2026-08-31T11:34:14.153063 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL The curve \(C\) has equation \[y = \frac{2x + 3}{...

Ten marks on the rational curve \(y = \dfrac{2x + 3}{x - 1}\). The figure shows part of \(C\) with its two asymptotes drawn as broken lines, and the curve crossing the axes at \(A\) and \(B\). Everything here follows from two habits: look at where the denominator vanishes, and look at what happens for numerically large \(x\).

(a) Equations of the two asymptotes [2 marks]

A vertical asymptote occurs where the denominator is zero but the numerator is not, because the fraction then grows without bound: \[x - 1 = 0 \ \Rightarrow\ x = 1 \quad \textbf{B1}\] For the horizontal asymptote, divide numerator and denominator by \(x\): \[y = \frac{2 + \dfrac{3}{x}}{1 - \dfrac{1}{x}} \to \frac{2}{1} = 2 \quad \text{as} \ x \to \pm\infty, \quad \text{so} \ y = 2 \quad \textbf{B1}\] Both are B marks, awarded for the correct statements with no method credit available. Each answer must be an equation: "\(x = 1\)", not "1", and "\(y = 2\)", not "2". A shortcut worth knowing is that for \(\dfrac{ax+b}{cx+d}\) the horizontal asymptote is \(y = \dfrac{a}{c}\), here \(y = \dfrac{2}{1}\).

(b) The coordinates of \(A\) and \(B\) [3 marks]

A curve crosses the \(x\)-axis where \(y = 0\). A fraction is zero only when its numerator is zero: \[2x + 3 = 0 \ \Rightarrow\ x = -\tfrac{3}{2}, \qquad A\left(-\tfrac{3}{2},\, 0\right) \quad \textbf{M1 A1}\] It crosses the \(y\)-axis where \(x = 0\): \[y = \frac{2(0) + 3}{0 - 1} = \frac{3}{-1} = -3, \qquad B(0,\, -3) \quad \textbf{A1}\] The M1 is for setting the numerator to zero, which is the idea being tested; the two A marks are for the two points. Watch the sign in \(B\): the denominator at \(x = 0\) is \(-1\), so \(y\) is negative, which agrees with the figure.

(c) Show that no value of \(x\) gives \(y = 2\) [2 marks]

This part explains why \(y = 2\) is an asymptote rather than an ordinary value of the function. Suppose such an \(x\) existed and derive a contradiction: \[\frac{2x + 3}{x - 1} = 2 \ \Rightarrow\ 2x + 3 = 2(x - 1) = 2x - 2 \quad \textbf{M1}\] \[3 = -2, \ \text{which is impossible; hence no such } x \text{ exists.} \quad \textbf{A1}\] The M1 is for clearing the fraction, the A1 for reaching an impossible statement and saying so. Because the target is given, the marks are entirely for the argument; asserting "the curve never reaches its asymptote" without algebra earns nothing. Note that the \(2x\) terms cancel, which is precisely the algebraic signature of a horizontal asymptote at \(y = 2\).

(d) The set of values of \(x\) for which \(y \ge 3\) [3 marks]

Do not multiply by \((x - 1)\), whose sign is unknown. Record \(x \ne 1\), then multiply by \((x - 1)^2\), which is positive for all \(x \ne 1\) and so preserves the inequality: \[(2x + 3)(x - 1) \ge 3(x - 1)^2 \quad \textbf{M1}\] \[2x^2 + x - 3 \ge 3x^2 - 6x + 3 \ \Rightarrow\ 0 \ge x^2 - 7x + 6 \quad \textbf{A1}\] \[(x - 1)(x - 6) \le 0 \ \Rightarrow\ 1 \le x \le 6, \quad \text{and excluding } x = 1: \ 1 \lt x \le 6 \quad \textbf{A1}\] A positive quadratic is at or below the axis between its roots, which is why this part gives a single bounded interval rather than two pieces. The final A1 depends on the strict inequality at \(x = 1\), the point where the curve is undefined.

Common wrong turns on this question

• Multiplying by \((x - 1)\) in part (d), which reverses the inequality on one side of \(x = 1\) and typically produces \(x \ge 6\) or \(x \le 1\).
• Including \(x = 1\) in the final set. The curve has no value there at all.
• Setting the denominator to zero to find \(A\). That gives the asymptote, not the \(x\)-intercept.
• Giving the asymptotes as \(x = -1\) and \(y = 3\) by reading the constants off the fraction rather than solving \(x - 1 = 0\) and taking the ratio of the \(x\)-coefficients.
• In part (c), substituting \(y = 2\) and "solving" to get \(x = \) something. The point is that the equation has no solution; the contradiction must be exhibited.

How to check

Check the intercepts against the figure: \(A\) lies to the left of the origin and \(B\) below it, which matches \(\left(-\tfrac{3}{2}, 0\right)\) and \((0, -3)\). For part (d), test the ends and the middle: at \(x = 2\), \(y = \dfrac{7}{1} = 7 \ge 3\), so the interval is right; at \(x = 6\), \(y = \dfrac{15}{5} = 3\) exactly, confirming the inclusive bound; at \(x = 7\), \(y = \dfrac{17}{6} = 2.83\ldots\), which is less than 3, confirming the interval stops at 6. As \(x\) increases beyond 6 the curve is descending towards \(y = 2\) from above, which is consistent with part (c).