Stationary Points and Curve Sketching

Find and classify the stationary point of $y = x^2 - 6x + 5$.

• Minimum at $(3, -4)$

  Correct. dy/dx = 2x-6 = 0 gives x = 3; y(3) = 9-18+5 = -4. d²y/dx² = 2 > 0, confirming a minimum.

• BMaximum at $(3,-4)$

  The coordinates are right and the nature is backwards: d²y/dx² = 2, which is POSITIVE, and a positive second derivative signals a minimum, not a maximum.

• CMinimum at $(0, 5)$

  $(0,5)$ is the y-intercept — read off the constant term, c = 5 — not the turning point. The turning point comes from solving dy/dx = 0, a completely different calculation that only afterwards involves substituting back into y.

• DMinimum at $(6, 5)$

  This comes from solving $2x-6=0$ as though it were $x-6=0$ — dropping the coefficient of x rather than dividing by it. $2x=6$ gives $x=3$, not $x=6$; substituting the wrong x-value of 6 back into y does give a real point on the curve, $(6,5)$, which is exactly what makes this distractor plausible — it just is not the stationary point.

Traps tested: Min max sign inverted · Y intercept confused with turning point · Coefficient dropped when solving derivative equation

A curve has $\dfrac{dy}{dx} = (x-1)(x-4)$. For which values of $x$ is the curve decreasing?

• $1 < x < 4$

  Correct. (x-1)(x-4) is an upward-opening quadratic in x with roots 1 and 4, so it is negative BETWEEN them — checked at x = 2: (1)(-2) = -2 < 0 — and decreasing is exactly where dy/dx < 0.

• B$x<1$ or $x>4$

  This is where dy/dx is POSITIVE — outside the two roots — which is where the curve is increasing, not decreasing. The two regions are exact opposites of each other.

• C$x<1$

  This captures only where the first factor, (x-1), is negative, ignoring that the sign of the product also depends on the second factor, (x-4). Testing an actual value is the reliable check: at x = 2, (2-1)(2-4) = (1)(-2) = -2 < 0, and x = 2 is not even inside this answer's range.

• D$x=1$ and $x=4$ only

  x = 1 and x = 4 are the stationary points themselves — where dy/dx = 0, momentarily flat, neither increasing nor decreasing. The decreasing region is the whole open interval BETWEEN them, not the two isolated endpoints.

Traps tested: Increasing region mistaken for decreasing · Only one branch of the region identified · Stationary points confused with decreasing interval

A stationary point is found, and the second-derivative test gives $\dfrac{d^2y}{dx^2}=0$ there — inconclusive. Which method correctly determines the point's nature and is confirmed, on a real WMA12 mark scheme, to earn full credit?

• Test the sign of $dy/dx$ at a value just below, and a value just above, the stationary point's x-coordinate

  Correct. A real WMA12 examiner report confirms this exact method — testing the gradient's sign either side of the point — earns "full credit as it used further calculus" (Jan 2021 Q2), precisely because it goes back to the definition of increasing/decreasing rather than relying on the (here, unhelpful) second derivative.

• BCompare the curve's y-values at a value just below, and a value just above, the stationary point's x-coordinate

  This looks similar to the correct method — 'either side' of the point — but it is not the same calculation. A real WMA12 examiner report confirms this exact approach "was not given credit as it did not use further calculus as asked for in the question" (Jan 2021 Q2): comparing heights is not the same as testing gradients, and only the gradient test uses calculus at all.

• CConclude it must be a point of inflection, since the second-derivative test failed to give an answer

  d²y/dx² = 0 means the test has not decided — it does not mean the point IS a point of inflection. The point could still turn out to be a genuine maximum or minimum once tested properly; 'inconclusive' is not itself a conclusion.

• DRecompute the second derivative with more decimal places, in case the zero was a rounding artefact

  d²y/dx² = 0 here is an exact result, not a rounding error to chase — this course's questions are solved in exact form throughout, and no amount of extra precision turns an exact zero into something else. The issue is which METHOD to use next, not how precisely to repeat the same one.

Traps tested: Y values eyeballed instead of gradient tested · Inconclusive case overclaimed as a definite conclusion · Inconclusive result treated as a precision problem

Using the stationary points found in this lesson's marked-solution above — $(-1,7)$, a maximum, and $(3,-25)$, a minimum, on the cubic $f(x)=x^3-3x^2-9x+2$ — which correctly describes the curve's behaviour as $x\to-\infty$ and $x\to+\infty$?

• $y\to-\infty$ as $x\to-\infty$; $y\to+\infty$ as $x\to+\infty$

  Correct. The leading term, $x^3$, has a positive coefficient and an odd power, so for very negative x it dominates and drags y very negative, and for very positive x it drags y very positive — exactly the shape needed to connect '-∞ up to the maximum, down to the minimum, up to +∞' into one continuous curve.

• B$y\to+\infty$ at both ends

  Both-ends-up is the shape of a positive EVEN-degree curve (a quartic, or $x^2$), not an odd-degree one. A cubic's two ends always go in opposite directions, whatever its coefficients.

• C$y\to-\infty$ at both ends

  Both-ends-down is the shape of a NEGATIVE even-degree curve. This cubic's leading coefficient is positive, and in any case an odd-degree curve's two ends never match — they always point in opposite directions.

• D$y\to+\infty$ as $x\to-\infty$; $y\to-\infty$ as $x\to+\infty$

  This is the shape a NEGATIVE leading coefficient would produce. Here the coefficient of $x^3$ is $+1$, positive, so the curve's ends point the other way: down on the left, up on the right.

Traps tested: Even degree end behaviour applied to a cubic · End behaviour reversed

The general marking principles for WMA12 define the method mark for an attempt at differentiation as...

• the power of at least one term decreased by 1

  Correct, and it is the exact mirror of the integration method mark used elsewhere on this paper ('the power of at least one term increased by 1') — the mark is earned by the structure of the attempt, independent of whether the working that follows is correct.

• Bthe power of every term decreased by 1

  The real condition only requires AT LEAST ONE term to show a genuine attempt at differentiation — not that every single term must already be correctly handled. Requiring all of them would turn a method mark into an accuracy mark in disguise.

• Cthe constant term correctly differentiated to zero

  Correctly dropping a constant term is one small consequence of differentiating correctly, not the definition of the method mark itself — a function with no constant term at all can still earn this mark from its other terms.

• Dthe final answer simplified into a single fraction

  How neatly the final expression is written has nothing to do with whether a valid method was attempted — the method mark is about the differentiation step itself, not the tidiness of what comes out of it.

Traps tested: Method mark condition overstated · Single term rule mistaken for the general condition · Presentation mistaken for method

$\dfrac{dy}{dx}=3x^2-4x-7=(3x-7)(x+1)$ for the curve $y=x^3-2x^2-7x+4$. What are the exact x-coordinates of its stationary points?

• $x=\dfrac{7}{3}$ and $x=-1$

  Correct. $(3x-7)(x+1)=0$ gives $3x-7=0$ or $x+1=0$, so $x=\dfrac{7}{3}$ or $x=-1$. The quadratic factorises cleanly, with no formula needed — but that does not make $\dfrac{7}{3}$ an integer, and it must be left as an exact fraction.

• B$x\approx 2.33$ and $x=-1$

  2.33 is $\dfrac{7}{3}$ correctly rounded to 2 decimal places — but the general marking principles are explicit that where an exact answer is required, marks are normally lost for resorting to a rounded decimal, however precisely it is rounded. The method mark for solving the quadratic is unaffected; it is the accuracy mark for this coordinate that a decimal answer forfeits.

• C$x=7$ and $x=-1$

  This comes from solving $3x-7=0$ as though it were $x-7=0$ — dropping the coefficient of x rather than dividing by it. $3x=7$ gives $x=\dfrac{7}{3}$, not $x=7$.

• D$x=\dfrac{7}{3}$ and $x=1$

  The second factor is $(x+1)$, not $(x-1)$: setting $x+1=0$ gives $x=-1$, not $x=1$. Flipping the sign inside a factor silently changes the sign of the root.

Traps tested: Stationary point x coordinate rounded instead of exact · Coefficient dropped when solving derivative equation · Factor sign flipped