A-Level Graphs and their transformations: worked solution

Question

For the recurrence u_{n+1}=(−2u_n+6)/(u_n−1), sketch y=(−2x+6)/(x−1)−x, giving the asymptote equations and the coordinates where the graph meets the axes. [3]

Method: combine into a single fraction, then divide out. The quotient gives the oblique asymptote and the zero of the denominator gives the vertical one. Step 1: Put y = (-2x + 6)/(x - 1) - x over a common denominator. y = [(-2x + 6) - x(x - 1)]/(x - 1) = (-2x + 6 - x² + x)/(x - 1) = (-x² - x + 6)/(x - 1) Step 2: Vertical asymptote where the denominator is zero. x - 1 = 0, so x = 1. Step 3: Divide -x² - x + 6 by x - 1 to find the oblique asymptote. -x² ÷ x = -x; -x(x - 1) = -x² + x; subtracting leaves -x - x + 6 = -2x + 6 -2x ÷ x = -2; -2(x - 1) = -2x + 2; subtracting leaves 6 - 2 = 4 So y = -x - 2 + 4/(x - 1). As x → ±∞ the term 4/(x - 1) → 0, so the oblique asymptote is y = -x - 2. Step 4: x-axis crossings, where y = 0, i.e. the numerator is zero. -x² - x + 6 = 0 Multiply by -1: x² + x - 6 = 0 Factorise (two numbers with product -6 and sum +1 are +3 and -2): (x + 3)(x - 2) = 0 x = -3 or x = 2, giving the points (-3, 0) and (2, 0). Step 5: y-axis crossing, where x = 0. y = (-2(0) + 6)/(0 - 1) - 0 = 6/(-1) = -6, giving the point (0, -6). Step 6: Which side of the oblique asymptote each branch sits on, from the sign of 4/(x - 1). For x < 1 it is negative, so the left branch lies BELOW y = -x - 2, and y → -∞ as x → 1⁻. For x > 1 it is positive, so the right branch lies ABOVE y = -x - 2, and y → +∞ as x → 1⁺. Sketch: two branches separated by x = 1. Left branch: comes down along y = -x - 2 from the upper left, passes through (-3, 0) and (0, -6), and falls to -∞ at x = 1⁻. Right branch: drops from +∞ at x = 1⁺, passes through (2, 0), and settles down onto y = -x - 2 from above. Asymptotes: x = 1 and y = -x - 2. Axis crossings: (-3, 0), (2, 0) and (0, -6).