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Admissions tests / TMUA / Paper 1 / Graphs and transformations

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Graphs and transformations, set 2

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

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

    A function is always positive, decreases as x increases over its entire domain, and gets arbitrarily close to zero (without ever reaching it) as x tends to infinity. Which of the following could be this function?

    1. A y = 2^x
    2. B y = |x|
    3. C y = 2^(-x)
    4. D y = sqrt(x)
  2. 21 mark

    The point (4, 5) lies on the graph of y = f(x). Which point must lie on the graph of y = -2f(x) + 3?

    1. A (4, -7)
    2. B (4, 13)
    3. C (4, -16)
    4. D (-5, 5)
  3. 31 mark

    The graph of y = f(x) has a single maximum point at (6, 10). What are the coordinates of the maximum point on the graph of y = f(2x)?

    1. A (12, 10)
    2. B (6, 5)
    3. C (3, 5)
    4. D (3, 10)
  4. 41 mark

    f(x) = x^2 + x and g(x) = 5 - 2x. What is the value of f(g(1))?

    1. A 6
    2. B 12
    3. C 1
    4. D 9
  5. 51 mark

    A straight line y = mx + c passes through the points (2, 11) and (5, 20). What is the value of c?

    1. A 5
    2. B -7
    3. C -5
    4. D 3
  6. 61 mark

    By completing the square, find the coordinates of the turning point of the graph of y = x^2 - 6x + 11.

    1. A (-3, 2)
    2. B (6, -25)
    3. C (3, 2)
    4. D (3, 11)
  7. 71 mark

    Find the x-coordinate of the stationary point of the curve y = x^2 - 10x + 7, and state whether it is a maximum or a minimum.

    1. A x = 10, minimum
    2. B x = 5, maximum
    3. C x = -5, minimum
    4. D x = 5, minimum
  8. 81 mark

    For what values of x is the curve y = x^3 - 12x decreasing?

    1. A x < 2
    2. B -2 < x < 2
    3. C x < -2 or x > 2
    4. D -12 < x < 12
  9. 91 mark

    A cubic equation ax^3 + bx^2 + cx + d = 0 has real coefficients, with a != 0. Which of the following is NOT a possible number of real roots (counting a repeated root once) for such an equation?

    1. A 1
    2. B 2
    3. C 0
    4. D 3
  10. 101 mark

    The equation x^4 - 5x^2 + 4 = 0 has how many real roots, and what are they?

    1. A x = -2, -1, 1, 2
    2. B x = 1, 4
    3. C x = 1, 2
    4. D x = -4, -1, 1, 4
  11. 111 mark

    For what values of k does the line y = kx fail to intersect the curve y = x^2 + 4 at all?

    1. A k = 4 or k = -4
    2. B -4 < k < 4
    3. C k < -4 or k > 4
    4. D -16 < k < 16
  12. 121 mark

    Solve the equation |2x - 3| = 7.

    1. A x = 5 only
    2. B x = -5 or x = 5
    3. C x = -4 or x = 10
    4. D x = -2 or x = 5

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 description requires three things at once: always positive, decreasing as x increases, and approaching (never reaching) zero as x becomes large.
    2. y = 2^(-x) can be rewritten as (1/2)^x: each time x increases by 1, the value is halved, so it decreases as x increases.
    3. Since a positive number raised to any real power stays positive, y = 2^(-x) is always positive.
    4. As x tends to infinity, (1/2)^x gets closer and closer to 0 but never reaches it, since no power of a positive number equals zero.
    5. So y = 2^(-x) satisfies every condition in the description, option C.
    • Why not A: y = 2^x is always positive, but it INCREASES as x increases (doubling each time x goes up by 1), the opposite of the required decreasing behaviour; picking any 'exponential' function without checking the sign of the exponent is the error here.
    • Why not B: y = |x| is never negative, but it is not decreasing over its whole domain: it decreases for x < 0 and then increases for x > 0, and it grows without bound rather than approaching zero as x tends to infinity.
    • Why not D: y = sqrt(x) is only defined for x >= 0, increases rather than decreases, and grows without bound; a candidate may wrongly associate 'gets close to zero' with the start of its domain near x = 0 rather than with the required behaviour as x tends to infinity.

  2. Question 2Answer: A

    1. A point (a, b) lies on y = f(x) exactly when f(a) = b, so here f(4) = 5.
    2. The transformed graph is y = -2f(x) + 3, so its y-value at x = 4 is -2f(4) + 3.
    3. Substituting f(4) = 5: -2(5) + 3 = -10 + 3 = -7.
    4. The x-coordinate is unaffected, since nothing here changes what is substituted into f: the point (4, -7) lies on the transformed graph, option A.
    • Why not B: Drops the negative sign on the coefficient, computing 2(5) + 3 = 13 instead of -2(5) + 3 = -7.
    • Why not C: Applies the '+3' before the multiplication, computing -2(5 + 3) = -2(8) = -16, as if the +3 were inside the same bracket as f(x) rather than added on afterwards.
    • Why not D: Transforms the x-coordinate instead of the y-coordinate, computing a new value -2(4) + 3 = -5 and pairing it with the unchanged y-value of 5; the transformation y = -2f(x) + 3 acts on the OUTPUT of f, so it changes the y-coordinate of a point on the graph, not the x-coordinate.

  3. Question 3Answer: D

    1. For y = f(2x), the maximum occurs where the INPUT to f equals 6, since that is where f is known to be at its maximum value of 10.
    2. So 2x = 6, which gives x = 3.
    3. The output value itself is unchanged: at this x, y = f(2x) = f(6) = 10.
    4. So the maximum point of y = f(2x) is (3, 10), option D.
    5. In general, replacing x with ax inside a function compresses the graph horizontally by scale factor 1/a (for a > 1), leaving y-values exactly as they were.
    • Why not A: Doubles the x-coordinate instead of halving it, as if replacing x with 2x inside a function stretched the graph horizontally by scale factor 2; in fact it compresses the graph horizontally towards the y-axis by scale factor 1/2, so the maximum's x-coordinate should be divided by 2, not multiplied.
    • Why not B: Applies the factor of 2 to the y-coordinate instead of the x-coordinate, halving 10 to get 5 while leaving x unchanged at 6; the '2' inside the bracket in f(2x) acts on the input to f (the x-values), not on the output (the y-values).
    • Why not C: Combines both errors above, halving both the x-coordinate and the y-coordinate.

  4. Question 4Answer: B

    1. f(g(1)) means: first evaluate g(1), then substitute that result into f.
    2. g(1) = 5 - 2(1) = 5 - 2 = 3.
    3. f(3) = 3^2 + 3 = 9 + 3 = 12.
    4. So f(g(1)) = 12, option B.
    • Why not A: Evaluates g(1) using 2x - 5 instead of 5 - 2x, reversing the subtraction: this gives -3 instead of the correct 5 - 2(1) = 3, and then f(-3) = (-3)^2 + (-3) = 9 - 3 = 6.
    • Why not C: Computes g(f(1)) instead of f(g(1)), applying the functions in the wrong order: f(1) = 1^2 + 1 = 2, then g(2) = 5 - 2(2) = 1.
    • Why not D: Drops the linear term of f, using f(x) = x^2 only instead of f(x) = x^2 + x, giving f(3) = 3^2 = 9 instead of 3^2 + 3 = 12.

  5. Question 5Answer: A

    1. The gradient of the line is the change in y divided by the change in x between the two points: m = (20 - 11) / (5 - 2) = 9 / 3 = 3.
    2. To find c, substitute one point and the gradient into y = mx + c and rearrange: 11 = 3(2) + c, so c = 11 - 6 = 5.
    3. Checking with the other point confirms it: 20 = 3(5) + c gives c = 20 - 15 = 5, the same value.
    4. So c = 5, option A.
    • Why not B: Uses the difference in y-coordinates (20 - 11 = 9) directly as the gradient, forgetting to divide by the difference in x-coordinates (5 - 2 = 3); with this wrong gradient of 9, c = 11 - 9(2) = -7.
    • Why not C: Correctly finds the gradient m = 3, but rearranges y = mx + c incorrectly as c = mx - y instead of c = y - mx, giving c = 3(2) - 11 = -5, the negative of the correct value.
    • Why not D: Correctly computes the gradient m = 3, but reports it as the final answer, confusing the gradient with the y-intercept c that the question actually asks for.

  6. Question 6Answer: C

    1. To complete the square on x^2 - 6x + 11, take half of the coefficient of x: half of -6 is -3.
    2. (x - 3)^2 expands to x^2 - 6x + 9, so 9 must be subtracted back off to leave the original expression unchanged: x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2.
    3. Since (x - 3)^2 is never negative and equals zero exactly when x = 3, and the coefficient of x^2 is positive, this is a minimum turning point.
    4. So the turning point is (3, 2), option C.
    • Why not A: Writes the completed square as (x + 3)^2 + 2 instead of (x - 3)^2 + 2, a sign confusion, and reads the vertex's x-coordinate off as -3 instead of 3.
    • Why not B: Forgets to halve the coefficient of x before forming the bracket, using (x - 6)^2 instead of (x - 3)^2 (treating the coefficient -6 itself, rather than half of it, as the number inside the bracket), giving (x - 6)^2 - 36 + 11 = (x - 6)^2 - 25 and turning point (6, -25).
    • Why not D: Correctly forms the bracket (x - 3)^2 but forgets to subtract the 9 that must be taken back off to compensate for expanding it, keeping the original constant 11 unchanged instead of adjusting it to 11 - 9 = 2.

  7. Question 7Answer: D

    1. A stationary point occurs where the gradient of the curve is zero, so differentiate and set dy/dx = 0.
    2. Differentiating y = x^2 - 10x + 7 term by term: dy/dx = 2x - 10.
    3. Setting this to zero: 2x - 10 = 0, so 2x = 10, so x = 5.
    4. The coefficient of x^2 in the original equation is +1, which is positive, so the curve is an upward-opening parabola with a single minimum turning point.
    5. So the stationary point is at x = 5 and it is a minimum, option D.
    • Why not A: Correctly sets up 2x - 10 = 0, but then treats 2x = 10 as giving x = 10 directly, forgetting to divide both sides by the coefficient of x (2).
    • Why not B: Correctly finds x = 5, but classifies it as a maximum because the original expression contains a '-10x' term with a negative coefficient; the type of stationary point is decided by the sign of the x^2 coefficient (positive here, so the curve is an upward-opening parabola and x = 5 is a minimum), not by the sign of the linear term.
    • Why not C: Rearranges 2x - 10 = 0 incorrectly as 2x = -10 (a sign error moving -10 to the other side), giving x = -5 instead of the correct x = 5.

  8. Question 8Answer: B

    1. A curve is decreasing wherever its gradient is negative, so find where dy/dx < 0.
    2. Differentiating: dy/dx = 3x^2 - 12.
    3. Find where dy/dx = 0: 3x^2 - 12 = 0, so 3x^2 = 12, so x^2 = 4, so x = 2 or x = -2.
    4. Testing a value between these, x = 0: dy/dx = 3(0) - 12 = -12, which is negative, so the curve is decreasing between x = -2 and x = 2.
    5. Testing a value outside, x = 3: dy/dx = 3(9) - 12 = 15, which is positive, confirming the curve is increasing (not decreasing) outside this interval.
    6. So the curve is decreasing for -2 < x < 2, option B.
    • Why not A: Solves 3x^2 - 12 = 0 as x^2 = 4 but takes only the positive square root, x = 2, missing the negative root x = -2, and then wrongly treats every value below the one root found as part of the decreasing region.
    • Why not C: Correctly finds the two boundary values x = -2 and x = 2, but reports the region OUTSIDE them as decreasing, when in fact the derivative 3x^2 - 12 is negative BETWEEN the roots and positive outside them, so the curve is increasing (not decreasing) for x < -2 or x > 2.
    • Why not D: Reads the coefficients directly out of the derivative expression 3x^2 - 12 as the interval bounds (using the number 12 itself) instead of solving 3x^2 - 12 = 0 for x, which requires dividing by 3 and taking a square root to reach the true bounds of -2 and 2.

  9. Question 9Answer: C

    1. A cubic with real coefficients has a graph that is continuous and runs from one infinite extreme to the other as x runs from -infinity to +infinity (for example, from -infinity up to +infinity if a > 0).
    2. Because the graph must pass from one extreme to the other without any jumps, it is guaranteed to cross the x-axis at least once, so a cubic can never have zero real roots.
    3. A cubic can have exactly one real root (the other two roots then form a complex conjugate pair), exactly two distinct real roots (when one root is repeated), or three distinct real roots.
    4. So the only number in the list that a cubic can never achieve is 0, option C.
    • Why not A: 1 real root is entirely possible: for example x^3 + x + 1 = 0 has exactly one real root, its other two roots forming a complex conjugate pair. A candidate who assumes a cubic must always factorise into three real linear factors, simply because its degree is 3, would wrongly rule this out.
    • Why not B: 2 distinct real roots is possible when the cubic has a repeated root: for example (x - 1)^2(x + 1) = x^3 - x^2 - x + 1 has roots x = 1 (repeated) and x = -1, which is two distinct values. Forgetting that a repeated root still counts once towards the number of distinct roots would wrongly rule this out.
    • Why not D: 3 real roots is certainly possible, as in x^3 - x = 0, which has roots -1, 0 and 1. Picking this confuses 'not possible' with 'not guaranteed': not every cubic has 3 real roots, but 3 is still an achievable count, so it is not the answer to what is impossible.

  10. Question 10Answer: A

    1. This is a quadratic in disguise: substituting u = x^2 turns it into u^2 - 5u + 4 = 0.
    2. Factorising: u^2 - 5u + 4 = (u - 1)(u - 4), so u = 1 or u = 4.
    3. Since u = x^2, each value of u gives two values of x (its positive and negative square roots): u = 1 gives x = 1 or x = -1, and u = 4 gives x = 2 or x = -2.
    4. So the four real roots are x = -2, -1, 1, 2, option A.
    • Why not B: Pattern-matches the equation to a quadratic in x directly, treating x^4 - 5x^2 + 4 as if it were x^2 - 5x + 4 = (x - 1)(x - 4), and never substitutes back to solve for x; this ignores that the powers here are x^4 and x^2, not x^2 and x.
    • Why not C: Correctly substitutes u = x^2 to get u^2 - 5u + 4 = (u - 1)(u - 4) = 0, giving u = 1 or u = 4, but then takes only the positive square root of each (x = 1 and x = 2), forgetting that x^2 = 1 also gives x = -1, and x^2 = 4 also gives x = -2.
    • Why not D: Correctly substitutes to get u = 1 or u = 4, and correctly takes both square roots for u = 1 (giving x = 1 and x = -1), but for u = 4 mistakenly uses the value of u itself as a root (x = 4 and x = -4) instead of taking its square root (x = 2 and x = -2).

  11. Question 11Answer: B

    1. The line and curve meet where kx = x^2 + 4, i.e. where x^2 - kx + 4 = 0.
    2. This quadratic in x has no real solutions exactly when its discriminant is negative: (-k)^2 - 4(1)(4) < 0, i.e. k^2 - 16 < 0.
    3. Rearranging: k^2 < 16, so -4 < k < 4.
    4. So the line fails to meet the curve at all exactly when -4 < k < 4, option B.
    5. Geometrically, no real solutions to the simultaneous equations means no point lies on both graphs at once, i.e. the line and curve do not cross or touch.
    • Why not A: Sets the discriminant equal to zero rather than less than zero, finding the boundary values where the line TOUCHES the curve at exactly one point (a repeated root), rather than the range of k for which there is no intersection at all.
    • Why not C: Solves k^2 - 16 < 0 with the inequality reversed, reporting the region where the discriminant is actually positive (two real intersections) as if it were the region with no real intersections.
    • Why not D: Correctly reaches the condition k^2 < 16 but forgets to take the square root of 16, reporting the bound as 16 itself instead of sqrt(16) = 4.

  12. Question 12Answer: D

    1. |2x - 3| = 7 means 2x - 3 is exactly 7 away from zero, so it must equal either 7 or -7.
    2. Case 1: 2x - 3 = 7, so 2x = 10, so x = 5.
    3. Case 2: 2x - 3 = -7, so 2x = -7 + 3 = -4, so x = -2.
    4. So the solutions are x = -2 or x = 5, option D.
    • Why not A: Solves only the branch 2x - 3 = 7 (giving x = 5) and drops the second branch 2x - 3 = -7 entirely, missing that a modulus equation |...| = 7 splits into two separate cases.
    • Why not B: Handles the branch 2x - 3 = -7 with a sign slip, moving the -3 to the other side without changing its sign: 2x = -7 - 3 = -10, giving x = -5, instead of the correct 2x = -7 + 3 = -4, giving x = -2.
    • Why not C: Correctly forms both branches, 2x - 3 = 7 and 2x - 3 = -7, and correctly isolates 2x = 10 and 2x = -4, but then forgets to divide by the coefficient of x (2) in both cases, reporting x = 10 and x = -4 directly.

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