O-Level Functions together with their graphs

Complete the square (part a): y = x² + 4x + 7 = (x + 2)² + 3. Turning point is a minimum at (-2, 3). y-intercept: set x = 0 to get y = 7, so (0, 7). x-intercepts: discriminant = 4² - 4(1)(7) = 16 - 28 = -12 < 0, so there are no real x-intercepts; the curve lies entirely above the x-axis. Sketch: a U-shaped (positive) parabola with minimum vertex (-2, 3), passing through (0, 7), never touching the x-axis.

(i) The equation is already in the completed-square form y = (x - a)² + b, so read the sketch straight off it rather than plotting points. The coefficient of (x - 2)² is +1, which is positive, so the curve is a parabola opening UPWARDS (a U shape). A square is never negative, so (x - 2)² ≥ 0 and therefore y ≥ 3. The smallest value of y is 3, and it happens when x - 2 = 0, that is when x = 2. So the lowest point of the curve is (2, 3). Find the y-intercept by putting x = 0: y = (0 - 2)² + 3 = (-2)² + 3 = 4 + 3 = 7, so the curve passes through (0, 7). Check whether it cuts the x-axis by putting y = 0: (x - 2)² + 3 = 0 would need (x - 2)² = -3, and a square can never be negative, so there are NO x-intercepts and the curve lies entirely above the x-axis. Sketch: a U-shaped parabola whose lowest point is at (2, 3), passing through (0, 7) on the y-axis, never touching the x-axis, and symmetrical about the vertical line through its lowest point. (ii) From the completed-square form, the minimum point is (2, 3). (iii) A parabola is symmetrical about the vertical line through its vertex. The vertex is at x = 2, so the line of symmetry is x = 2.

(a) y = x(x − 2) expands to y = x² − 2x. The coefficient of x² is positive, so the curve is a parabola that opens upwards. Find where it crosses the x-axis by putting y = 0: x(x − 2) = 0 A product is zero when one factor is zero, so x = 0 or x − 2 = 0, giving x = 0 or x = 2. The curve crosses the x-axis at (0, 0) and (2, 0). Find where it crosses the y-axis by putting x = 0: y = 0 × (0 − 2) = 0, so the y-intercept is (0, 0) as well. Sketch: an upward-opening parabola cutting the x-axis at (0, 0) and (2, 0), dipping below the axis between them. (b) Complete the square to find the lowest point: y = x² − 2x Halve the coefficient of x (that is −2 ÷ 2 = −1) and use it inside the bracket, then subtract its square: y = (x − 1)² − 1 (x − 1)² is a square, so it is never negative, and it equals 0 when x = 1. The smallest possible value of y is therefore 0 − 1 = −1, reached at x = 1. Minimum value = −1 (c) y = p is a horizontal line, so the number of solutions is the number of times that line cuts the parabola in the sketch. The lowest point of the parabola is at y = −1. If p < −1 the line lies entirely below the curve and misses it: 0 points. If p = −1 the line just touches the curve at the vertex (1, −1): 1 point only. If p > −1 the line is above the vertex and cuts the curve once on the left branch and once on the right branch: exactly 2 points. So exactly two points of intersection requires p > −1.