A-Level Graphs and their transformations: worked solution

Question

A function has f(x) = x + 4 for -4 <= x < -1 and f(x) = x^2 + 2 for -1 <= x < 2, with f(x) = f(x + 6) for all real x. Sketch y = f(x) over -6 <= x < 6.

The two given rules cover -4 <= x < 2, an interval of length 6, and the condition f(x) = f(x + 6) says the graph repeats every 6 units. So draw that one period first, then copy it 6 units left and 6 units right to fill -6 <= x < 6. One period, -4 <= x < 2: On -4 <= x < -1 the graph is the straight line y = x + 4, running from (-4, 0) up to (-1, 3). On -1 <= x < 2 the graph is the parabola y = x² + 2. At x = -1 it gives (-1)² + 2 = 3, the same value as the line, so the two pieces join with no break at x = -1. It falls to its minimum at (0, 2) and rises to an open endpoint at (2, 6), since x = 2 is not included. There IS a break at the ends of the period: f(2) = f(2 - 6) = f(-4) = 0, so at x = 2 the graph jumps from 6 straight down to 0, and the same jump happens at x = -4. Now repeat the period: For -6 <= x < -4: f(x) = f(x + 6) = (x + 6)² + 2, a parabola with minimum at (-6, 2) rising to an open endpoint at (-4, 6). For -4 <= x < -1: the line y = x + 4 from the closed point (-4, 0) to (-1, 3). For -1 <= x < 2: the parabola y = x² + 2 through (-1, 3), minimum (0, 2), up to the open endpoint (2, 6). For 2 <= x < 5: f(x) = f(x - 6) = (x - 6) + 4 = x - 2, the line from the closed point (2, 0) to (5, 3). For 5 <= x < 6: f(x) = f(x - 6) = (x - 6)² + 2, joining (5, 3) and falling towards (6, 2) at the right-hand edge. On the sketch mark: the minimum (-6, 2); the jump at x = -4, with a filled circle at (-4, 0) and an open circle at (-4, 6); the join (-1, 3); the minimum (0, 2); the jump at x = 2, with a filled circle at (2, 0) and an open circle at (2, 6); and (5, 3).