Σ TMUA Practice
Home›Solutions›Graphs of functions

Graphs of functions — solutions

18 full solutions · spec MM8.1, MM8.5-MM8.7, M4.15

Section 1 Part 1: AS-level pure mathematics18 questions
✎ Practise these questions📖 Revision notesAll solution banks

Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. GRA-01Graphs of functions · spec MM8.6 · Paper 1 style · ★☆☆

How many times does the graph of \(y = x^3 - x^2 - 6x\) cross or touch the \(x\)-axis?

  1. A\(3\)
  2. B\(1\)
  3. C\(0\)
  4. D\(2\)
  5. E\(4\)
Answer: A. \(x(x^2 - x - 6) = x(x - 3)(x + 2)\): three distinct real roots, so three crossings.
Full method — how and why
Axis crossings are roots. Factorise fully and count distinct roots.
  1. Factor out \(x\): \(x^3 - x^2 - 6x = x(x^2 - x - 6)\).
    There is no constant term, so \(x\) is a factor - always check this first.
  2. Factorise the quadratic: \(x^2 - x - 6 = (x-3)(x+2)\).
    Two numbers multiplying to \(-6\), adding to \(-1\).
  3. Roots: \(x = 0, 3, -2\) - three distinct values, each a simple root, so the graph crosses three times.
    Distinct simple roots mean genuine crossings; a repeated root would be a touch.
Where it goes wrong
  • Missing the root \(x = 0\) and answering \(2\).
  • Counting the \(y\)-intercept as an extra crossing of the \(x\)-axis.
Takeaway Factor out \(x\) when there is no constant term; count distinct roots for the number of axis meetings.
2. GRA-02Graphs of functions · spec MM8.7 · Paper 1 style · ★☆☆

The graphs of \(y = x^2 - 2x\) and \(y = k\) meet at exactly two points. Find the complete set of values of \(k\).

  1. A\(k > -1\)
  2. B\(k \geq -1\)
  3. C\(k > 0\)
  4. D\(k = -1\)
  5. E\(k < -1\)
  6. Fall real \(k\)
Answer: A. \(y = (x-1)^2 - 1\) has minimum \(-1\); a horizontal line above the minimum meets the parabola twice.
Full method — how and why
A horizontal line meeting a parabola twice: the line must be above the minimum.
  1. Complete the square: \(x^2 - 2x = (x-1)^2 - 1\), so the minimum value is \(-1\) at \(x = 1\).
    The turning point governs how many times a horizontal line can cut the curve.
  2. A line \(y = k\) meets the parabola twice when \(k\) is strictly above the minimum.
    At \(k = -1\) the line is tangent at the vertex: exactly one point. Below it: none.
  3. So \(k > -1\).
    Strict inequality, because equality gives only one intersection.
Where it goes wrong
  • Including \(k = -1\) and choosing \(k \geq -1\) - offered as a distractor.
  • Using the \(y\)-intercept \(0\) instead of the minimum value.
Takeaway Count intersections with a horizontal line by sliding it against the turning point; equality gives tangency.
3. GRA-03Graphs of functions · spec MM8.1 · Paper 1 style · ★☆☆

Which of the following could be the equation of a cubic curve that touches the \(x\)-axis at \(x = 2\) and crosses it at \(x = -1\), and has a negative \(y\)-intercept?

  1. A\(y = -(x - 2)^2(x + 1)\)
  2. B\(y = (x - 2)(x + 1)^2\)
  3. C\(y = (x + 2)^2(x - 1)\)
  4. D\(y = -(x + 2)^2(x - 1)\)
  5. E\(y = (x - 2)^2(x + 1)\)
Answer: A. Touching at \(2\) needs the factor \((x - 2)^2\) and crossing at \(-1\) the factor \((x + 1)\). \(y = (x-2)^2(x+1)\) has \(y\)-intercept \(4 > 0\), so the sign is reversed: \(y = -(x-2)^2(x+1)\) with intercept \(-4\).
Full method — how and why
Build the cubic from its root behaviour, then use the \(y\)-intercept to fix the sign.
  1. 'Touches at \(x = 2\)' means a repeated factor \((x-2)^2\); 'crosses at \(x = -1\)' means a simple factor \((x+1)\).
    Even-power factors touch the axis; odd-power factors cross it.
  2. So the cubic is \(y = a(x-2)^2(x+1)\) for some constant \(a \neq 0\).
    Degree \(2 + 1 = 3\) ✓.
  3. Compute the \(y\)-intercept: at \(x = 0\), \(y = a(4)(1) = 4a\).
    Substitute \(x = 0\) into the factorised form.
  4. A negative intercept needs \(a < 0\), so \(y = -(x-2)^2(x+1)\) works (intercept \(-4\)).
    The option with \(a = +1\) has intercept \(+4\) and is the trap.
Where it goes wrong
  • Choosing \((x-2)^2(x+1)\) and ignoring the sign of the intercept.
  • Swapping which root is repeated, e.g. \((x-2)(x+1)^2\).
Takeaway Repeated factor \(\Rightarrow\) touch; simple factor \(\Rightarrow\) cross; the \(y\)-intercept fixes the leading sign.
4. GRA-05Graphs of functions · spec MM8.1 · Paper 2 style · ★☆☆

Which of the following statements about the graph of \(y = |2x - 4|\) is/are true?

  1. IIt meets the \(x\)-axis at exactly one point.
  2. IIIt is symmetric about the line \(x = 2\).
  3. IIIIt lies entirely on or above the line \(y = 0\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: H. \(|2x - 4| = 0\) only at \(x = 2\) (I). \(|2x - 4| = 2|x - 2|\) is symmetric about \(x = 2\) (II). A modulus is never negative (III).
Full method — how and why
Three properties of a modulus graph. Rewriting \(|2x-4|\) as \(2|x-2|\) makes all three obvious.
  1. Rewrite: \(|2x - 4| = 2|x - 2|\).
    Factoring the \(2\) out of the modulus is legitimate because \(|2u| = 2|u|\).
  2. I: the graph meets \(y = 0\) when \(|x-2| = 0\), i.e. only at \(x = 2\). True.
    A V-shape touches the axis at its single vertex.
  3. II: \(|x - 2|\) depends only on the distance from \(2\), so the graph is symmetric about \(x = 2\). True.
    Replacing \(x\) by \(4 - x\) leaves the value unchanged.
  4. III: a modulus is never negative, so the graph lies on or above the \(x\)-axis. True.
    \(|u| \geq 0\) by definition.
  5. All three.
    Match to the option list.
Where it goes wrong
  • Thinking the graph crosses the axis twice because \(|x^2 - 4|\)-type graphs do - this one is linear inside the modulus.
  • Reading the line of symmetry as \(x = 4\).
Takeaway \(|ax + b| = |a|\left|x + \frac ba\right|\): a V-shape with its vertex at the root, never below the axis.
5. GRA-06Graphs of functions · spec MM8.6 · Paper 1 style · ★☆☆

The curve \(y = x^4 - 5x^2 + 4\) meets the coordinate axes at how many points in total?

  1. A\(5\)
  2. B\(3\)
  3. C\(6\)
  4. D\(1\)
  5. E\(2\)
  6. F\(4\)
Answer: A. \(x\)-axis: \((x^2 - 1)(x^2 - 4) = 0\), four points. \(y\)-axis: \((0, 4)\), one point. Total \(5\).
Full method — how and why
Count both axes. The \(x\)-axis meetings come from the roots; the \(y\)-axis always contributes exactly one point.
  1. Factorise: \(x^4 - 5x^2 + 4 = (x^2-1)(x^2-4)\).
    Quadratic in \(x^2\): two numbers multiplying to \(4\) and adding to \(-5\).
  2. Solve: \(x = \pm1\) and \(x = \pm2\) - four \(x\)-axis crossings.
    Each positive value of \(x^2\) gives two values of \(x\).
  3. \(y\)-axis: substitute \(x = 0\) to get \(y = 4\) - one point.
    A function meets the \(y\)-axis exactly once.
  4. Total: \(4 + 1 = 5\).
    The \(y\)-intercept is a separate point (it is not one of the roots here, since \(y(0) \neq 0\)).
Where it goes wrong
  • Forgetting the \(y\)-axis and answering \(4\).
  • Counting \(x^2 = 1, 4\) as only two points.
Takeaway 'Meets the coordinate axes' means both axes: roots plus the single \(y\)-intercept.
6. GRA-09Graphs of functions · spec MM8.1 · Paper 1 style · ★☆☆

The graph of \(y = \log_2 x\) is reflected in the line \(y = x\). The resulting graph has equation

  1. A\(y = 2^{x}\)
  2. B\(y = \log_{\frac12} x\)
  3. C\(y = \log_2(-x)\)
  4. D\(y = 2^{-x}\)
  5. E\(y = x^2\)
  6. F\(y = -\log_2 x\)
Answer: A. Reflection in \(y = x\) gives the inverse function: \(x = \log_2 y\), i.e. \(y = 2^x\).
Full method — how and why
Reflection in \(y = x\) produces the inverse function - swap \(x\) and \(y\) and re-solve.
  1. Reflecting in \(y = x\) swaps the roles of \(x\) and \(y\): the new graph satisfies \(x = \log_2 y\).
    Every point \((a,b)\) maps to \((b,a)\).
  2. Rewrite in exponential form: \(y = 2^x\).
    \(\log_2 y = x\) means \(y = 2^x\) - logs and exponentials are inverse functions.
Where it goes wrong
  • Choosing \(y = \log_2(-x)\) (a \(y\)-axis reflection) or \(-\log_2 x\) (an \(x\)-axis reflection).
  • Thinking the inverse of a log is \(x^2\) because of the base \(2\).
Takeaway Reflection in \(y = x\) gives the inverse; \(y = a^x\) and \(y = \log_a x\) are mirror images in that line.
7. GRA-10Graphs of functions · spec MM8.6 · Paper 1 style · ★☆☆

A polynomial of degree \(5\) with real coefficients has how many real roots at least, and at most?

  1. Aat least \(1\), at most \(5\)
  2. Bexactly \(5\)
  3. Cat least \(1\), at most \(4\)
  4. Dat least \(0\), at most \(4\)
  5. Eat least \(0\), at most \(5\)
  6. Fat least \(2\), at most \(5\)
Answer: A. An odd-degree polynomial tends to opposite infinities at each end, so it crosses the axis at least once; a degree-\(5\) polynomial has at most \(5\) roots.
Full method — how and why
Two standard facts about polynomial roots: the maximum is the degree, and odd degree forces at least one real root.
  1. At most: a degree-\(5\) polynomial has at most \(5\) roots, since each root contributes a linear factor.
    More than \(5\) factors would raise the degree above \(5\).
  2. At least: as \(x \to +\infty\) and \(x \to -\infty\), an odd-degree polynomial tends to opposite infinities.
    The leading term \(ax^5\) dominates and changes sign with \(x\).
  3. A continuous function that is positive somewhere and negative somewhere must cross zero, so there is at least one real root.
    Intermediate value reasoning; this is why every odd-degree polynomial has a real root.
  4. At least \(1\), at most \(5\).
    Match to the options.
Where it goes wrong
  • Saying 'at least \(0\)' by analogy with even degrees like \(x^2 + 1\), which genuinely have none.
  • Saying 'exactly \(5\)', which counts complex roots or ignores repeats.
Takeaway Odd degree \(\Rightarrow\) at least one real root; degree \(n\) \(\Rightarrow\) at most \(n\) real roots.
8. GRA-12Graphs of functions · spec MM8.1 · Paper 1 style · ★☆☆

Which of the following functions is undefined at exactly two real values of \(x\)?

  1. A\(f(x) = \dfrac{1}{(x - 2)^2}\)
  2. B\(f(x) = \dfrac{x}{x^2}\)
  3. C\(f(x) = \sqrt{x - 2}\)
  4. D\(f(x) = \dfrac{1}{x^2 + 4}\)
  5. E\(f(x) = \dfrac{1}{x^2 - 4}\)
Answer: E. \(x^2 - 4 = 0\) at \(x = \pm 2\): two values. \(x^2 + 4\) is never zero; \(\frac{x}{x^2}\) and \(\frac{1}{(x-2)^2}\) are undefined at one value; \(\sqrt{x - 2}\) is undefined for all \(x < 2\).
Full method — how and why
'Undefined' means division by zero (or a negative under a square root). Count the offending values for each option.
  1. \(\dfrac{1}{x^2-4}\): the denominator vanishes at \(x = 2\) and \(x = -2\) - two values.
    Difference of two squares gives two distinct roots.
  2. \(\dfrac{1}{x^2+4}\): the denominator is always at least \(4\), so it is never zero - defined everywhere.
    A sum of a square and a positive constant is never zero.
  3. \(\dfrac{x}{x^2}\) and \(\dfrac{1}{(x-2)^2}\): each has a single problem value (\(0\) and \(2\) respectively).
    A repeated factor still vanishes at only one point.
  4. \(\sqrt{x-2}\): undefined for all \(x < 2\) - infinitely many values, not two.
    Square roots need a non-negative argument.
Where it goes wrong
  • Counting \((x-2)^2\) as two values because of the power.
  • Thinking \(x^2 + 4\) has roots \(\pm 2\).
Takeaway Count the distinct values that break the expression: repeated factors give one, sums of squares give none.
9. GRA-16Graphs of functions · spec MM8.7 · Paper 1 style · ★☆☆

The line \(y = mx\) meets the curve \(y = x^3 - x\) at three distinct points. Find the complete set of values of \(m\).

  1. A\(m < -1\)
  2. B\(-1 < m < 0\)
  3. C\(m > 1\)
  4. D\(m > 0\)
  5. E\(m > -1\)
  6. Fall real \(m\)
Answer: E. \(x^3 - x - mx = x(x^2 - (1 + m)) = 0\): three distinct roots iff \(1 + m > 0\), i.e. \(m > -1\).
Full method — how and why
Intersections of a line through the origin with a cubic. Factor out the common \(x\) and use the discriminant on what remains.
  1. Set them equal: \(mx = x^3 - x\), so \(x^3 - x - mx = 0\).
    Bring everything to one side.
  2. Factor out \(x\): \(x\left(x^2 - (1 + m)\right) = 0\).
    \(x = 0\) is always a solution, because both graphs pass through the origin.
  3. The other roots satisfy \(x^2 = 1 + m\), which has two distinct non-zero solutions exactly when \(1 + m > 0\).
    If \(1 + m = 0\) the root coincides with \(x = 0\); if negative, there are no more real roots.
  4. So \(m > -1\).
    Three distinct points in total.
Where it goes wrong
  • Forgetting the root \(x = 0\) and requiring three roots from the quadratic.
  • Including \(m = -1\), where all three roots coincide at \(x = 0\).
Takeaway A line through the origin always meets \(y = x^3 - x\) there; factor out \(x\) and analyse the quadratic that remains.
10. GRA-04Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

The line \(y = x + c\) and the curve \(y = \dfrac{1}{x}\) have no points of intersection. Find the complete set of values of \(c\).

  1. A\(c < -2\) or \(c > 2\)
  2. B\(-1 < c < 1\)
  3. C\(c > 2\)
  4. D\(c < 2\)
  5. E\(-2 < c < 2\)
  6. Fno such values
Answer: F. \(x^2 + cx - 1 = 0\) always has discriminant \(c^2 + 4 > 0\): two intersections for every \(c\). So there are no values of \(c\) with no intersection.
Full method — how and why
A 'find the values' question whose honest answer is 'none'. Substituting shows the discriminant can never be negative.
  1. Set the expressions equal: \(x + c = \dfrac1x\).
    Intersections satisfy both equations.
  2. Multiply by \(x\) (noting \(x \neq 0\), since \(\frac1x\) is undefined there): \(x^2 + cx - 1 = 0\).
    Clearing the fraction gives a quadratic whose roots are the intersection \(x\)-values.
  3. Compute the discriminant: \(c^2 - 4(1)(-1) = c^2 + 4\).
    The \(-1\) constant is what makes the discriminant unavoidably positive.
  4. \(c^2 + 4 > 0\) for every real \(c\), so there are always two intersections: no value of \(c\) avoids them.
    Also check the roots are non-zero: their product is \(-1 \neq 0\), so neither is the excluded \(x = 0\).
Where it goes wrong
  • Assuming a line and a hyperbola must sometimes miss, and choosing an interval.
  • Sign error making the discriminant \(c^2 - 4\), which would give \(-2 < c < 2\) - an offered option.
Takeaway Do the algebra even when the answer 'feels' like an interval; \(b^2 - 4ac\) with \(ac < 0\) is always positive.
11. GRA-07Graphs of functions · spec MM8.5 · Paper 1 style · ★★☆

The graph of \(y = f(x)\) has \(f'(x) > 0\) for \(x < 1\), \(f'(1) = 0\), \(f'(x) > 0\) for \(1 < x < 3\), \(f'(x) < 0\) for \(x > 3\). Which of the following is true?

  1. A\(f\) has local maxima at \(x = 1\) and \(x = 3\)
  2. B\(f\) has a local minimum at \(x = 1\) and a local maximum at \(x = 3\)
  3. C\(f\) has a local maximum at \(x = 3\) and a stationary point of inflexion at \(x = 1\)
  4. D\(f\) has a local maximum at \(x = 1\) and a local minimum at \(x = 3\)
  5. E\(f\) is increasing for all \(x\)
Answer: C. At \(x = 1\) the derivative is zero but does not change sign: a stationary point that is not a turning point (an inflexion). At \(x = 3\) the derivative changes from positive to negative: local maximum.
Full method — how and why
Reading a function's shape from the sign of its derivative. A zero derivative without a sign change is an inflexion, not a turning point.
  1. Before and after \(x = 1\): \(f' > 0\) on both sides, with \(f'(1) = 0\).
    The gradient dips to zero and comes back positive - the curve flattens momentarily but keeps rising.
  2. So \(x = 1\) is a stationary point that is not a turning point: a stationary point of inflexion.
    Turning points require a change of sign in \(f'\).
  3. At \(x = 3\): \(f' > 0\) just before and \(f' < 0\) just after, so the curve rises then falls - a local maximum.
    A \(+ \to -\) sign change is exactly a local maximum.
  4. So: local maximum at \(x = 3\), stationary inflexion at \(x = 1\).
    Match to the options.
Where it goes wrong
  • Assuming \(f'(1) = 0\) must be a maximum or minimum.
  • Reading the \(+ \to -\) change at \(x = 3\) as a minimum.
Takeaway \(f' = 0\) with no sign change \(=\) inflexion; \(+ \to -\) is a maximum, \(- \to +\) is a minimum.
12. GRA-08Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

How many real solutions does the equation \(x^3 = 3x + 1\) have?

  1. A\(2\)
  2. B\(0\)
  3. C\(1\)
  4. D\(3\)
  5. Einfinitely many
Answer: D. Sketch \(y = x^3 - 3x - 1\): turning points at \(x = \pm1\) with \(y(-1) = 1 > 0\) and \(y(1) = -3 < 0\). A cubic whose local maximum is positive and local minimum is negative crosses the axis three times.
Full method — how and why
Counting roots of a cubic. Find the turning values and see whether they straddle the axis.
  1. Rearrange to \(f(x) = x^3 - 3x - 1 = 0\).
    Everything on one side, so the question becomes 'how many roots?'.
  2. Differentiate: \(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\), so the turning points are at \(x = \pm1\).
    A cubic with two turning points can have one, two or three roots.
  3. Evaluate: \(f(-1) = -1 + 3 - 1 = 1 > 0\) (local maximum) and \(f(1) = 1 - 3 - 1 = -3 < 0\) (local minimum).
    The values at the turning points decide the count.
  4. The local maximum is above the axis and the local minimum below, so the curve crosses three times.
    It comes up from \(-\infty\), crosses, comes down past the axis, crosses, then rises again.
Where it goes wrong
  • Trying to factorise - this cubic has no rational roots, so factorising wastes time.
  • Answering \(1\) because 'a cubic always has at least one root' without checking the turning values.
Takeaway Cubic root count: find the turning values; three roots iff the local max and min lie on opposite sides of the axis.
13. GRA-11Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

The curves \(y = 2^{x}\) and \(y = x^2\) intersect at how many points?

  1. A\(3\)
  2. B\(0\)
  3. C\(2\)
  4. D\(4\)
  5. E\(1\)
Answer: A. They meet at \(x = 2\) and \(x = 4\) (\(4 = 4\), \(16 = 16\)), and once for negative \(x\) (at \(x = 0\): \(1 > 0\); at \(x = -1\): \(\frac12 < 1\)). For \(x > 4\), \(2^x\) grows faster. Three intersections.
Full method — how and why
Exponential versus quadratic. Test the obvious integer points and use end behaviour to be sure nothing is missed.
  1. Try small integers: \(x = 2\) gives \(4 = 4\) ✓ and \(x = 4\) gives \(16 = 16\) ✓ - two intersections already.
    \(2^x = x^2\) has these two well-known solutions.
  2. Check negative \(x\): at \(x = 0\), \(2^0 = 1 > 0 = x^2\); at \(x = -1\), \(2^{-1} = \frac12 < 1 = x^2\). The order swaps, so there is a crossing between \(-1\) and \(0\).
    A sign change in the difference \(2^x - x^2\) guarantees a crossing.
  3. Check beyond \(x = 4\): \(2^x\) grows faster than \(x^2\) (e.g. at \(x = 5\), \(32 > 25\)), so they never meet again.
    Exponential growth eventually outpaces any polynomial.
  4. Total: three intersections.
    One negative, and \(x = 2\), \(x = 4\).
Where it goes wrong
  • Finding only \(x = 2\) and \(x = 4\) and answering \(2\).
  • Forgetting that \(x^2\) is positive for negative \(x\) while \(2^x\) stays small but positive.
Takeaway Test integer points, then check the two ends: a sign change in the difference means an extra crossing.
14. GRA-13Graphs of functions · spec MM8.7 · Paper 1 style · ★★☆

For how many values of \(k\) does the equation \(|x^2 - 4| = k\) have exactly three real solutions?

  1. A\(1\)
  2. B\(0\)
  3. C\(2\)
  4. D\(3\)
  5. Einfinitely many
Answer: A. The graph of \(y = |x^2 - 4|\) is a W-shape with minima \(0\) at \(x = \pm2\) and a local maximum \(4\) at \(x = 0\). A horizontal line gives three solutions only when it passes through the local maximum: \(k = 4\). One value.
Full method — how and why
Sliding a horizontal line across a modulus graph. Three solutions happen only at the height of the local maximum.
  1. Sketch \(y = |x^2 - 4|\): the parabola \(x^2 - 4\) with the part between \(x = -2\) and \(x = 2\) flipped above the axis.
    Reflecting the negative part gives a W-shape with two zeros and a hump.
  2. Identify the key heights: zeros at \(x = \pm2\), and a local maximum of \(4\) at \(x = 0\).
    At \(x = 0\), \(|0 - 4| = 4\) - the top of the central hump.
  3. Count intersections with \(y = k\): for \(0 < k < 4\) there are four; for \(k > 4\) there are two; for \(k = 0\) there are two; for \(k = 4\) the line passes through the hump's peak and cuts the outer arms: three.
    At the peak, two of the four intersections merge into one.
  4. So exactly one value of \(k\) (namely \(4\)) gives three solutions.
    The question asks how many values of \(k\), not how many solutions.
Where it goes wrong
  • Answering \(4\) (the value of \(k\)) instead of \(1\) (the number of such \(k\)).
  • Forgetting \(k = 0\) gives two solutions, not three.
Takeaway Slide the horizontal line and note the heights where the count changes: those are the turning values.
15. GRA-14Graphs of functions · spec MM8.1 · Paper 1 style · ★★☆

The graph of \(y = \sqrt{x}\) is transformed to \(y = \sqrt{4 - x}\). Which sequence of transformations achieves this?

  1. Atranslate \(4\) units in the positive \(x\)-direction, then reflect in the \(y\)-axis
  2. Btranslate \(4\) units in the negative \(x\)-direction, then reflect in the \(y\)-axis
  3. Creflect in the \(x\)-axis, then translate \(4\) units in the positive \(y\)-direction
  4. Dstretch parallel to the \(x\)-axis with factor \(4\)
  5. Ereflect in the \(y\)-axis, then translate \(4\) units in the positive \(x\)-direction
Answer: E. \(\sqrt{4 - x} = \sqrt{-(x - 4)}\): reflect \(y = \sqrt x\) in the \(y\)-axis to get \(\sqrt{-x}\), then translate \(4\) to the right. (Translating first then reflecting gives \(\sqrt{-x - 4}\).)
Full method — how and why
Order matters when combining a reflection with a translation. Rewrite the target to reveal the correct sequence.
  1. Rewrite the target: \(\sqrt{4 - x} = \sqrt{-(x - 4)}\).
    Factoring out the \(-1\) exposes both operations: a negation of \(x\) and a shift.
  2. Reading it as \(f(-(x-4))\) where \(f(x) = \sqrt x\): first reflect in the \(y\)-axis (giving \(\sqrt{-x}\)), then translate \(4\) right (replacing \(x\) by \(x - 4\)).
    For nested transformations, the operation applied directly to \(x\) (here the shift) is done last.
  3. Check with a point: \(y = \sqrt{4-x}\) passes through \((4, 0)\); \(\sqrt{-x}\) passes through \((0,0)\), which moves to \((4,0)\) after shifting right ✓.
    A point check settles order questions decisively.
  4. Verify the other order fails: translating first gives \(\sqrt{x-4}\), then reflecting gives \(\sqrt{-x-4}\) - not the target.
    Different sequences give genuinely different graphs.
Where it goes wrong
  • Choosing 'translate then reflect', which gives \(\sqrt{-x-4}\).
  • Reflecting in the \(x\)-axis instead of the \(y\)-axis.
Takeaway Rewrite as \(f(\pm(x - h))\) and test a point; with mixed reflections and shifts the order is not interchangeable.
16. GRA-15Graphs of functions · spec MM8.6 · Paper 2 style · ★★☆

The cubic \(y = x^3 + ax + b\) has three distinct real roots. Which of the following statements must be true?

  1. I\(a < 0\)
  2. II\(b \neq 0\)
  3. IIIThe curve has two turning points.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: F. Three distinct real roots require two turning points (III), which needs \(y' = 3x^2 + a\) to have two roots, so \(a < 0\) (I). II is false: \(x^3 - x\) has roots \(0, \pm1\) and \(b = 0\).
Full method — how and why
Conditions for a cubic to have three distinct roots. Two claims follow from the shape; the third needs a counterexample.
  1. III: three distinct real roots force the curve to turn twice (up, down, up), so it must have two turning points. True.
    Between consecutive roots the curve must turn around.
  2. I: \(y' = 3x^2 + a\) has two real roots only when \(a < 0\) (so that \(x^2 = -\frac a3\) has solutions). True.
    This follows directly from III: two turning points require \(y'\) to have two roots.
  3. II: test \(y = x^3 - x = x(x-1)(x+1)\): three distinct roots \(0, \pm1\), and here \(b = 0\). False.
    One counterexample is enough; \(b = 0\) simply means the origin is a root.
  4. I and III only.
    Match to the options.
Where it goes wrong
  • Assuming \(b \neq 0\) because 'otherwise the cubic factorises' - it does, and that is fine.
  • Failing to link I to III and guessing the sign of \(a\).
Takeaway Three distinct roots \(\Leftrightarrow\) two turning points with values of opposite sign; that forces \(a < 0\) but says nothing about \(b\).
17. X2-09Graphs of functions · spec MM8.7 · Paper 2 style · ★★☆

The graphs of \(y = f(x)\) and \(y = g(x)\) meet at exactly three points. Which of the following statements must be true?

  1. IThe equation \(f(x) = g(x)\) has exactly three real solutions.
  2. IIThe graph of \(y = f(x) - g(x)\) crosses the \(x\)-axis exactly three times.
  3. III\(f(x) - g(x)\) is a polynomial of degree at least \(3\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: B. I is the definition of an intersection. II fails because the difference could touch the axis rather than cross it at one of the points. III fails because \(f\) and \(g\) need not be polynomials at all (e.g. \(\sin x\) and a line).
Full method — how and why
Intersections and roots. Distinguish a root from a crossing, and resist assuming the functions are polynomials.
  1. I: the graphs meet where \(f(x)=g(x)\), so three intersection points means exactly three solutions. True - it is the definition.
    Each intersection point has a distinct \(x\)-value, since a graph has one \(y\) per \(x\).
  2. II: at a root of \(f-g\) the graph could touch the \(x\)-axis and turn back rather than cross - a repeated root.
    Meeting and crossing are different: \(y = x^2\) meets the axis but does not cross it.
  3. So the number of crossings could be fewer than three. False.
    For instance \(f-g = x^2(x-1)\) meets the axis at \(0\) and \(1\) but crosses only at \(1\).
  4. III: nothing says \(f\) and \(g\) are polynomials. Take \(f(x)=\sin x\) and a suitable straight line \(g\), meeting three times. False.
    Read the hypotheses: only 'the graphs of \(y=f(x)\) and \(y=g(x)\)' is given.
  5. So I only.
    Match to the list.
Where it goes wrong
  • Accepting II by treating 'crosses' as a synonym for 'meets'.
  • Accepting III by assuming a degree just because three intersections are mentioned.
Takeaway Intersections \(=\) solutions of \(f=g\); but crossing the axis needs a sign change, and never assume a function type that was not stated.
18. X2-21Graphs of functions · spec MM8.6 · Paper 2 style · ★★☆

\(p(x)\) is a polynomial of degree \(4\) with real coefficients. Which of the following statements must be true?

  1. I\(p(x) = 0\) has at least one real root.
  2. II\(p(x) = 0\) has at most four real roots.
  3. IIIThe graph of \(y = p(x)\) has at least one turning point.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: G. I fails: \(x^4 + 1\) has no real roots. II: a degree-4 polynomial has at most 4 roots. III: an even-degree polynomial tends to the same infinity at both ends, so it must turn at least once.
Full method — how and why
Three claims about a quartic. Two follow from degree; the first needs a counterexample.
  1. I: must a degree-\(4\) polynomial have a real root? No: \(x^4+1 \geq 1 > 0\) for all real \(x\). False.
    Even-degree polynomials can stay entirely above the axis - unlike odd-degree ones.
  2. II: a polynomial of degree \(n\) has at most \(n\) real roots, so at most \(4\). True.
    Each root supplies a linear factor; more than \(4\) would exceed the degree.
  3. III: as \(x \to \pm\infty\), a quartic tends to the same infinity at both ends (the sign of the leading coefficient).
    Even powers dominate and are positive at both extremes.
  4. A continuous function going to the same infinity both ways must turn round somewhere, so it has at least one turning point. True.
    It has a global minimum (or maximum) at a finite point.
  5. So II and III.
    Match to the list.
Where it goes wrong
  • Accepting I by analogy with odd-degree polynomials, which must have a real root.
  • Reading II as 'exactly four roots' - repeated roots and non-real roots reduce the count.
Takeaway Odd degree guarantees a real root; even degree guarantees a turning point. Degree \(n\) caps the real roots at \(n\).