A-Level Graphs and their transformations: worked solution

Question

Region R is enclosed by the curve y=3x²/[(x+1)(3x²+x+1)], the line x=4 and the x-axis. For x≥0, sketch this curve, marking any axis intercepts and asymptote equations, and shade R. [3]

Step 1 - check for vertical asymptotes. A vertical asymptote occurs where the denominator is zero. (x + 1) = 0 gives x = -1, which is outside the required range x ≥ 0. For 3x² + x + 1 the discriminant is b² - 4ac = 1² - 4(3)(1) = 1 - 12 = -11, which is negative, so this quadratic is never zero. So on x ≥ 0 the denominator is never zero and the curve has NO vertical asymptote. Step 2 - find the axis intercepts. Put x = 0: y = 3(0)²/[(0 + 1)(0 + 0 + 1)] = 0/1 = 0. The curve passes through the origin (0, 0). The numerator 3x² is zero only at x = 0, so (0, 0) is the only intercept in x ≥ 0. Step 3 - find the sign of y. For x > 0 the numerator 3x² is positive, and both (x + 1) and 3x² + x + 1 are positive, so y > 0. The curve lies above the x-axis for all x > 0. Step 4 - find the horizontal asymptote. Multiply out the denominator: (x + 1)(3x² + x + 1) = 3x³ + x² + x + 3x² + x + 1 = 3x³ + 4x² + 2x + 1. So y = 3x²/(3x³ + 4x² + 2x + 1). Divide the top and the bottom by the highest power, x³: y = (3/x) / (3 + 4/x + 2/x² + 1/x³). As x → ∞ every term 3/x, 4/x, 2/x², 1/x³ tends to 0, so y → 0/3 = 0. Horizontal asymptote: y = 0, approached from above. Step 5 - get the shape right. The curve leaves the origin going upwards, rises to a single maximum near x = 1.12 (where y ≈ 0.302), then turns and decreases, flattening off towards the x-axis. At the right-hand boundary, x = 4: y = 3(4²)/(3(4³) + 4(4²) + 2(4) + 1) = 48/(192 + 64 + 8 + 1) = 48/265 ≈ 0.181, so the curve is still just above the axis where the line x = 4 cuts it. Step 6 - the sketch and the region R. Draw the curve for x ≥ 0 with that shape. Mark the intercept (0, 0), label the horizontal asymptote y = 0, and draw the vertical line x = 4. R is bounded above by the curve, below by the x-axis and on the right by x = 4, and it closes up at the origin on the left. Shade the whole area between x = 0 and x = 4 that lies under the curve and above the x-axis.