حل السؤال 20، 4MA1/1H يناير 2020: صورة إكمال المربع ورأس المنحنى

1. 1y = 4(x − 1)² − a
2. 2Find turning point: minimum when (x − 1)² = 0 => x = 1
3. 3Substitute x = 1: y = 4(0)² − a = −a => turning point (1, −a)
4. 4Find y-intercept: set x = 0 => y = 4(0 − 1)² − a = 4(1) − a = 4 − a
5. 5So y-intercept is (0, 4 − a)
6. 6Find x-intercepts: set y = 0 => 4(x − 1)² − a = 0
7. 74(x − 1)² = a
8. 8(x − 1)² = a/4
9. 9x − 1 = ±(√a)/2
10. 10x = 1 ± (√a)/2
11. 11So x-intercepts are (1 + (√a)/2, 0) and (1 − (√a)/2, 0)
12. 12Since a > 4, √a > 2, so 1 − (√a)/2 < 0 => one intercept on negative x-axis, one on positive
13. 13Minimum at (1, −a), and a > 4 => y-coordinate of turning point < −4 => in fourth quadrant
14. 14Sketch parabola opening upwards, vertex in fourth quadrant, crossing y-axis at (0, 4 − a), x-axis at both points found

AnswerSee graph with turning point at (1, −a), x-axis intercepts at (1 + (√a)/2, 0) and (1 − (√a)/2, 0), y-axis intercept at (0, 4 − a)