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Admissions tests / ESAT / Maths 2 / Graphs of functions

Test standard. 15 questions, 15 marks, about 22 minutes.

ESAT Mathematics 2: Graphs of functions, set 1

Sketching the common function families, transformations, asymptotes, and reading the number and location of solutions off a sketch.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    What are the coordinates of the vertex (minimum point) of the graph of y = |x - 3|?

    1. A (-3, 0)
    2. B (0, 3)
    3. C (3, 0)
    4. D (0, -3)
  2. 21 mark

    Which of the following correctly describes the graph of y = sqrt(x)?

    1. A The graph is defined for all real values of x, including negative ones
    2. B The graph passes through the origin and becomes steeper as x increases
    3. C The graph decreases once x exceeds 1
    4. D The graph passes through the origin and is always increasing, but becomes less steep as x increases
  3. 31 mark

    The graph of y = f(x) passes through the point (2, 5). What point must lie on the graph of y = f(x) + 3?

    1. A (2, 2)
    2. B (5, 5)
    3. C (2, 8)
    4. D (-1, 5)
  4. 41 mark

    The graph of y = f(x) passes through the point (2, 5). What point must lie on the graph of y = f(x + 3)?

    1. A (5, 5)
    2. B (2, 8)
    3. C (-1, 5)
    4. D (2, 2)
  5. 51 mark

    Given f(x) = x^2 and g(x) = x - 3, evaluate f(g(5)).

    1. A 22
    2. B 25
    3. C 2
    4. D 4
  6. 61 mark

    A line has equation y = -3x + 7. Which statement about this line is correct?

    1. A It has gradient -3 and crosses the y-axis at (0, 7)
    2. B It has gradient 7 and crosses the y-axis at (0, -3)
    3. C It has gradient -3 and crosses the y-axis at (0, -7)
    4. D It has gradient 3 and crosses the y-axis at (0, 7)
  7. 71 mark

    As the value of m increases from 1 to 5, keeping c fixed at 2, how does the graph of y = mx + c change?

    1. A The line moves to a higher position on the page but keeps the same steepness
    2. B The line becomes steeper, still passing through (0, 2)
    3. C The y-intercept moves to (0, 5)
    4. D The line becomes less steep
  8. 81 mark

    The graph of y = 2(x - 3)^2 + 4 has a minimum point at:

    1. A (3, 4)
    2. B (3, -4)
    3. C (-3, 4)
    4. D (-3, -4)
  9. 91 mark

    For the graph of y = a(x + b)^2 + c, which statement is correct when a < 0?

    1. A The graph has a minimum point, because a negative value of a always produces a minimum
    2. B The graph becomes a straight line, since a < 0 removes the quadratic term
    3. C The graph has a maximum point, since a negative coefficient reflects the parabola so it opens downward
    4. D The turning point's coordinates, (-b, c), both change sign compared with when a > 0
  10. 101 mark

    Find the x-coordinate of the stationary point of the curve y = x^2 - 6x + 5.

    1. A 5
    2. B 6
    3. C -3
    4. D 3
  11. 111 mark

    The curve y = x^3 - 3x has dy/dx = 3x^2 - 3. For which values of x is the curve increasing (that is, dy/dx > 0)?

    1. A -1 < x < 1
    2. B x > 1 only
    3. C x < -1 or x > 1
    4. D x < -1 only
  12. 121 mark

    Find the x-coordinates of the points where the curve y = x^2 + 3x - 10 crosses the x-axis.

    1. A x = -5 or x = 2
    2. B x = 5 or x = 2
    3. C x = 5 or x = -2
    4. D x = 4 or x = -10
  13. 131 mark

    How many real roots does the equation x^3 - 4x = 0 have?

    1. A 3
    2. B 1
    3. C 2
    4. D 0
  14. 141 mark

    The curve y = x^2 and the line y = 3x - 2 intersect where x equals:

    1. A x = -1 or x = -2
    2. B x = 1 or x = 2
    3. C x = 1 or x = -2
    4. D There are no real intersection points
  15. 151 mark

    The line y = 5 is drawn on the same axes as the curve y = (x - 2)^2 + 1. How many points of intersection are there, and at what x-values do they occur?

    1. A 2 points, at x = 0 and x = 4
    2. B 0 points, since the curve's minimum value is 1, not 5
    3. C 3 points, including the curve's turning point
    4. D 1 point, where the line touches the curve at its turning point

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. The graph of y = |x - 3| is the graph of y = |x| translated. Inside the modulus, x is replaced by x - 3, which is a horizontal translation.
    2. y = f(x - 3) shifts the graph of y = f(x) to the RIGHT by 3 units (a common trap: the sign inside the bracket is the opposite of the direction of the shift).
    3. The graph of y = |x| has its vertex (minimum point) at (0, 0), so shifting it right by 3 moves the vertex to (3, 0).
    4. At this point y = |3 - 3| = 0, confirming (3, 0) lies on the graph, so the vertex is at (3, 0), option C.
    • Why not A: Solves x - 3 = 0 with a sign error, treating the vertex condition as x + 3 = 0, giving x = -3 instead of x = 3.
    • Why not B: Confuses a horizontal shift with a vertical one, treating y = |x - 3| as if it were y = |x| + 3, which has vertex (0, 3) rather than (3, 0).
    • Why not D: Combines both errors above: shifts the vertex vertically instead of horizontally, and gets the sign of that shift wrong too, landing on (0, -3).

  2. Question 2Answer: D

    1. y = sqrt(x) is only defined when x >= 0, since a negative number has no real square root, so the domain is x >= 0, not all real x.
    2. At x = 0, y = 0, so the graph passes through the origin, and for x > 0, y is always positive and increasing as x increases, so the graph never turns back down.
    3. Because sqrt(x) increases more slowly than x itself once x is large (for example sqrt(4) = 2 and sqrt(9) = 3: increasing x by 5 only increases y by 1), the graph gets flatter, i.e. less steep, as x increases.
    4. So the correct description is: passes through the origin, always increasing, becoming less steep as x increases, option D.
    • Why not A: Confuses y = sqrt(x) with a function like y = x^3, which is defined for every real x; the square root function requires a non-negative argument, so it is only defined for x >= 0.
    • Why not B: Confuses the shape of y = sqrt(x) with an increasing exponential or quadratic curve; sqrt(x) actually grows more slowly as x increases, not faster, because each extra unit of x adds a smaller amount to sqrt(x) than the last one did.
    • Why not C: Mistakes the flattening rate of increase (the graph rising more and more slowly) for the graph turning around and decreasing; sqrt(x) never decreases on its domain, it just increases at an ever-slower rate.

  3. Question 3Answer: C

    1. The transformation y = f(x) + 3 takes every point (x, y) on the graph of y = f(x) and moves it straight up by 3, to (x, y + 3). The x-coordinate is unchanged.
    2. Since (2, 5) lies on y = f(x), the corresponding point on y = f(x) + 3 has the same x-coordinate, 2, and a y-coordinate of 5 + 3 = 8.
    3. So the point (2, 8) lies on the graph of y = f(x) + 3, option C.
    • Why not A: Gets the direction of the vertical shift backwards, subtracting 3 from the y-coordinate instead of adding it, giving (2, 5 - 3) = (2, 2).
    • Why not B: Applies the '+3' to the x-coordinate instead of the y-coordinate, as though the transformation were horizontal rather than vertical, giving (2 + 3, 5) = (5, 5).
    • Why not D: Confuses y = f(x) + 3 (a vertical shift) with y = f(x + 3) (a horizontal shift), and additionally gets that horizontal shift's direction backwards, giving (2 - 3, 5) = (-1, 5).

  4. Question 4Answer: C

    1. For a transformation y = f(x + a), the graph of y = f(x) is shifted horizontally by a, and the direction is the OPPOSITE of what the sign of a suggests: y = f(x + 3) shifts the graph LEFT by 3 (a common trap, since +3 inside the bracket might look like a shift to the right).
    2. Formally, if (p, q) lies on y = f(x), then the point on y = f(x + 3) with the same y-coordinate satisfies x + 3 = p, i.e. x = p - 3.
    3. Since (2, 5) lies on y = f(x), the corresponding point on y = f(x + 3) has x = 2 - 3 = -1 and the same y-coordinate, 5.
    4. So the point (-1, 5) lies on the graph of y = f(x + 3), option C.
    • Why not A: Gets the direction of the shift backwards: y = f(x + 3) shifts the graph of y = f(x) to the LEFT by 3, not to the right, so treats (2, 5) as moving to (2 + 3, 5) = (5, 5) instead of (2 - 3, 5).
    • Why not B: Confuses the horizontal shift y = f(x + 3) with the vertical shift y = f(x) + 3, adding 3 to the y-coordinate instead of adjusting the x-coordinate.
    • Why not D: Applies an unrelated operation to the y-coordinate, subtracting 3 from it, instead of recognising that a horizontal shift leaves the y-coordinate unchanged and adjusts x instead.

  5. Question 5Answer: D

    1. f(g(x)) means: first apply g to x, then apply f to the result. Working from the inside out, first evaluate g(5).
    2. g(5) = 5 - 3 = 2.
    3. Now apply f to this result: f(2) = 2^2 = 4.
    4. So f(g(5)) = 4, option D.
    • Why not A: Computes the composition in the wrong order, evaluating g(f(5)) instead of f(g(5)): f(5) = 25, then g(25) = 25 - 3 = 22.
    • Why not B: Ignores g entirely and applies f directly to 5, giving f(5) = 25, instead of first evaluating g(5) as the notation f(g(x)) requires.
    • Why not C: Stops after evaluating the inner function g(5) = 5 - 3 = 2, and forgets to then apply f to that result.

  6. Question 6Answer: A

    1. For a line written as y = mx + c, the coefficient of x is the gradient, m, and the constant term is the y-intercept, c.
    2. Here y = -3x + 7 is already in this form, so m = -3 and c = 7.
    3. So the line has gradient -3 and crosses the y-axis at (0, 7), option A.
    • Why not B: Swaps the roles of m and c, reading off the coefficient of x as the y-intercept and the constant term as the gradient.
    • Why not C: Reads the gradient correctly but takes the y-intercept with the wrong sign, giving (0, -7) instead of (0, 7).
    • Why not D: Reads the y-intercept correctly but drops the negative sign on the gradient, giving 3 instead of -3.

  7. Question 7Answer: B

    1. In y = mx + c, m controls the steepness (gradient) of the line, and c controls where it crosses the y-axis (the y-intercept); these two effects are independent of each other.
    2. Increasing m from 1 to 5, with c unchanged, makes the line rise more steeply for each step to the right, without moving where it crosses the y-axis.
    3. Since c is fixed at 2 throughout, the line still passes through (0, 2).
    4. So the correct description is that the line becomes steeper while still passing through (0, 2), option B.
    • Why not A: Describes the effect of increasing c (a vertical shift) rather than increasing m; with c fixed at 2, the line's position on the y-axis does not change at all as m varies.
    • Why not C: Confuses the roles of m and c, treating the new value of m (5) as though it were a new y-intercept.
    • Why not D: Gets the direction of the effect backwards: increasing the gradient m from 1 to 5 makes the line steeper, not less steep.

  8. Question 8Answer: A

    1. For a quadratic written in the form y = a(x + b)^2 + c, the turning point is at (-b, c).
    2. Here y = 2(x - 3)^2 + 4 matches this form with b = -3 (since x - 3 = x + (-3)) and c = 4, so the turning point's x-coordinate is -b = -(-3) = 3.
    3. The y-coordinate of the turning point is c = 4, and since a = 2 > 0 the parabola opens upwards, so this is a minimum point.
    4. So the minimum point is at (3, 4), option A.
    • Why not B: Reads the vertical shift with the wrong sign, treating the '+ 4' at the end as though it lowered the graph to y = -4 rather than raising it to y = 4.
    • Why not C: Reads the horizontal shift with the wrong sign, treating x - 3 as though it gave a turning point at x = -3 rather than x = 3.
    • Why not D: Makes both sign errors at once: reads the turning point's x-coordinate as -3 instead of 3, and its y-coordinate as -4 instead of 4.

  9. Question 9Answer: C

    1. In y = a(x + b)^2 + c, the squared term (x + b)^2 is always greater than or equal to zero, so its overall sign is controlled entirely by a.
    2. When a > 0, adding a non-negative amount (scaled by a) to c means y >= c always, so the turning point (-b, c) is the lowest point on the graph, a minimum.
    3. When a < 0, that same squared term is multiplied by a negative number, so a(x + b)^2 <= 0 always, meaning y <= c: the turning point becomes the HIGHEST point on the graph, a maximum.
    4. So when a < 0 the graph has a maximum point, because the negative coefficient reflects the parabola to open downward, option C.
    • Why not A: Gets the effect of a's sign backwards: a positive value of a gives an upward-opening parabola with a MINIMUM turning point; a negative value gives a downward-opening parabola with a MAXIMUM, not the other way round.
    • Why not B: Confuses the sign of the coefficient with its size; a < 0 changes the direction the parabola opens, it does not remove the squared term or turn the graph into a straight line.
    • Why not D: Confuses the effect of the sign of a with the roles of b and c; changing the sign of a alone does not change the turning point's coordinates, (-b, c), at all, it only flips whether that point is a maximum or a minimum.

  10. Question 10Answer: D

    1. Differentiate y = x^2 - 6x + 5: dy/dx = 2x - 6.
    2. At a stationary point, dy/dx = 0, so 2x - 6 = 0.
    3. Solving: 2x = 6, so x = 3, option D.
    • Why not A: Confuses the stationary point's x-coordinate with the constant term of the original function, 5, rather than differentiating and solving dy/dx = 0.
    • Why not B: Sets the derivative's coefficient of x equal to the constant term without solving properly, treating 2x - 6 = 0 as x = 6 instead of correctly dividing both sides by 2 to get x = 3.
    • Why not C: Makes a sign error solving 2x - 6 = 0, treating it as 2x + 6 = 0 and getting x = -3.

  11. Question 11Answer: C

    1. The curve is increasing wherever dy/dx > 0, so solve 3x^2 - 3 > 0.
    2. Divide by 3: x^2 - 1 > 0, i.e. x^2 > 1.
    3. This is satisfied when x is further from 0 than 1 in either direction, i.e. when x < -1 or when x > 1 (not when -1 < x < 1, which is where x^2 < 1 and the curve is decreasing).
    4. So the curve is increasing for x < -1 or x > 1, option C.
    • Why not A: Solves 3x^2 - 3 > 0 correctly as far as finding the critical values x = -1 and x = 1, but then picks the interval BETWEEN them, which is actually where dy/dx < 0 (the curve is decreasing), rather than the two outer intervals where dy/dx > 0.
    • Why not B: Finds one boundary value x = 1 correctly but forgets that x^2 > 1 is also satisfied by sufficiently negative x, missing the symmetric branch x < -1.
    • Why not D: Finds one boundary value x = -1 correctly but forgets the symmetric branch x > 1, where x^2 > 1 is also satisfied.

  12. Question 12Answer: A

    1. Factorise x^2 + 3x - 10: look for two numbers that multiply to -10 and add to 3.
    2. Trying 5 and -2: 5 x (-2) = -10 and 5 + (-2) = 3, so x^2 + 3x - 10 = (x + 5)(x - 2).
    3. Set each factor to zero: x + 5 = 0 gives x = -5; x - 2 = 0 gives x = 2.
    4. So the curve crosses the x-axis at x = -5 and x = 2, option A.
    • Why not B: Correctly spots that 5 and 2 multiply to give 10, but does not check that the pair also needs to sum to +3 with the right signs; solving x - 5 = 0 and x - 2 = 0 gives 5 and 2, but neither is actually a correct factor of this expression.
    • Why not C: Factorises with both signs flipped, as (x - 5)(x + 2), which expands to x^2 - 3x - 10, the wrong sign on the middle term, and reads off roots 5 and -2 instead of -5 and 2.
    • Why not D: Uses the quadratic formula but forgets to divide by 2a at the final step, computing x = -3 +- sqrt(49) = -3 +- 7 directly as 4 or -10, instead of dividing that result by 2a = 2 to get 2 or -5.

  13. Question 13Answer: A

    1. Factorise x^3 - 4x by taking out the common factor x: x^3 - 4x = x(x^2 - 4).
    2. The bracket is a difference of two squares: x^2 - 4 = (x - 2)(x + 2), so the equation is x(x - 2)(x + 2) = 0.
    3. Setting each factor to zero gives three solutions: x = 0, x = 2 and x = -2.
    4. So the equation x^3 - 4x = 0 has 3 real roots, option A.
    • Why not B: Factors out x to get x(x^2 - 4) = 0 and stops after reading off x = 0 as a solution, forgetting that the remaining factor x^2 - 4 = 0 also needs to be solved.
    • Why not C: Solves x^2 - 4 = 0 by taking only the positive square root, x = 2, and forgets the negative root x = -2 that also satisfies it, missing one of the three roots.
    • Why not D: Assumes a cubic could have no real roots at all, the way a quadratic can when its discriminant is negative; but a cubic with real coefficients always crosses the x-axis at least once, so it always has at least one real root, and this equation in fact has three.

  14. Question 14Answer: B

    1. To find where y = x^2 and y = 3x - 2 meet, set the two expressions for y equal: x^2 = 3x - 2.
    2. Rearrange so everything is on one side: x^2 - 3x + 2 = 0.
    3. Factorise: two numbers that multiply to 2 and add to -3 are -1 and -2, so x^2 - 3x + 2 = (x - 1)(x - 2).
    4. Setting each factor to zero gives x = 1 or x = 2, so the graphs meet where x = 1 and x = 2, option B.
    • Why not A: Rearranges x^2 = 3x - 2 with a sign error, moving both terms to the same side as x^2 + 3x + 2 = 0 instead of x^2 - 3x + 2 = 0, and factorises that instead to get x = -1 or x = -2.
    • Why not C: Factorises x^2 - 3x + 2 correctly as far as spotting the numbers 1 and 2, but assigns the wrong sign to one of them, giving (x - 1)(x + 2) instead of (x - 1)(x - 2).
    • Why not D: Rearranges the equation incorrectly as x^2 - 3x - 2 = 0 (a sign error on the constant term), which has an irrational discriminant, and wrongly concludes from this that the original curves do not actually meet.

  15. Question 15Answer: A

    1. To find the intersections, set the two expressions for y equal: (x - 2)^2 + 1 = 5.
    2. Rearrange: (x - 2)^2 = 4.
    3. Take square roots of both sides: x - 2 = 2 or x - 2 = -2, giving x = 4 or x = 0.
    4. So the line and the curve meet at two points, x = 0 and x = 4, matching the two real solutions of the equation, option A.
    • Why not B: Correctly notes the curve's minimum value is 1, but wrongly concludes that because 1 is not equal to 5 the line must miss the curve altogether, rather than recognising that any height greater than the minimum, including 5, is reached at two different x-values.
    • Why not C: Counts the turning point of the curve as an extra intersection alongside the two genuine crossing points, even though the line y = 5 does not pass anywhere near the turning point, which is at height 1.
    • Why not D: Assumes the line y = 5 just touches the curve at its turning point, as though the minimum value of the curve were 5; the curve's actual minimum value is 1 (at x = 2), so a horizontal line at height 5 crosses it twice rather than touching it once.

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