ESAT Mathematics 2: Graphs of functions, set 2

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

1. 11 mark

   The graph of y = 2^x is transformed to give the graph of y = 2^x - 4. What is the y-intercept of the transformed graph?

   1. A (0, 1)
   2. B (0, 1/16)
   3. C (0, -3)
   4. D (0, 5)
2. 21 mark

   Which of the following correctly describes the graph of y = log_2(x)?

   1. A It is defined only for x > 0, passes through the point (1, 0), and increases without bound as x increases
   2. B It is defined for all real values of x, including negative ones
   3. C It passes through the point (0, 1)
   4. D It decreases as x increases
3. 31 mark

   For 0 <= x <= 360, where x is measured in degrees, for how many values of x does sin(x) = 0?

   1. A 2
   2. B 4
   3. C 1
   4. D 3
4. 41 mark

   The graph of y = f(x) passes through the point (6, 7). What point must lie on the graph of y = f(2x)?

   1. A (12, 7)
   2. B (3, 7)
   3. C (6, 14)
   4. D (3, 3.5)
5. 51 mark

   Given f(x) = sqrt(x) and g(x) = x + 5, evaluate g(f(4)).

   1. A 3
   2. B 2
   3. C 7
   4. D 9
6. 61 mark

   A line passes through the point (0, -4) and has gradient 5. What is the equation of the line?

   1. A y = 5x - 4
   2. B y = -4x + 5
   3. C y = 5x + 4
   4. D y = -5x - 4
7. 71 mark

   The line y = 4x + c is redrawn after c is increased from 2 to 9. Which statement about the effect on the line is correct?

   1. A The line becomes steeper, since increasing c increases how quickly y grows as x increases
   2. B The line is translated vertically upwards; its gradient stays at 4, but its y-intercept moves from (0, 2) to (0, 9)
   3. C The x-intercept stays fixed, since only the y-intercept can move when c changes
   4. D The line rotates about the point where it originally crossed the x-axis
8. 81 mark

   The graph of y = a(x + b)^2 + c has a minimum point at (5, -2). What are the values of b and c?

   1. A b = -5, c = -2
   2. B b = 5, c = -2
   3. C b = -5, c = 2
   4. D b = 5, c = 2
9. 91 mark

   For the curve y = 12x - x^3, what is the value of dy/dx at x = 1, and is the curve increasing or decreasing there?

   1. A dy/dx = 9, so the curve is decreasing at x = 1
   2. B dy/dx = 11, so the curve is increasing at x = 1
   3. C dy/dx = 15, so the curve is increasing at x = 1
   4. D dy/dx = 9, so the curve is increasing at x = 1
10. 101 mark

    Find the x-coordinates of the stationary points of the curve y = x^3 - 12x + 1.

    1. A x = 4
    2. B x = 2 only
    3. C x = 2 or x = -2
    4. D x = 2sqrt(3) or x = -2sqrt(3)
11. 111 mark

    Find the x-coordinates of the points where the curve y = 2x^2 - 5x - 3 crosses the x-axis.

    1. A x = -1/2 or x = 3
    2. B x = -1 or x = 3
    3. C x = 1/2 or x = -3
    4. D x = -1/2 or x = -3
12. 121 mark

    Find the y-intercept of the curve y = (x - 2)(x + 5)(x - 1).

    1. A -10
    2. B -2
    3. C 2
    4. D 10
13. 131 mark

    A polynomial has degree 4 (a quartic) and real coefficients, with no repeated roots. Which of the following is a possible total number of real roots for this polynomial?

    1. A 1
    2. B 2
    3. C 5
    4. D 3
14. 141 mark

    The line y = 2x + c is tangent to the curve y = x^2 - 6x + 10. Find the value of c.

    1. A c = 6
    2. B c = 12
    3. C c = -6
    4. D c = -24
15. 151 mark

    The line y = x - 7 and the curve y = x^2 - 3x + 9 are sketched on the same axes. Which statement about their points of intersection is correct?

    1. A They do not intersect, since setting the two expressions equal gives x^2 - 4x + 16 = 0, which has discriminant 16 - 64 = -48, a negative number
    2. B They intersect at exactly one point, since the discriminant of x^2 - 4x + 16 = 0 is negative, which means the line just touches the curve
    3. C They intersect at two points, since combining the two equations always produces a quadratic, and a quadratic equation always has two real solutions
    4. D They intersect at two points, found from x^2 - 4x + 2 = 0, obtained by treating -(x - 7) as -x - 7 instead of -x + 7 when rearranging

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 = 2^x - 4 is the graph of y = 2^x translated vertically, since the '-4' is applied outside the exponent, to the whole output of 2^x.
   2. At x = 0, 2^0 = 1, since any nonzero number raised to the power 0 equals 1.
   3. Applying the vertical shift: y = 2^0 - 4 = 1 - 4 = -3.
   4. So the transformed graph crosses the y-axis at (0, -3), option C.
   • Why not A: Finds the y-intercept of the original graph y = 2^x, which is (0, 1), and forgets to apply the vertical shift of -4 at all.
   • Why not B: Confuses the vertical shift y = 2^x - 4 with a horizontal shift y = 2^(x - 4), and evaluates 2^(0 - 4) = 2^(-4) = 1/16 instead of subtracting 4 from 2^0.
   • Why not D: Gets the direction of the shift backwards, adding 4 to 2^0 instead of subtracting it, giving 1 + 4 = 5 instead of 1 - 4 = -3.

2. Question 2Answer: A

   1. A logarithm log_2(x) is only defined when its argument is positive, so the domain of y = log_2(x) is x > 0; there is no y-intercept because x = 0 is excluded from the domain.
   2. At x = 1, log_2(1) = 0, since 2^0 = 1, so the graph passes through (1, 0).
   3. As x increases beyond 1, log_2(x) keeps increasing, though more and more slowly, and it increases without bound as x grows large.
   4. So the correct description is: defined only for x > 0, passing through (1, 0), and increasing without bound, option A.
   • Why not B: Confuses y = log_2(x) with a function like y = x^3 that is defined for every real x; a logarithm's argument must be positive, so the graph exists only for x > 0.
   • Why not C: Swaps the roles of the axes, describing the y-intercept of the exponential graph y = 2^x (which does pass through (0, 1)) rather than the logarithm graph, which has no y-intercept at all since x = 0 is not in its domain.
   • Why not D: Gets the direction of the graph backwards; log_2(x) is an increasing function throughout its domain, growing (slowly) without bound as x increases, never decreasing.

3. Question 3Answer: D

   1. The graph of y = sin(x) crosses the x-axis (sin(x) = 0) at every multiple of 180 degrees.
   2. Within the closed interval 0 <= x <= 360, the multiples of 180 are x = 0, x = 180 and x = 360, and all three lie within the given range, including both endpoints.
   3. So there are 3 values of x for which sin(x) = 0 in this interval, option D.
   • Why not A: Correctly finds x = 180 and x = 360 but overlooks that x = 0 also satisfies sin(x) = 0 and lies within the given closed interval.
   • Why not B: Mistakenly believes sin(x) = 0 also holds at x = 90, confusing this with the point where sin(x) reaches its maximum value of 1, and so counts x = 0, 90, 180, 360 as four solutions.
   • Why not C: Forgets that both endpoints of the closed interval, x = 0 and x = 360, are valid solutions, and counts only the interior zero at x = 180.

4. Question 4Answer: B

   1. For a transformation y = f(ax), the graph of y = f(x) is compressed horizontally towards the y-axis by a factor of 1/a; the y-coordinate of each point is unchanged.
   2. To land on the known point (6, 7), the input to f must be 6, so 2x = 6, giving x = 3.
   3. The y-coordinate is unaffected by this transformation, so it stays at 7.
   4. So the point (3, 7) lies on the graph of y = f(2x), option B.
   • Why not A: Gets the direction of the horizontal scaling backwards: y = f(2x) compresses the graph of y = f(x) towards the y-axis (by a factor of 1/2), but this option doubles the x-coordinate instead of halving it, as though the transformation stretched the graph instead.
   • Why not C: Confuses the horizontal scaling y = f(2x) with the vertical scaling y = 2f(x), and doubles the y-coordinate instead of adjusting the x-coordinate.
   • Why not D: Correctly halves the x-coordinate to 3, but then also incorrectly halves the y-coordinate, treating the transformation as though it affected both coordinates rather than only x.

5. Question 5Answer: C

   1. g(f(4)) means: first apply f to 4, then apply g to the result. Working from the inside out, first evaluate f(4).
   2. f(4) = sqrt(4) = 2.
   3. Now apply g to this result: g(2) = 2 + 5 = 7.
   4. So g(f(4)) = 7, option C.
   • Why not A: Computes the composition in the wrong order, evaluating f(g(4)) instead of g(f(4)): g(4) = 4 + 5 = 9, then f(9) = sqrt(9) = 3.
   • Why not B: Stops after evaluating the inner function f(4) = sqrt(4) = 2, and forgets to then apply g to that result, as the notation g(f(4)) requires.
   • Why not D: Ignores f entirely and applies g directly to 4, giving g(4) = 9, instead of first evaluating f(4) as the inner function.

6. Question 6Answer: A

   1. A line with gradient m and y-intercept (0, c) has equation y = mx + c.
   2. Here the gradient is m = 5, and the line passes through (0, -4), so its y-intercept is c = -4.
   3. Substituting gives y = 5x - 4, option A.
   • Why not B: Swaps the roles of the gradient and the y-intercept, using the gradient value as the y-intercept and the y-intercept value as the gradient.
   • Why not C: Reads the gradient correctly as 5 but gets the sign of the y-intercept wrong, using +4 instead of -4.
   • Why not D: Reads the y-intercept correctly as -4 but gets the sign of the gradient wrong, using -5 instead of 5.

7. Question 7Answer: B

   1. In y = mx + c, m controls the gradient (steepness) of the line and c controls where it crosses the y-axis; these two roles are independent.
   2. Increasing c from 2 to 9, with m fixed at 4, shifts every point on the line vertically upwards by 9 - 2 = 7 units, without changing the gradient.
   3. The y-intercept moves from (0, 2) to (0, 9), and because the gradient is unchanged the line looks identical except for this vertical shift.
   4. So the line is translated vertically upwards, keeping gradient 4 but moving its y-intercept from (0, 2) to (0, 9), option B.
   • Why not A: Confuses the role of c with the role of m; in y = mx + c it is m, not c, that controls the steepness of the line, so changing c alone cannot make the line steeper.
   • Why not C: Assumes a vertical shift only ever moves the y-intercept, but for a line with nonzero gradient the x-intercept, at x = -c/m, also changes whenever c changes: here it moves from x = -1/2 to x = -9/4.
   • Why not D: Misdescribes a vertical translation as a rotation; increasing c shifts every point on the line straight upwards by the same amount, keeping the gradient (and so the line's direction) unchanged, rather than pivoting the line about a fixed point.

8. Question 8Answer: A

   1. For a quadratic in the form y = a(x + b)^2 + c, the turning point is at (-b, c).
   2. The turning point here is given as (5, -2), so -b = 5, which gives b = -5, and c = -2 directly, since c is the y-coordinate of the turning point.
   3. So b = -5 and c = -2, option A.
   • Why not B: Forgets the sign change: the turning point of y = a(x + b)^2 + c is at (-b, c), so an x-coordinate of 5 means -b = 5 and b = -5, not b = 5.
   • Why not C: Reads b correctly but gets the sign of c wrong, using c = 2 instead of c = -2, even though c is simply the turning point's y-coordinate with no sign change involved.
   • Why not D: Makes both sign errors at once: uses b = 5 instead of -5, and c = 2 instead of -2.

9. Question 9Answer: D

   1. Differentiate y = 12x - x^3 term by term: the derivative of 12x is 12, and the derivative of -x^3 is -3x^2, so dy/dx = 12 - 3x^2.
   2. At x = 1: dy/dx = 12 - 3(1)^2 = 12 - 3 = 9.
   3. Since dy/dx = 9 > 0, the curve is increasing at x = 1.
   4. So dy/dx = 9 and the curve is increasing at x = 1, option D.
   • Why not A: Finds the correct derivative value, dy/dx = 9, but misreads what a positive derivative means, treating it as a sign that the curve is decreasing rather than increasing.
   • Why not B: Differentiates -x^3 incorrectly as -x^2, dropping the factor of 3 that the power rule brings down, giving dy/dx = 12 - x^2, which equals 11 at x = 1 instead of the correct 9.
   • Why not C: Differentiates -x^3 with the wrong sign, treating it as +3x^2 instead of -3x^2, giving dy/dx = 12 + 3x^2, which equals 15 at x = 1 instead of the correct 9.

10. Question 10Answer: C

    1. Differentiate y = x^3 - 12x + 1: dy/dx = 3x^2 - 12.
    2. At a stationary point, dy/dx = 0, so 3x^2 - 12 = 0, giving x^2 = 4.
    3. Taking square roots of both sides gives x = 2 or x = -2.
    4. So the stationary points occur at x = 2 and x = -2, option C.
    • Why not A: Correctly reaches x^2 = 4 but then forgets to take the square root, stating x = 4 (the value of x^2) as if it were the value of x itself.
    • Why not B: Correctly solves x^2 = 4 as far as taking a square root, but only keeps the positive root x = 2 and forgets the negative root x = -2, which also satisfies x^2 = 4.
    • Why not D: Differentiates x^3 incorrectly as x^2, dropping the factor of 3 that the power rule brings down, so solves x^2 - 12 = 0 instead of 3x^2 - 12 = 0, giving x^2 = 12 and x = +-2sqrt(3) instead of x^2 = 4 and x = +-2.

11. Question 11Answer: A

    1. Factorise 2x^2 - 5x - 3: look for a factorisation (2x + p)(x + q) where the constants multiply to give -3 and combine to give the middle term -5x.
    2. Trying p = 1, q = -3: (2x + 1)(x - 3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3, which matches.
    3. Set each factor to zero: 2x + 1 = 0 gives x = -1/2; x - 3 = 0 gives x = 3.
    4. So the curve crosses the x-axis at x = -1/2 and x = 3, option A.
    • Why not B: Factorises correctly as (2x + 1)(x - 3) but then solves 2x + 1 = 0 as though the coefficient of x were 1, treating it as x + 1 = 0 and getting x = -1 instead of dividing by 2 to get x = -1/2.
    • Why not C: Gets the sign of both roots backwards, as though the factorisation were (2x - 1)(x + 3) instead of (2x + 1)(x - 3).
    • Why not D: Finds the correct first root but makes a sign error on the second factor, as though the factorisation were (2x + 1)(x + 3) instead of (2x + 1)(x - 3), giving x = -3 instead of x = 3.

12. Question 12Answer: D

    1. The y-intercept of a curve is its value when x = 0, so substitute x = 0 into y = (x - 2)(x + 5)(x - 1).
    2. y = (0 - 2)(0 + 5)(0 - 1) = (-2)(5)(-1).
    3. (-2) x 5 = -10, and -10 x (-1) = 10.
    4. So the y-intercept is 10, option D.
    • Why not A: Misreads the sign of the first factor at x = 0, treating (0 - 2) as +2 instead of -2, giving (2)(5)(-1) = -10 instead of (-2)(5)(-1) = 10.
    • Why not B: Evaluates only the first factor at x = 0 and stops, reading off -2 as if it were the whole y-intercept, forgetting to multiply by the other two factors.
    • Why not C: Confuses the y-intercept with one of the graph's x-intercepts, giving the root x = 2 (from the factor x - 2) instead of evaluating the whole expression at x = 0.

13. Question 13Answer: B

    1. A polynomial with real coefficients has exactly as many roots (counted with multiplicity) as its degree, and any non-real roots must occur in complex conjugate pairs.
    2. For a quartic, this means the number of non-real roots is always even (0, 2 or 4), so the number of real roots is 4 minus an even number: 4, 2 or 0, never an odd number like 1 or 3, and never more than 4.
    3. A count of 2 real roots is consistent with this: for example, 2 real roots plus one conjugate pair of non-real roots accounts for all 4 roots.
    4. So 2 is a possible total number of real roots for this quartic, option B.
    • Why not A: Ignores the fact that non-real roots of a polynomial with real coefficients always occur in conjugate pairs; having exactly 1 real root among 4 total roots would leave 3 non-real roots, which cannot pair up evenly, so this is impossible.
    • Why not C: Confuses the number of real roots with the total number of roots the polynomial can have; a quartic has exactly 4 roots in total (counted with multiplicity, and here none are repeated), so it cannot have 5 distinct real roots at all.
    • Why not D: Makes the same parity error as option A: 3 real roots among 4 total roots would leave exactly 1 non-real root, which cannot occur on its own without a conjugate partner, so this is also impossible.

14. Question 14Answer: C

    1. Setting the line equal to the curve: x^2 - 6x + 10 = 2x + c, which rearranges to x^2 - 8x + (10 - c) = 0.
    2. A line is tangent to a curve exactly when this combined equation has a repeated root, which happens when its discriminant is zero: (-8)^2 - 4(1)(10 - c) = 0.
    3. This gives 64 - 40 + 4c = 0, so 24 + 4c = 0, and c = -6.
    4. So c = -6, option C.
    • Why not A: Makes a sign error rearranging the equation, writing x^2 - 8x + (10 + c) = 0 instead of x^2 - 8x + (10 - c) = 0, and so solves 64 - 4(10 + c) = 0 instead of 64 - 4(10 - c) = 0.
    • Why not B: Forgets to square b in the discriminant formula, using b = -8 directly instead of b^2 = 64, and solves -8 - 4(10 - c) = 0 instead of 64 - 4(10 - c) = 0.
    • Why not D: Distributes the 4 across the bracket only partially, treating 4(10 - c) as (40 - c) instead of (40 - 4c), and so solves 64 - (40 - c) = 0 instead of 64 - 40 + 4c = 0.

15. Question 15Answer: A

    1. To find where the line and curve meet, set the two expressions for y equal: x^2 - 3x + 9 = x - 7.
    2. Rearranging correctly: x^2 - 3x + 9 - x + 7 = 0, which simplifies to x^2 - 4x + 16 = 0.
    3. The discriminant of this quadratic is (-4)^2 - 4(1)(16) = 16 - 64 = -48, which is negative, so the equation has no real solutions.
    4. So the line and curve do not intersect anywhere, option A.
    • Why not B: Correctly computes a negative discriminant but misreads what that means: a discriminant of exactly zero indicates a repeated root (the line just touching the curve), while a negative discriminant means there is no real solution at all, so the graphs do not meet.
    • Why not C: Overgeneralises, assuming every quadratic equation has two real solutions; in fact the number of real solutions depends on the sign of the discriminant, and can be zero when the discriminant is negative, as it is here.
    • Why not D: Makes a sign error distributing the negative sign across the bracket when rearranging x^2 - 3x + 9 = x - 7, writing -(x - 7) as -x - 7 instead of -x + 7, which changes the combined equation to x^2 - 4x + 2 = 0 (with a positive discriminant of 8) instead of the correct x^2 - 4x + 16 = 0.