persamaan garis singgung kurva y = x^2 – 2x – 3 yang tegak lurus dengan garis x + 2y – 3 = 0 adalah…

Y = x² – 2x – 3
y' = m = 2x – 2
x + 2y – 3 = 0
2y = -x + 3
y = -(1/2)x + (3/2) → m = -1/2
tgk lurus → m1 = -1/m = -1/(-1/2) = 2
substitusi pd m
2 = 2x – 2 → 2x = 4
x = 2 → substitusi pd y
y = 2² – 2(2) – 3 = -3
titik (2, -3)
y – y1 = m1(x – x1)
y + 3 = 2(x – 2)
y = 2x – 7 ← PGS