A-Level Graphs and their transformations: worked solution

Question

For y = f(x) with f(x) = 1 - sqrt(q^2 - x^2), q > 1 (a semicircle), sketch y = 1/f(x), giving any asymptote equations and the coordinates of the end-points.

Step 1 - recall what taking a reciprocal does to a graph. Wherever f(x) = 0, the value 1/f(x) is undefined and the graph shoots off: those x-values give VERTICAL asymptotes. Wherever f(x) = 1 or f(x) = -1, 1/f(x) takes the same value, so the two graphs meet there. The sign is unchanged: f negative gives 1/f negative, f positive gives 1/f positive. A minimum of f at a negative value becomes a maximum of 1/f. Step 2 - find the domain, because it fixes the end-points. f(x) = 1 - √(q² - x²) needs q² - x² ≥ 0, i.e. -q ≤ x ≤ q, so the graph starts and stops at x = ±q. At those two ends, √(q² - q²) = 0, so f(±q) = 1 - 0 = 1. Therefore 1/f(±q) = 1/1 = 1. End-points: (-q, 1) and (q, 1). Step 3 - find the vertical asymptotes by solving f(x) = 0. 1 - √(q² - x²) = 0 √(q² - x²) = 1 Square both sides: q² - x² = 1 x² = q² - 1 x = ±√(q² - 1). Because q > 1 we have q² - 1 > 0, so both values are real, and since q² - 1 < q² they lie strictly between -q and q. Vertical asymptotes: x = -√(q² - 1) and x = √(q² - 1). Step 4 - find the turning point in the middle. At x = 0, √(q² - 0) = q, so f(0) = 1 - q, which is NEGATIVE because q > 1. This is the smallest (most negative) value f takes, so the reciprocal has a maximum there: 1/f(0) = 1/(1 - q). So the middle branch has its highest point at (0, 1/(1 - q)), which lies below the x-axis. Step 5 - put the sketch together (it is symmetric about the y-axis). Between the asymptotes, |x| < √(q² - 1): here f is negative, so this branch lies entirely below the x-axis. It comes up from -∞ beside x = -√(q² - 1), peaks at (0, 1/(1 - q)), then drops back to -∞ beside x = √(q² - 1). Outside the asymptotes but still inside the domain: here f is positive and never exceeds 1, so 1/f ≥ 1. Two branches, each falling steeply from +∞ next to an asymptote down to the end-point at height 1. The curve stops dead at x = -q and x = q, so mark those end-points clearly and do not draw arrows there.