2026-08-31T11:34:02.070959 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 1 shows the curve \(C\) with equation \(y ...

This question runs one idea through six parts: the number of times a line meets a curve is the number of real roots of the quadratic you get by eliminating \(y\), and that number is controlled by the discriminant. A tangent touches once, so the discriminant is zero; two distinct crossings need it positive. Once part (c) has produced the quadratic in \(x\), parts (d), (e) and (f) are three different questions about the same discriminant, which is why they are worth so many marks between them.

(a) Express \(x^2 - 6x + 10\) in the form \((x - p)^2 + q\) [2]

Half of \(-6\) is \(-3\), so \((x - 3)^2 = x^2 - 6x + 9\) and nine must be given back:

\[x^2 - 6x + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1 \qquad \textbf{[M1]}\] \[p = 3, \quad q = 1 \qquad \textbf{[A1]}\]

The [M1] is for the completing-the-square structure \((x - 3)^2 - 9 + 10\); the [A1] is for both constants. The requested form has a minus sign inside the bracket, so \(p = 3\) here, positive; watch the sign convention in each question because it changes between papers.

(b) Coordinates of the minimum point of \(C\) [1]

\((x - 3)^2 \ge 0\) always, and is zero only at \(x = 3\), so the smallest value of \(y\) is \(1\), reached at \(x = 3\). The minimum point is \((3,\, 1)\) [B1]. This is written down from the completed square, not found by differentiation, which is what "hence" signals. A very common slip is to give \((-3,\, 1)\): the vertex sits where the bracket vanishes, so the \(x\)-coordinate is \(+3\).

(c) Show that the intersections satisfy \(x^2 - 8x + 10 - k = 0\) [2]

The target is printed, so all the credit is in the elimination. At any common point of \(C\) and \(l\) the two \(y\)-values are equal:

\[x^2 - 6x + 10 = 2x + k \qquad \textbf{[M1]}\] \[x^2 - 6x - 2x + 10 - k = 0 \quad\Rightarrow\quad x^2 - 8x + 10 - k = 0 \qquad \textbf{[A1]}\]

The [M1] is for equating the two expressions, the [A1] for the tidy printed form. Subtracting \(2x\) gives \(-8x\), not \(-4x\), and the constant term is \(10 - k\), a single quantity that must be carried as such into the discriminant later.

(d) Find \(k\) and the coordinates of \(P\) [5]

The line is a tangent, so it meets \(C\) exactly once, so the quadratic from (c) has a repeated root, so its discriminant is zero:

\[b^2 - 4ac = 0 \qquad \textbf{[M1]}\] \[(-8)^2 - 4 \times 1 \times (10 - k) = 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k = 0 \quad\Rightarrow\quad 4k = -24 \quad\Rightarrow\quad k = -6 \qquad \textbf{[A1]}\]

The first [M1] is for knowing that tangency means zero discriminant; the second is for substituting \(a = 1\), \(b = -8\), \(c = 10 - k\) correctly. Expanding \(-4(10 - k)\) as \(-40 - 4k\) is the standard error and loses the [A1], giving \(k = 6\).

For the point of contact, put \(k = -6\) back into the quadratic:

\[x^2 - 8x + 16 = 0 \quad\Rightarrow\quad (x - 4)^2 = 0 \quad\Rightarrow\quad x = 4 \qquad \textbf{[M1]}\] \[y = 2(4) + (-6) = 2, \quad\text{so } P(4,\, 2) \qquad \textbf{[A1]}\]

The [M1] here is follow-through: a candidate with the wrong \(k\) can still earn it by solving their own quadratic and finding \(y\) from the line. The [A1] needs the correct pair. The repeated-root shortcut \(x = -\dfrac{b}{2a} = 4\) is quicker and avoids a factorising slip.

(e) Set of values of \(k\) for two distinct points [3]

Two distinct intersections means two distinct real roots, so the same discriminant must now be strictly positive:

\[64 - 4(10 - k) \gt 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k \gt 0 \quad\Rightarrow\quad 24 + 4k \gt 0 \qquad \textbf{[M1]}\] \[k \gt -6 \qquad \textbf{[A1]}\]

The first [M1] is for using \(b^2 - 4ac \gt 0\), the second for correct expansion, the [A1] for the final set. Two traps: writing \(\ge\) instead of \(\gt\) admits the tangent case, where the two points coincide, so it is wrong; and dividing an inequality by a negative number without reversing it. Notice the structural sense of the answer: \(k = -6\) is the tangent found in (d), and any larger \(k\) lifts the line clear of the touching position so that it cuts twice.

(f) Given \(k = 10\), find the two intersection points [3]

Substituting \(k = 10\) into the quadratic from (c) makes the constant term vanish:

\[x^2 - 8x + 10 - 10 = 0 \quad\Rightarrow\quad x^2 - 8x = 0 \qquad \textbf{[M1]}\] \[x(x - 8) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 8 \qquad \textbf{[A1]}\]

Using \(y = 2x + 10\) gives the points \((0,\, 10)\) and \((8,\, 26)\) [A1].

The wrong turn this part is designed to catch is dividing through by \(x\) to get \(x = 8\) only. Dividing by a quantity that may be zero destroys the root \(x = 0\), and half the answer with it; always factorise instead. The second [A1] also requires the \(y\)-values, since the question asks for coordinates.

Check: both points must lie on \(C\) as well as on \(l\). At \(x = 0\), \(y = 0 - 0 + 10 = 10\); at \(x = 8\), \(y = 64 - 48 + 10 = 26\). Both agree. Consistency with (e) is a further check: \(k = 10 \gt -6\), so two distinct points were expected, and two were found.