Diketahui kurva y = √x+1 dengan koordinat 2 (y) .Tentukan persamaan garis singgung

jawab

y = √(x+1)
ordinat 2 → y = 2
√(x+1) = 2
x+1 = 4
x = 3
titik (x1,y1) = (3, 2)

y = √(x+1) = (x + 1)^(1/2)
m  = y' = 1/2 (x+1)^(-1/2)
m = 1/(2 √(x+1))
x = 3 → m = 1/ (2 √4) = 1/4

y - y1  = m (x - x1)
y - 2 = 1/4 (x - 4)
4(y-2) =1 (x - 4)
4y - 8 = x - 4
4y = x - 4 + 8
4y = x + 4
x - 4y = - 4
x - 4y + 4= 0