garis yang melalui P (2,11) dan memotong kurva y = -x²-2x+3 di 2 titik, gradiennya berada dalam interval
Jawaban Terkonfirmasi
Y – b = m(x – a)
y – 11 = m(x – 2)
y – 11 = mx – 2m
y = mx – 2m + 11
mx – 2m + 11 = -x² – 2x + 3
-x² – 2x – mx + 3 + 2m – 11 = 0
-x² – (2 + m)x – (8 – 2m) = 0
x² + (2 + m)x + (8 – 2m) = 0
a = 1
b = 2 + m
c = 8 – 2m
b² – 4ac > 0
(2 + m)² – 4(1)(8 – 2m) > 0
(m² + 4m + 4) – 4(8 – 2m) > 0
m² + 4m + 8m + 4 – 32 > 0
m² + 12m – 28 > 0
(m – 2)(m + 14) > 0
m < -14 atau m > 2
= (-∞, -14) U (2, ∞)