A-Level Graphs and their transformations: worked solution

Question

The curve y=(ax²+bx+c)/(x+d) has vertical asymptote x=−1 and oblique asymptote y=2x+2, with c≠2. State d and prove that a=2 and b=4.

Step 1: find d from the vertical asymptote. A vertical asymptote of y = (ax²+bx+c)/(x+d) occurs where the denominator is zero, i.e. x + d = 0, so x = −d. We are told this asymptote is x = −1, so −d = −1 and therefore d = 1. The curve is y = (ax²+bx+c)/(x+1). Step 2: to see the oblique asymptote, split the fraction by algebraic long division. Divide ax² + bx + c by x + 1: ax² ÷ x = ax; ax(x + 1) = ax² + ax; subtracting leaves (b − a)x + c. (b − a)x ÷ x = (b − a); (b − a)(x + 1) = (b − a)x + (b − a); subtracting leaves c − b + a. So y = ax + (b − a) + (c − b + a)/(x + 1). Step 3: read off the oblique asymptote. As x → ±∞ the remainder term (c − b + a)/(x + 1) → 0, so the curve approaches the line y = ax + (b − a). This is the oblique asymptote. Step 4: compare with the given asymptote y = 2x + 2. Two lines are identical only if their gradients match and their intercepts match, so gradient: a = 2 intercept: b − a = 2, and since a = 2 this gives b = 2 + 2 = 4. Hence a = 2 and b = 4, as required. (The condition c ≠ 2 matters: the remainder is c − b + a = c − 4 + 2 = c − 2, and c ≠ 2 makes it non-zero, so the curve is genuinely a curve approaching the line rather than being the line y = 2x + 2 itself.)