x − 4)² − 5. Since the square term is never negative, the minimum value is −5, occurring at x = 4.

Question 32Functions and graphs

A quadratic function has its vertex at (2, −3) and passes through the point (0, 5). What is its equation?

  1. y = 2(x + 2)² − 3
  2. y = (x − 2)² − 3
  3. y = 2(x − 2)² − 3 Correct answer.
  4. y = −2(x − 2)² − 3

Explanation

The vertex form is y = a(x − 2)² − 3. Substituting (0, 5) gives 5 = 4a − 3, so a = 2, producing y = 2(x − 2)² − 3.

Question 33Functions and graphs

The parabola y = x² + 4x + 7 is transformed by a reflection in the x-axis. What is the equation of the resulting graph?

  1. y = −x² − 4x − 7 Correct answer.
  2. y = −x² + 4x + 7
  3. y = x² − 4x − 7
  4. y = x² + 4x − 7

Explanation

Reflecting in the x-axis replaces y with −y, so the new equation is −y = x² + 4x + 7, that is y = −x² − 4x − 7. Every output changes sign while the x-coordinates stay fixed.

Question 34Functions and graphs

The functions f(x) = x + 3 and g(x) = x² are given. What is f(g(x))?

  1. (x + 3)²
  2. x² + 3 Correct answer.
  3. x² + 3x + 9
  4. x² − 3

Explanation

In f(g(x)) the inner function g is applied first, producing x², and then f adds 3 to that result. Hence f(g(x)) = x² + 3, which differs from g(f(x)).

Question 35Functions and graphs

The function f(x) = (2x − 1)/(x + 4) is defined for x ≠ −4. What is the equation of the vertical asymptote of y = f(x)?

  1. x = −4 Correct answer.
  2. x = 4
  3. y = 2
  4. x = 1/2

Explanation

A vertical asymptote occurs where the denominator is zero while the numerator is not. Solving x + 4 = 0 gives x = −4, and the numerator there is −9, so the asymptote is x = −4.

Question 36Functions and graphs

The equation x² + (k − 2)x + 9 = 0 has exactly one real solution. What are the possible values of k?

  1. k = 8 only
  2. k = 8 or k = −4 Correct answer.
  3. k = 2 only
  4. k = −4 only

Explanation

A repeated root requires the discriminant to be zero: (k − 2)² − 36 = 0. Thus k − 2 = ±6, giving k = 8 or k = −4, and both values produce a perfect square.

Question 37Functions and graphs

The function f(x) = x² − 4x + 6 is restricted to the domain x ≥ 2, making it one-to-one. What is f⁻¹(x)?

  1. f⁻¹(x) = 2 − √(x − 2)
  2. f⁻¹(x) = √(x + 2) + 2
  3. f⁻¹(x) = 2 + √(x − 2) Correct answer.
  4. f⁻¹(x) = (x + 4)/2

Explanation

Completing the square gives f(x) = (x − 2)² + 2. Setting y = (x − 2)² + 2 and solving with x ≥ 2 gives x = 2 + √(y − 2), so f⁻¹(x) = 2 + √(x − 2), where the positive root matches the restricted domain.

Question 38Functions and graphs

The graph of y = f(x) has a vertical asymptote at x = 3 and passes through (1, 0). The transformation y = f(x + 2) − 1 is applied. Where are the vertical asymptote and the image of the point (1, 0)?

  1. Asymptote x = 3; image point (−1, −1)
  2. Asymptote x = 1; image point (−1, −1) Correct answer.
  3. Asymptote x = 5; image point (3, −1)
  4. Asymptote x = 1; image point (3, 1)

Explanation

In y = f(x + 2) − 1 the graph shifts 2 units left and 1 unit down. The asymptote x = 3 moves to x = 1, and the point (1, 0) becomes (1 − 2, 0 − 1) = (−1, −1).

Question 39Functions and graphs

The function f(x) = 2ˣ is transformed to g(x) = 2^(x − 1) + 3. Which statement about g is correct?

  1. g has range y > 3 and passes through (1, 4). Correct answer.
  2. g has range y > 3 and passes through (0, 5).
  3. g has range y > 0 and passes through (0, 4).
  4. g has range y > 1 and passes through (1, 5).

Explanation

The base exponential 2ˣ has range y > 0; shifting up 3 gives range y > 3. At x = 1 the exponent is 0, so g(1) = 1 + 3 = 4, meaning the graph passes through (1, 4).

Question 40Functions and graphs

The function f(x) = x³ − 3x is defined for all real x. Which statement about f is correct?

  1. f is an even function, so its graph is symmetric about the y-axis.
  2. f is one-to-one on its whole domain, so f⁻¹ is a function.
  3. f is one-to-one only for x ≥ 0.
  4. f is not one-to-one, so f has no inverse function on its whole domain. Correct answer.

Explanation

A cubic with two turning points fails the horizontal line test: for example f(0) = 0 and f(√3) = 0, so different inputs share an output. Therefore f has no inverse function unless its domain is restricted.

Question 41Trigonometric functions

What is the exact value of sin(π/6)?

  1. √2/2
  2. 1/2 Correct answer.
  3. √3/2
  4. 1

Explanation

On the unit circle, the angle π/6 (30°) gives the point (√3/2, 1/2), so the sine, which is the y-coordinate, equals 1/2. This is one of the standard exact values worth memorising.

Question 42Trigonometric functions

The period of the function y = cos(3x) is:

  1. 6π
  2. 3π
  3. 2π/3 Correct answer.
  4. π/3

Explanation

For y = cos(nx), the period equals 2π divided by the coefficient n. Here n = 3, so the period is 2π/3.

Question 43Trigonometric functions

What is the amplitude of y = -4 sin(2x) + 1?

  1. 1
  2. 2
  3. -4
  4. 4 Correct answer.

Explanation

The amplitude is the absolute value of the coefficient in front of the sine function. Since that coefficient is -4, the amplitude is 4; the negative sign only reflects the graph, and the +1 shifts it vertically.

Question 44Trigonometric functions

Evaluate tan(π/4).

  1. 1 Correct answer.
  2. √3
  3. 0
  4. 1/√3

Explanation

At π/4 (45°) the sine and cosine are equal, so their ratio is 1. Hence tan(π/4) = 1.

Question 45Trigonometric functions

Which of the following is the range of y = sin(x)?

  1. (-∞, ∞)
  2. [-1, 1] Correct answer.
  3. [0, 1]
  4. (-1, 1)

Explanation

The sine of any real angle lies between -1 and 1 inclusive, and both endpoints are attained (at π/2 and 3π/2). So the range is the closed interval [-1, 1].

Question 46Trigonometric functions

The graph of y = sin(x) is shifted π/2 units to the left. Which function describes the new graph?

  1. y = cos(x) Correct answer.
  2. y = sin(x) - π/2
  3. y = sin(x - π/2)
  4. y = cos(x) - π/2

Explanation

A left shift by π/2 replaces x with x + π/2, giving sin(x + π/2). Using the identity sin(x + π/2) = cos(x), this simplifies to cos(x).

Question 47Trigonometric functions

What is the period of y = tan(2x)?

  1. π/2 Correct answer.
  2. π/4
  3. 2π
  4. π

Explanation

The tangent function has natural period π, and dividing by the coefficient of x gives π/2. So y = tan(2x) repeats every π/2.

Question 48Trigonometric functions

Solve sin(x) = 1/2 for 0 ≤ x ≤ 2π.

  1. x = π/6 or x = 7π/6
  2. x = π/6 only
  3. x = π/3 or x = 2π/3
  4. x = π/6 or x = 5π/6 Correct answer.

Explanation

The reference angle for sin(x) = 1/2 is π/6. Sine is positive in the first and second quadrants, so the solutions in the given interval are π/6 and π - π/6 = 5π/6.

Question 49Trigonometric functions

The expression sin²(x) + cos²(x) simplifies to:

  1. 0
  2. 2sin(x)cos(x)
  3. tan²(x)
  4. 1 Correct answer.

Explanation

This is the Pythagorean identity, which holds for all real x because the point (cos x, sin x) lies on the unit circle of radius 1. Thus the sum of the squares equals 1.

Question 50Trigonometric functions

Simplify sin(π - x).

  1. cos(x)
  2. -cos(x)
  3. -sin(x)
  4. sin(x) Correct answer.

Explanation

The angle π - x is the supplement of x, and sine of supplementary angles is equal. So sin(π - x) = sin(x), consistent with the second-quadrant symmetry of the sine graph.