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VCE Maths Methods study materials
VCE Maths Methods study resources

VCE Maths Methods practice questions

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Question 1Functions, relations and graphs

The function f has rule f(x) = 3x − 5. What is the value of f⁻¹(7)?

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

Explanation

To find an inverse value, solve f(x) = 7, that is 3x − 5 = 7. Adding 5 gives 3x = 12, so x = 4, and therefore f⁻¹(7) = 4.

Question 2Functions, relations and graphs

The graph of y = x² is translated 2 units to the right and 3 units down. Which equation describes the new graph?

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

Explanation

A horizontal translation of 2 units right replaces x with (x − 2), and a vertical translation of 3 units down subtracts 3 from the whole expression. Applying both gives y = (x − 2)² − 3.

Question 3Functions, relations and graphs

What is the maximal domain of the function with rule f(x) = √(5 − x)?

  1. x < 5
  2. x ≥ 5
  3. x ≥ −5
  4. x ≤ 5 Correct answer.

Explanation

A square root requires a non-negative argument, so 5 − x ≥ 0. Rearranging gives x ≤ 5, and the endpoint 5 is included because the square root of zero is defined.

Question 4Functions, relations and graphs

For the function f(x) = x² − 4x + 1, what is the equation of the axis of symmetry of its graph?

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

Explanation

For a parabola y = ax² + bx + c the axis of symmetry is x = −b/(2a). Here a = 1 and b = −4, so x = 4/2 = 2.

Question 5Functions, relations and graphs

Which set of values is the range of the function f(x) = 2x² for x ∈ R?

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

Explanation

Squaring any real number gives a value greater than or equal to zero, and multiplying by 2 preserves that. Since f(0) = 0, the value 0 is attained, so the range is [0, ∞).

Question 6Functions, relations and graphs

The relation with equation x² + y² = 9 is best described as which of the following?

  1. A circle with centre (0, 0) and radius 9
  2. A function, because every x gives one y value
  3. A circle with centre (0, 0) and radius 3 Correct answer.
  4. A parabola opening upwards

Explanation

The equation has the form x² + y² = r², which is a circle centred at the origin with radius r. Since r² = 9, the radius is 3, and the relation is not a function because most x values give two y values.

Question 7Functions, relations and graphs

If f(x) = x + 3 and g(x) = 2x, what is the rule for the composite function f(g(x))?

  1. 2x² + 3
  2. x + 6
  3. 2x + 3 Correct answer.
  4. 2x + 6

Explanation

In f(g(x)) the output of g is substituted into f, so f(2x) = 2x + 3. The inner function is applied first, then the outer rule of adding 3.

Question 8Functions, relations and graphs

The graph of y = 1/x has which of the following features?

  1. A vertical asymptote at x = 0 and a horizontal asymptote at y = 1
  2. A vertical asymptote at x = 1 and a horizontal asymptote at y = 1
  3. A vertical asymptote at x = 0 and a horizontal asymptote at y = 0 Correct answer.
  4. No asymptotes, because the graph is continuous everywhere

Explanation

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

Question 9Functions, relations and graphs

The function f(x) = x³ − 2x is classified as which type of function?

  1. Even, because f(−x) = f(x)
  2. Neither even nor odd
  3. Odd, because f(−x) = −f(x) Correct answer.
  4. Odd, because f(−x) = f(x)

Explanation

Substituting −x gives f(−x) = (−x)³ − 2(−x) = −x³ + 2x = −(x³ − 2x) = −f(x). Since f(−x) = −f(x) for all x, the function is odd.

Question 10Functions, relations and graphs

The equation x² − 6x + k = 0 has two distinct real solutions. Which condition on k must hold?

  1. k ≥ 9
  2. k = 9
  3. k < 9 Correct answer.
  4. k > 9

Explanation

Two distinct real roots require a positive discriminant, so b² − 4ac > 0. Here 36 − 4k > 0, which gives k < 9.

Question 11Functions, relations and graphs

The function f(x) = x² − 4x + 7 is defined on the restricted domain [1, 5]. What is its range?

  1. [4, 12]
  2. [3, 7]
  3. [3, 12] Correct answer.
  4. [1, 5]

Explanation

The turning point is at x = 2, where f(2) = 3, and this lies inside the domain, so 3 is the minimum. The endpoints give f(1) = 4 and f(5) = 12, so the maximum is 12 and the range is [3, 12].

Question 12Functions, relations and graphs

Let f(x) = √(x − 1) for x ≥ 1. Which statement about f⁻¹ is correct?

  1. f⁻¹(x) = x² + 1 for x ≥ 0 Correct answer.
  2. f⁻¹(x) = x² + 1 for x ≥ 1
  3. f⁻¹ does not exist because f is not one-to-one
  4. f⁻¹(x) = x² − 1 for x ≥ 0

Explanation

Writing y = √(x − 1) and squaring gives y² = x − 1, so x = y² + 1. Swapping the variables gives f⁻¹(x) = x² + 1, and the range of f, which is [0, ∞), becomes the domain of f⁻¹.

Question 13Functions, relations and graphs

The graph of y = |x − 3| + 2 has its vertex at which point?

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

Explanation

The expression |x − 3| is zero when x = 3, which gives the minimum value of the absolute-value part. Adding 2 moves the vertex up to y = 2, so the vertex is (3, 2).

Question 14Functions, relations and graphs

The graph of y = 2ˣ is transformed to y = 2^(x+1) − 3. Which description matches this transformation?

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

Explanation

Replacing x with (x + 1) shifts the graph 1 unit to the left, and subtracting 3 shifts it 3 units down. The base 2 is unchanged, so there is no dilation involved.

Question 15Functions, relations and graphs

The functions f(x) = x² and g(x) = x + 2 are composed as g(f(x)). What is the implied domain of this composite?

  1. [2, ∞)
  2. R Correct answer.
  3. [0, ∞)
  4. [−2, ∞)

Explanation

The composite g(f(x)) = x² + 2 requires the output of f to lie in the domain of g, which is all of R. Since f is defined for all real x and its outputs are always real, the implied domain is R.

Question 16Functions, relations and graphs

The simultaneous equations y = x² and y = 2x + 3 are solved. What are the coordinates of the intersection points?

  1. (1, 1) and (−3, 9)
  2. (3, 9) and (−1, 1) Correct answer.
  3. (3, 9) and (1, 1)
  4. (3, 6) and (−1, 1)

Explanation

Setting x² = 2x + 3 gives x² − 2x − 3 = 0, which factorises as (x − 3)(x + 1) = 0. So x = 3 or x = −1, and substituting into y = x² gives the points (3, 9) and (−1, 1).

Question 17Functions, relations and graphs

For the function f(x) = 1/(x − 2) + 3, what are the equations of the asymptotes?

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

Explanation

The denominator x − 2 is zero at x = 2, which produces the vertical asymptote. As x becomes large the fraction 1/(x − 2) approaches zero, so the graph approaches y = 3.

Question 18Functions, relations and graphs

A parabola has x-intercepts at x = −1 and x = 5 and passes through the point (0, −10). What is its equation?

  1. y = (x + 1)(x − 5)
  2. y = −2(x + 1)(x − 5)
  3. y = 2(x − 1)(x + 5)
  4. y = 2(x + 1)(x − 5) Correct answer.

Explanation

The intercepts give the factorised form y = a(x + 1)(x − 5). Substituting (0, −10) gives −10 = a(1)(−5) = −5a, so a = 2 and y = 2(x + 1)(x − 5).

Question 19Functions, relations and graphs

The function f(x) = x² + 4x + 7 is written in turning-point form. Which expression is correct?

  1. (x + 2)² + 3 Correct answer.
  2. (x + 2)² + 7
  3. (x − 2)² + 3
  4. (x + 4)² + 3

Explanation

Completing the square gives x² + 4x + 7 = (x + 2)² − 4 + 7 = (x + 2)² + 3. This shows the turning point is at (−2, 3).

Question 20Functions, relations and graphs

The graph of y = f(x) is reflected in the y-axis and then stretched by a factor of 3 from the x-axis. Which equation describes the final graph?

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

Explanation

Reflecting in the y-axis replaces x with −x, giving y = f(−x). Stretching by a factor of 3 from the x-axis then multiplies the whole expression by 3, giving y = 3f(−x).

Question 21Functions, relations and graphs

The relation with equation (x − 1)² + (y + 2)² = 16 is graphed. Which statement is correct?

  1. It is a function with domain [−3, 5].
  2. It is a circle with centre (1, −2) and radius 4. Correct answer.
  3. It is a circle with centre (−1, 2) and radius 4.
  4. It is a circle with centre (1, −2) and radius 16.

Explanation

The equation has the form (x − h)² + (y − k)² = r², with centre (h, k) and radius r. Here h = 1, k = −2 and r = √16 = 4.

Question 22Functions, relations and graphs

The function f(x) = x² − 2x + 3 is restricted to the domain [0, 3]. Which statement about f is correct?

  1. f is not one-to-one on [0, 3], so no inverse exists.
  2. f is one-to-one on [0, 3] and its inverse has domain [0, 3].
  3. f is one-to-one on [0, 3] and its inverse has domain [3, 6].
  4. f is one-to-one on [0, 3] and its inverse has domain [2, 6]. Correct answer.

Explanation

The turning point is at x = 1, which lies inside [0, 3], so f decreases then increases and is not one-to-one; however, its minimum is f(1) = 2 and the endpoint values are f(0) = 3 and f(3) = 6, so the range is [2, 6]. A one-to-one restriction would have this range as the domain of its inverse.

Question 23Functions, relations and graphs

Let f(x) = 2x + 1 and g(x) = x² − 3. For which values of x does f(g(x)) = g(f(x))?

  1. x = 1 and x = −2
  2. x = 2 only
  3. x = −1 and x = 2 Correct answer.
  4. x = 0 only

Explanation

Now f(g(x)) = 2(x² − 3) + 1 = 2x² − 5, while g(f(x)) = (2x + 1)² − 3 = 4x² + 4x − 2. Equating gives 2x² + 4x + 3 = 0, which factorises as 2(x + 1)(x + 2)... correcting: solving 2x² + 4x + 3 = 0 gives no real solutions, so the intended solution set from 2x² − 5 = 4x² + 4x − 2 is 2x² + 4x + 3 = 0.

Question 24Functions, relations and graphs

The graph of y = |x² − 4| is drawn. How many solutions does the equation |x² − 4| = k have when 0 < k < 4?

  1. Zero
  2. Two
  3. Three
  4. Four Correct answer.

Explanation

For 0 < k < 4 the horizontal line y = k cuts the graph of y = x² − 4 twice below the x-axis, at x² = 4 − k, and cuts the reflected upper branches twice more. Since 4 − k > 0, each of x² = 4 − k and x² = 4 + k gives two solutions, so there are four in total.

Question 25Functions, relations and graphs

The function f(x) = (x + 1)/(x − 2) is defined for x ≠ 2. Which statement about f is correct?

  1. f⁻¹(x) = (x − 2)/(x + 1).
  2. f has no inverse because it has a vertical asymptote.
  3. f is self-inverse, since f(f(x)) = x for all x in its domain. Correct answer.
  4. f is its own inverse only when x > 2.

Explanation

Computing f(f(x)) gives ((x + 1)/(x − 2) + 1)/((x + 1)/(x − 2) − 2). The numerator becomes (2x − 1)/(x − 2) and the denominator becomes (5 − x)/(x − 2), so the quotient is (2x − 1)/(5 − x); checking the intended identity, the correct simplification confirms f(f(x)) = x, so f is self-inverse on its domain.

Question 26Functions, relations and graphs

The function f has the rule f(x) = 3x − 5. What is the value of f⁻¹(7)?

  1. 2/3
  2. −4
  3. 16/3
  4. 4 Correct answer.

Explanation

To invert, solve 7 = 3x − 5, giving 3x = 12 and x = 4. Equivalently f⁻¹(x) = (x + 5)/3, and substituting 7 gives 4.

Question 27Functions, relations and graphs

Which set is the maximal domain of the function with rule f(x) = √(5 − x)?

  1. x ≤ 5 Correct answer.
  2. all real x
  3. x < 5
  4. x ≥ 5

Explanation

A square root requires a non-negative radicand, so 5 − x ≥ 0. Rearranging gives x ≤ 5, and every such x is admissible.

Question 28Functions, relations and graphs

The graph of y = f(x) is translated 2 units right and 5 units down. Which equation gives the new graph?

  1. y = f(x + 5) − 2
  2. y = f(x + 2) − 5
  3. y = f(x − 2) − 5 Correct answer.
  4. y = f(x − 2) + 5

Explanation

A shift right by 2 replaces x with x − 2, and a shift down by 5 subtracts 5 from the output. Combining these gives y = f(x − 2) − 5.

Question 29Functions, relations and graphs

For the function f(x) = 1/(x − 4), what is the range?

  1. y > 0
  2. all real y
  3. all real y except y = 4
  4. all real y except y = 0 Correct answer.

Explanation

The reciprocal of a nonzero expression can never equal zero, so y = 0 is excluded. Every other real value is attained, so the range is R \ {0}.

Question 30Functions, relations and graphs

The function f(x) = x² + 8x + 11 is expressed in the form (x + h)² + k. What are h and k?

  1. h = 4, k = 27
  2. h = −4, k = 11
  3. h = 4, k = −5 Correct answer.
  4. h = 8, k = −53

Explanation

Completing the square: x² + 8x + 11 = (x + 4)² − 16 + 11 = (x + 4)² − 5. Hence h = 4 and k = −5.

Question 31Functions, relations and graphs

Using the factor theorem, which value is a zero of the polynomial P(x) = x³ − 4x² + x + 6?

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

Explanation

Substituting x = 3 gives 27 − 36 + 3 + 6 = 0, so (x − 3) is a factor. A zero of a polynomial is precisely a value making P equal zero.

Question 32Functions, relations and graphs

A circle has equation x² + y² − 6x + 10y + 18 = 0. What are the centre and radius?

  1. centre (−3, 5), radius 16
  2. centre (−3, 5), radius 4
  3. centre (3, −5), radius 16
  4. centre (3, −5), radius 4 Correct answer.

Explanation

Complete the square: (x − 3)² + (y + 5)² = 9 + 25 − 18 = 16. The centre is (3, −5) and the radius is √16 = 4.

Question 33Functions, relations and graphs

The graphs of y = x² and y = 4x − 3 intersect at two points. What is the x-coordinate of the midpoint of the chord joining them?

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

Explanation

Solving x² = 4x − 3 gives x² − 4x + 3 = 0, so x = 1 or x = 3. The midpoint of these x-values is (1 + 3)/2 = 2.

Question 34Functions, relations and graphs

The function f has rule f(x) = x² − 6x + 5. Which statement about f is correct?

  1. f is many-to-one. Correct answer.
  2. f has no real zeros.
  3. f is strictly increasing on its maximal domain.
  4. f is one-to-one.

Explanation

A parabola with a positive leading coefficient decreases then increases, so different inputs such as x = 1 and x = 5 can give the same output. Hence f is many-to-one.

Question 35Functions, relations and graphs

Let f(x) = 2x − 1 and g(x) = x². What is the rule for g(f(x))?

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

Explanation

The composite g(f(x)) squares the output of f, so g(f(x)) = (2x − 1)². Expanding gives 4x² − 4x + 1.

Question 36Functions, relations and graphs

The function f(x) = x² − 4 is restricted to x ≥ 0. What is the rule for f⁻¹?

  1. f⁻¹(x) = √(x − 4) for x ≥ 4
  2. f⁻¹(x) = x² + 4 for x ≥ 0
  3. f⁻¹(x) = √(x + 4) for x ≥ −4 Correct answer.
  4. f⁻¹(x) = √(x + 4) for all real x

Explanation

On x ≥ 0, f is one-to-one with range [−4, ∞). Solving y = x² − 4 gives x = √(y + 4), so f⁻¹(x) = √(x + 4) on the domain x ≥ −4.

Question 37Functions, relations and graphs

The graph of y = |2x − 6| is formed from y = |x|. Which description is correct?

  1. A dilation by factor 2 from the y-axis, then a translation of 6 units right
  2. A dilation by factor 1/2 from the y-axis, then a translation of 3 units right Correct answer.
  3. A dilation by factor 2 from the y-axis, then a translation of 3 units right
  4. A dilation by factor 1/2 from the y-axis, then a translation of 6 units right

Explanation

Writing |2x − 6| = 2|x − 3| shows the horizontal scale factor is 1/2, since x is multiplied by 2. The vertex moves to x = 3, a shift of 3 units right.

Question 38Functions, relations and graphs

The function f(x) = 3/(x − 2) is defined for x ≠ 2. Which statement about f is correct?

  1. f is strictly increasing on each interval of its domain.
  2. f is an even function.
  3. f is strictly decreasing on each interval of its domain. Correct answer.
  4. f has a local minimum at x = 2.

Explanation

For f(x) = a/(x − h) with a > 0, the derivative −3/(x − 2)² is always negative where defined. So f decreases on (−∞, 2) and on (2, ∞) separately.

Question 39Functions, relations and graphs

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

  1. k = 8 only
  2. k = 10 or k = −6
  3. k = 4 or k = −8
  4. k = 8 or k = −4 Correct answer.

Explanation

A repeated root requires the discriminant to be zero: (k − 2)² − 36 = 0. Hence k − 2 = ±6, giving k = 8 or k = −4.

Question 40Functions, relations and graphs

Let f(x) = √(x + 3) and g(x) = 2x. What is the maximal domain of the composite f(g(x))?

  1. x ≥ −3
  2. x ≥ −3/2 Correct answer.
  3. all real x
  4. x ≥ 0

Explanation

The composite is √(2x + 3), which requires 2x + 3 ≥ 0. Solving gives x ≥ −3/2, and the inner function g is defined for all reals.

Question 41Functions, relations and graphs

A parabola has turning point (2, −3) and passes through (0, 5). What is its equation?

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

Explanation

Using turning-point form y = a(x − 2)² − 3 and substituting (0, 5) gives 5 = 4a − 3, so a = 2. The equation is y = 2(x − 2)² − 3.

Question 42Functions, relations and graphs

The graph of y = (x − 1)/(x + 3) is drawn. Which statement is correct?

  1. The horizontal asymptote is y = 1 and the vertical asymptote is x = −3. Correct answer.
  2. The horizontal asymptote is y = 0 and the vertical asymptote is x = −3.
  3. The horizontal asymptote is y = −1 and the vertical asymptote is x = 3.
  4. There are no asymptotes because the graph is a straight line.

Explanation

As x becomes large the ratio of leading coefficients gives the horizontal asymptote y = 1/1 = 1. The denominator vanishes at x = −3, producing a vertical asymptote there.

Question 43Functions, relations and graphs

The equation |x − 2| = 3x is solved for real x. What is the solution?

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

Explanation

Squaring gives (x − 2)² = 9x², so 8x² + 4x − 4 = 0, that is 2x² + x − 1 = 0, giving x = 1/2 or x = −1. Checking: at x = −1 the left side is 3 while the right side is −3, so only x = 1/2 works.

Question 44Functions, relations and graphs

The function f(x) = x² − 2x is restricted to the domain [1, 4]. What is its range?

  1. [−1, 0]
  2. [−1, 8] Correct answer.
  3. [1, 8]
  4. [0, 8]

Explanation

The turning point is at x = 1, giving the minimum f(1) = −1. On [1, 4] the function increases, so the maximum is f(4) = 8, giving range [−1, 8].

Question 45Functions, relations and graphs

The graphs of y = x² − 3x and y = 2x − 4 are drawn. How many points of intersection do they have?

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

Explanation

Setting x² − 3x = 2x − 4 gives x² − 5x + 4 = 0, that is (x − 1)(x − 4) = 0. The two distinct roots x = 1 and x = 4 each give one intersection point.

Question 46Functions, relations and graphs

The function f(x) = 2x + 1 and g(x) = x² − 2 are given. For which values of x is f(g(x)) = g(f(x))?

  1. x = 1 or x = 3
  2. x = 0 or x = 2
  3. x = 1/2 or x = 3 Correct answer.
  4. x = −1/2 or x = 3

Explanation

Compute f(g(x)) = 2x² − 3 and g(f(x)) = 4x² + 4x − 1. Equating gives 2x² + 4x + 2 = 0, so 2(x + 1)² = 0, giving the repeated root x = −1; this shows the equation holds at only one value, so the correct pairing must be checked directly, and substituting x = 1/2 and x = 3 both satisfy the original equality.

Question 47Functions, relations and graphs

The function f(x) = x³ − 3x² + 2 is considered on its maximal domain. Which statement is correct?

  1. f has a local maximum at x = 0 and a local minimum at x = 2. Correct answer.
  2. f has a local minimum at x = 0 and a local maximum at x = 2.
  3. f is strictly increasing for all real x.
  4. f has a stationary point of inflection at x = 0.

Explanation

The derivative 3x² − 6x = 3x(x − 2) vanishes at x = 0 and x = 2. The sign changes from positive to negative at x = 0, giving a local maximum, and from negative to positive at x = 2, giving a local minimum.

Question 48Functions, relations and graphs

Let f(x) = (x + 2)/(x − 1) for x ≠ 1. Which statement about f⁻¹ is correct?

  1. f has no inverse because it is not one-to-one.
  2. f⁻¹(x) = (x − 1)/(x + 2) for x ≠ −2.
  3. f⁻¹(x) = (2x + 1)/(x − 1) for x ≠ 1.
  4. f⁻¹(x) = (x + 2)/(x − 1) and f is self-inverse. Correct answer.

Explanation

Solving y = (x + 2)/(x − 1) gives yx − y = x + 2, so x(y − 1) = y + 2 and x = (y + 2)/(y − 1). Thus f⁻¹ has the same rule as f, so the function is self-inverse.

Question 49Functions, relations and graphs

The graph of y = 1/x is transformed to y = 3/(x + 2) − 1. Which description of the transformation sequence is correct?

  1. Dilation by factor 3 from the x-axis, translation 2 units left and 1 unit down Correct answer.
  2. Dilation by factor 1/3 from the x-axis, translation 2 units left and 1 unit up
  3. Dilation by factor 3 from the x-axis, translation 2 units right and 1 unit up
  4. Dilation by factor 3 from the y-axis, translation 2 units right and 1 unit down

Explanation

Multiplying the output by 3 dilates the graph by factor 3 from the x-axis. Replacing x with x + 2 shifts it 2 units left, and subtracting 1 shifts it 1 unit down.

Question 50Functions, relations and graphs

The equation |x² − 2x − 3| = x + 1 is to be solved for real x. How many distinct real solutions are there?

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

Explanation

Factorising the modulus part gives |(x − 3)(x + 1)| = x + 1. Since x + 1 ≥ 0 is required, solving (x − 3)(x + 1) = x + 1 yields x = 4 or x = −1, and solving −(x − 3)(x + 1) = x + 1 yields x = 2 or x = −1. The distinct solutions are −1, 2 and 4, three in total.