A-Level Equations and inequalities: worked solution

Question

On one set of axes, sketch y = (x+1)/(2x+3) and y = x^2 + x − 3, marking clearly the equation(s) of every asymptote and every intersection point. Use your sketch to solve the inequality (x+1)/(2x+3) ≥ x^2 + x − 3.

Step 1 - work out the features of the curve y = (x + 1)/(2x + 3) before sketching it. Vertical asymptote: the denominator is zero when 2x + 3 = 0, so x = -3/2. Horizontal asymptote: divide top and bottom by x to get (1 + 1/x)/(2 + 3/x). As x -> plus or minus infinity the 1/x and 3/x terms vanish, leaving y -> 1/2, so y = 1/2. Intercepts: y = 0 when x + 1 = 0, i.e. x = -1; and x = 0 gives y = 1/3. Sketch the two separate branches, one either side of x = -3/2, each approaching y = 1/2. Step 2 - work out the features of the parabola y = x² + x - 3. The x² coefficient is positive, so it is a U shape. Complete the square: x² + x - 3 = (x + 1/2)² - 1/4 - 3 = (x + 1/2)² - 13/4, so the minimum point is (-0.5, -3.25). It cuts the y-axis at (0, -3). Draw it on the SAME axes. Step 3 - find where the two graphs meet by solving them simultaneously. (x + 1)/(2x + 3) = x² + x - 3 Multiply both sides by (2x + 3): x + 1 = (x² + x - 3)(2x + 3) Expand the right-hand side term by term: (x²)(2x + 3) = 2x³ + 3x² (x)(2x + 3) = 2x² + 3x (-3)(2x + 3) = -6x - 9 Total: 2x³ + 5x² - 3x - 9 Bring everything to one side: 2x³ + 5x² - 3x - 9 - x - 1 = 0 2x³ + 5x² - 4x - 10 = 0 Factorise by grouping the first two terms and the last two terms: x²(2x + 5) - 2(2x + 5) = 0 (2x + 5)(x² - 2) = 0 So 2x + 5 = 0 giving x = -5/2, or x² = 2 giving x = -sqrt2 = -1.41 and x = sqrt2 = 1.41. Now find each y by substituting into y = (x + 1)/(2x + 3): x = -5/2: y = (-3/2)/(-2) = 3/4, so the point is (-2.5, 0.75) x = -sqrt2: y = (1 - sqrt2)/(3 - 2sqrt2) = -1 - sqrt2, so the point is (-1.41, -2.41) x = sqrt2: y = (1 + sqrt2)/(3 + 2sqrt2) = sqrt2 - 1, so the point is (1.41, 0.414) Mark all THREE intersection points on the sketch. Step 4 - read the inequality off the sketch. (x + 1)/(2x + 3) >= x² + x - 3 asks for the x values where the rational curve lies ON or ABOVE the parabola. The three intersections and the vertical asymptote cut the x-axis into five regions. Test one value in each: x < -2.5, test x = -3: curve = (-2)/(-3) = 0.67, parabola = 9 - 3 - 3 = 3. Curve is below. NOT included. -2.5 < x < -1.5, test x = -2: curve = (-1)/(-1) = 1, parabola = 4 - 2 - 3 = -1. Curve is above. INCLUDED. -1.5 < x < -1.41, test x = -1.45: curve = (-0.45)/(0.1) = -4.5, parabola = 2.10 - 1.45 - 3 = -2.35. Curve is below. NOT included. -1.41 < x < 1.41, test x = 0: curve = 1/3, parabola = -3. Curve is above. INCLUDED. x > 1.41, test x = 2: curve = 3/7 = 0.43, parabola = 4 + 2 - 3 = 3. Curve is below. NOT included. Step 5 - decide the endpoints. At the three intersection points the two sides are EQUAL, and the inequality allows equality, so x = -2.5, x = -sqrt2 and x = sqrt2 are all included. At x = -3/2 the curve is undefined (division by zero), so x = -3/2 must be EXCLUDED, with a strict inequality there. Solution: -5/2 <= x < -3/2 or -sqrt2 <= x <= sqrt2 i.e. -2.5 <= x < -1.5 or -1.41 <= x <= 1.41