TMUA Paper 1: Graphs and transformations, set 1

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

1. 11 mark

   Which of the following equations has a graph that passes through the origin, is defined for every real value of x, and is symmetric about the y-axis?

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

   The graph of y = log_2(x) is reflected in the line y = x. Which of the following is the equation of the resulting graph?

   1. A y = 2^x
   2. B y = -log_2(x)
   3. C y = log_2(-x)
   4. D y = 2^(-x)
3. 31 mark

   f(x) = 2x + 1 and g(x) = x^2. Which of the following equals g(f(x))?

   1. A 2x^2 + 1
   2. B 4x^2 + 1
   3. C 4x + 2
   4. D 4x^2 + 4x + 1
4. 41 mark

   The graph of y = f(x) has a single minimum point at (2, -5). What are the coordinates of the minimum point of the graph of y = f(x - 3) + 4?

   1. A (-1, -1)
   2. B (5, -1)
   3. C (5, -9)
   4. D (-1, -9)
5. 51 mark

   A straight line has equation y = mx + c. Its graph passes through the point (0, -4), and it is steeper than, but has the same sign of gradient as, the line y = 2x - 1. Which of the following could be the equation of the line?

   1. A y = 5x - 4
   2. B y = x - 4
   3. C y = -5x - 4
   4. D y = 5x - 1
6. 61 mark

   The graph of y = x^2 is transformed to the graph of y = 2(x + 3)^2 + 5. What are the coordinates of its vertex (turning point), and how does its width compare to the graph of y = x^2?

   1. A Vertex (3, 5); narrower than y = x^2
   2. B Vertex (-3, 5); wider than y = x^2
   3. C Vertex (-3, 5); narrower than y = x^2
   4. D Vertex (3, 5); wider than y = x^2
7. 71 mark

   The graph of y = x^2 is transformed by three steps, applied in order: first stretched vertically by scale factor 3, then translated 2 units in the negative x-direction, then translated 4 units in the negative y-direction. Which of the following is the resulting equation?

   1. A y = 3(x - 2)^2 - 4
   2. B y = 3(x + 2)^2 - 4
   3. C y = 3(x + 2)^2 + 4
   4. D y = 3(x - 2)^2 + 4
8. 81 mark

   The graph of y = (x - 2)(x + 4) is reflected in the x-axis. What are the roots of the resulting graph, and what is its y-intercept?

   1. A Roots x = 2 and x = -4; y-intercept 8
   2. B Roots x = -2 and x = 4; y-intercept 8
   3. C Roots x = 2 and x = -4; y-intercept -8
   4. D Roots x = -2 and x = 4; y-intercept -8
9. 91 mark

   The function f has domain -3 <= x <= 5 and range -2 <= f(x) <= 6. What is the range of g(x) = 2f(x) - 1?

   1. A -4 <= g(x) <= 12
   2. B -6 <= g(x) <= 10
   3. C -7 <= g(x) <= 9
   4. D -5 <= g(x) <= 11
10. 101 mark

    How many real roots does the equation x^3 - x = 0 have, and what are they?

    1. A x = -1, 0 and 1 (three real roots)
    2. B x = 0 only (one real root)
    3. C x = -1 and 1 (two real roots)
    4. D x = 0, 1 and 2 (three real roots)
11. 111 mark

    The graph of y = x^2 + kx + 9 touches the x-axis at exactly one point. Given that k > 0, what is the value of k?

    1. A k = -6
    2. B k = 36
    3. C k = 3
    4. D k = 6
12. 121 mark

    The line y = x + 1 intersects the curve y = x^2 - x - 1 at two points. What is the sum of the x-coordinates of these two points?

    1. A -2
    2. B 2
    3. C 1
    4. D 4

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. Passing through the origin means f(0) = 0. Being defined for every real x rules out any graph with a restricted domain.
   2. Symmetric about the y-axis means f(-x) = f(x) for every x in the domain (an even function), which is a different condition from being symmetric about the origin.
   3. y = x^3: f(0) = 0 and the domain is all reals, but f(-x) = -x^3 = -f(x), so it is odd, not symmetric about the y-axis.
   4. y = 2^x: f(0) = 1, not 0, so it fails at the first condition regardless of symmetry.
   5. y = sqrt(x): the domain is x >= 0 only, so it fails the 'defined for every real x' condition.
   6. y = |x|: f(0) = 0, the domain is all reals, and f(-x) = |-x| = |x| = f(x), so every condition holds. This is option C.
   • Why not A: y = x^3 does pass through the origin and is defined for every real x, but it is symmetric under a 180 degree turn about the origin (f(-x) = -f(x), an odd function), not a mirror reflection in the y-axis; point symmetry about the origin is being confused with axis symmetry.
   • Why not B: y = 2^x passes through (0, 1), not the origin, since any positive number raised to the power 0 equals 1, not 0; a candidate might wrongly assume every 'basic' graph passes through the origin.
   • Why not D: y = sqrt(x) is only defined for x >= 0, so it has no graph at all for negative x and cannot be symmetric about the y-axis; extending its domain to all real x by analogy with x^3 or x^2 is the error here.

2. Question 2Answer: A

   1. Reflecting the graph of y = f(x) in the line y = x swaps the roles of x and y, and this is exactly how the graph of the inverse function f^(-1)(x) is found.
   2. To find the inverse of y = log_2(x), swap x and y to get x = log_2(y), then rearrange: x = log_2(y) means y = 2^x.
   3. So reflecting y = log_2(x) in y = x gives y = 2^x, option A.
   4. This matches the general fact that a^x and log_a(x) are always reflections of one another in the line y = x, because they are inverse functions of each other.
   • Why not B: Reflects the graph in the x-axis (replacing y with -y) rather than in the line y = x, giving -log_2(x) instead of the true inverse function.
   • Why not C: Reflects the graph in the y-axis (replacing x with -x) rather than in the line y = x, giving log_2(-x), which is not even defined for any positive x.
   • Why not D: Correctly identifies that reflecting in y = x should undo the logarithm and produce an exponential, but then additionally reflects that exponential in the y-axis, introducing an unwanted extra step and landing on 2^(-x) instead of 2^x.

3. Question 3Answer: D

   1. g(f(x)) means: first work out f(x), then substitute that whole expression into g.
   2. f(x) = 2x + 1, so g(f(x)) = g(2x + 1).
   3. Since g(x) = x^2, this means squaring the whole bracket: g(2x + 1) = (2x + 1)^2.
   4. Expanding using (a + b)^2 = a^2 + 2ab + b^2, with a = 2x and b = 1: (2x)^2 + 2(2x)(1) + 1^2 = 4x^2 + 4x + 1.
   5. So g(f(x)) = 4x^2 + 4x + 1, option D.
   • Why not A: Computes f(g(x)) instead of g(f(x)), applying f to g(x) = x^2 first (giving 2x^2 + 1) rather than substituting f(x) into g; composition is not commutative, so f(g(x)) and g(f(x)) are generally different functions.
   • Why not B: Expands (2x + 1)^2 by squaring each term of the bracket separately, as (2x)^2 + 1^2, and drops the cross term; the correct expansion of (a + b)^2 is a^2 + 2ab + b^2, not a^2 + b^2.
   • Why not C: Misreads g(x) = x^2 as an instruction to double rather than square, so instead of squaring f(x) it doubles it, giving 2(2x + 1) = 4x + 2.

4. Question 4Answer: B

   1. y = f(x - 3) is a horizontal translation of y = f(x). Replacing x with (x - a) inside the function shifts the graph RIGHT by a, even though the sign inside the bracket is negative.
   2. So the minimum point's x-coordinate moves from 2 to 2 + 3 = 5.
   3. Adding 4 outside the function, y = [...] + 4, is a vertical translation UP by 4.
   4. So the minimum point's y-coordinate moves from -5 to -5 + 4 = -1.
   5. Combining both shifts, the minimum point of y = f(x - 3) + 4 is at (5, -1), option B.
   • Why not A: Treats y = f(x - 3) as a shift to the LEFT by 3, moving the x-coordinate to 2 - 3 = -1; in fact replacing x with (x - 3) inside a function shifts the graph RIGHT by 3, a well-known sign trap.
   • Why not C: Correctly shifts the minimum right by 3 to x = 5, but treats the '+4' outside the function as a downward shift rather than upward, giving y = -5 - 4 = -9 instead of -5 + 4 = -1.
   • Why not D: Combines both errors above: shifts the minimum left instead of right, and down instead of up, giving (-1, -9).

5. Question 5Answer: A

   1. In y = mx + c, the value c is the y-intercept, so passing through (0, -4) means c = -4.
   2. 'Steeper than y = 2x - 1' means the magnitude of the gradient exceeds 2, i.e. |m| > 2.
   3. 'Same sign of gradient' as y = 2x - 1 (whose gradient is +2) means m must be positive.
   4. Combining these, m must be a positive number greater than 2.
   5. Only y = 5x - 4 has m = 5 (positive and greater than 2) together with c = -4, satisfying every condition, option A.
   • Why not B: Picks a gradient less steep than 2 (m = 1), which satisfies 'same sign' but not 'steeper'; steeper means the magnitude of the gradient is greater, not merely that the gradient is positive.
   • Why not C: Chooses a gradient that is steeper in magnitude (5, since |-5| > 2) but has the wrong sign, giving a line that falls left to right instead of rising like y = 2x - 1.
   • Why not D: Uses the correct gradient m = 5, but takes the y-intercept from the comparison line y = 2x - 1 (c = -1) instead of from the point (0, -4) that was actually given for this line.

6. Question 6Answer: C

   1. For y = a(x + b)^2 + c, the squared term (x + b)^2 is zero exactly when x = -b, and is never negative, so its smallest possible value is 0.
   2. Here a = 2 > 0, so the graph has a minimum where (x + 3)^2 = 0, i.e. at x = -3 (since b = 3).
   3. At that x, y = 2(0) + 5 = 5, so the vertex is at (-3, 5).
   4. Comparing to y = x^2: multiplying by a = 2 makes every y-value above the vertex twice as large for the same x-displacement, so the graph rises more steeply and looks NARROWER, not wider.
   5. So the vertex is (-3, 5) and the graph is narrower than y = x^2, option C.
   • Why not A: Sets the vertex's x-coordinate equal to 3 (matching the number in the bracket) instead of -3; since the bracket is (x + 3), it equals zero when x = -3, not x = 3.
   • Why not B: Correctly locates the vertex, but reasons that multiplying the squared term by a factor of 2 stretches the graph outward, making it wider; multiplying y-values by a factor greater than 1 actually makes the graph rise more steeply, making it NARROWER.
   • Why not D: Combines both errors: uses x = 3 instead of x = -3 for the vertex, and treats the factor of 2 as widening rather than narrowing the graph.

7. Question 7Answer: B

   1. Stretching y = x^2 vertically by scale factor 3 multiplies every y-value by 3, giving y = 3x^2.
   2. Translating a graph in the negative x-direction (to the left) by 2 units is achieved by replacing x with x + 2 (moving left uses a plus sign inside the bracket, the opposite of what many candidates expect), giving y = 3(x + 2)^2.
   3. Translating in the negative y-direction (downward) by 4 units subtracts 4 from the whole expression, giving y = 3(x + 2)^2 - 4.
   4. So the final equation is y = 3(x + 2)^2 - 4, option B.
   • Why not A: Translates the graph 2 units in the negative x-direction (left) by replacing x with x - 2 instead of x + 2; shifting a graph left by an amount a is achieved by replacing x with x + a, and it is shifting RIGHT that uses x - a.
   • Why not C: Translates the graph 4 units in the negative y-direction (down) by adding 4 instead of subtracting it; a translation downward by 4 should subtract 4 from the expression, not add it.
   • Why not D: Combines both sign errors above: reverses the direction of the horizontal shift and the direction of the vertical shift.

8. Question 8Answer: A

   1. The original graph y = (x - 2)(x + 4) has roots where each bracket is zero, at x = 2 and x = -4.
   2. Its y-intercept is found by setting x = 0: y = (0 - 2)(0 + 4) = (-2)(4) = -8.
   3. Reflecting a graph in the x-axis replaces y with -y, so the new equation is y = -(x - 2)(x + 4).
   4. A root is a point where y = 0, and -0 = 0, so every root is unchanged by this reflection: the roots remain x = 2 and x = -4.
   5. The y-intercept, however, is a nonzero value that gets multiplied by -1: the new y-intercept is -(-8) = 8.
   6. So the reflected graph has roots x = 2 and x = -4 and y-intercept 8, option A.
   • Why not B: Moves the roots as though the graph had been reflected in the y-axis (which replaces x with -x) rather than in the x-axis (which replaces y with -y); a reflection in the x-axis cannot move any root, since a point with y = 0 satisfies -0 = 0 just as well.
   • Why not C: Correctly leaves the roots unchanged at x = 2 and x = -4, but keeps the original y-intercept of -8 instead of negating it; reflecting in the x-axis multiplies every y-value, including the y-intercept, by -1.
   • Why not D: Combines both errors: moves the roots as if the reflection were in the y-axis, and forgets to negate the y-intercept.

9. Question 9Answer: D

   1. A transformation applied outside a function, like 2f(x) - 1, acts on the OUTPUT values of f, so it transforms the RANGE of f, not its domain.
   2. f's range is -2 <= f(x) <= 6, so first multiply both endpoints by 2, since 2 is the coefficient directly on f(x): 2(-2) = -4 and 2(6) = 12, giving -4 <= 2f(x) <= 12.
   3. Then subtract 1 from both endpoints, since the '-1' is applied after the doubling: -4 - 1 = -5 and 12 - 1 = 11.
   4. So the range of g(x) = 2f(x) - 1 is -5 <= g(x) <= 11, option D.
   5. The domain of g is unchanged from f's domain, -3 <= x <= 5, since no transformation is applied to x itself here; only the range moves.
   • Why not A: Correctly doubles each end of f's range (2 x -2 = -4 and 2 x 6 = 12) but forgets the '-1' outside the doubling; multiplying by 2 stretches the range, but every value must then also have 1 subtracted.
   • Why not B: Applies the two operations to the range endpoints in the wrong order: subtracts 1 from each end of f's range first (giving -3 <= f(x) - 1 <= 5) and only then doubles (giving -6 <= g(x) <= 10); the '-1' is applied to 2f(x), not to f(x), so it must come after the multiplication by 2, not before it.
   • Why not C: Transforms the DOMAIN of f (-3 to 5) instead of its RANGE (-2 to 6), giving 2(-3) - 1 = -7 and 2(5) - 1 = 9; it is the set of output values, the range, that a transformation applied outside f actually stretches and shifts, not the set of input values.

10. Question 10Answer: A

    1. Factor out the common factor of x first: x^3 - x = x(x^2 - 1).
    2. Factorise the remaining quadratic using the difference of two squares: x^2 - 1 = (x - 1)(x + 1).
    3. So x^3 - x = x(x - 1)(x + 1) = 0, and a product of factors is zero exactly when at least one factor is zero.
    4. This gives three real roots: x = 0, x = 1 and x = -1, option A.
    • Why not B: Factorises x^3 - x = x(x^2 - 1) correctly, but then treats x^2 - 1 = 0 as having no real solutions, perhaps confusing it with x^2 + 1 = 0, which genuinely has none, leaving only the root from the factor of x.
    • Why not C: Divides the equation x^3 - x = 0 through by x to get x^2 - 1 = 0, giving x = -1 and x = 1; but dividing by x silently assumes x is not zero, which throws away the root x = 0 that the division divided out. An equation must never be divided by an expression that could itself be zero.
    • Why not D: Factorises x^2 - 1 incorrectly as (x - 1)(x - 2) rather than the correct difference of two squares (x - 1)(x + 1), giving the wrong second and third roots.

11. Question 11Answer: D

    1. A quadratic graph y = x^2 + kx + 9 touches the x-axis at exactly one point when x^2 + kx + 9 = 0 has a repeated root, which happens exactly when the discriminant b^2 - 4ac equals 0.
    2. Here a = 1, b = k, c = 9, so the condition is k^2 - 4(1)(9) = 0, i.e. k^2 - 36 = 0, i.e. k^2 = 36.
    3. This gives k = 6 or k = -6, since both square to 36.
    4. The question specifies k > 0, so the negative solution is rejected, leaving k = 6, option D.
    • Why not A: Solves k^2 = 36 correctly but selects the negative root, k = -6, ignoring the condition k > 0 given in the question.
    • Why not B: Correctly sets up k^2 - 4(1)(9) = 0 but then reports the value of k^2 itself, 36, as the answer, forgetting to take the square root to solve for k.
    • Why not C: Uses b^2 = c instead of the correct repeated-root condition b^2 = 4ac, dropping the factor of 4a. This gives k^2 = 9 and k = 3, rather than the correct k^2 = 36 and k = 6.

12. Question 12Answer: B

    1. The intersection points of y = x + 1 and y = x^2 - x - 1 satisfy both equations at once, so set the right-hand sides equal: x^2 - x - 1 = x + 1.
    2. Rearranging by moving every term to one side: x^2 - x - 1 - x - 1 = 0, i.e. x^2 - 2x - 2 = 0.
    3. This is a quadratic in x, so its two roots are exactly the x-coordinates of the two intersection points, which is the geometric meaning of solving two graphs' equations simultaneously.
    4. For a quadratic ax^2 + bx + c = 0, the sum of its roots is always -b/a, without needing to solve for the individual roots. Here a = 1, b = -2, so the sum is -(-2)/1 = 2.
    5. So the sum of the x-coordinates of the two intersection points is 2, option B.
    • Why not A: For a quadratic ax^2 + bx + c = 0, the sum of the roots is -b/a and the product is c/a; this computes the PRODUCT of the roots, c/a = -2/1 = -2, instead of the sum.
    • Why not C: Correctly finds that the two x-coordinates sum to 2, but then divides by 2 to report their average (mean), 1, rather than the sum the question actually asks for.
    • Why not D: Finds the sum of the y-coordinates of the two intersection points rather than the x-coordinates; since each y-coordinate equals x + 1, the two y-coordinates sum to the sum of the x-coordinates plus 2, giving 2 + 2 = 4, which answers a different question from the one asked.