Identify the oblique asymptote of f(x) = quantity x plus 4 over quantity 3 x squared plus 5 x minus 2.
2 answers:
3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x) + 3x(2) - 1(x) - 1(2) = 0
3x(x + 2) - 1(x + 2) = 0
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
+ 1 + 1 - 2 - 2
3x = 1 or x = -2
3 3 1 1
x = ¹/₃ or x = -2
f(x) = 3x² + 5x - 2
f(¹/₃) = 3(¹/₃)² + 5(¹/₃) - 2
f(¹/₃) = 3(¹/₉) + 1²/₃ - 2
f(¹/₃) = ¹/₃ - ¹/₃
f(¹/₃) = 0
(x, f(x)) = (¹/₃, 0)
f(x) = 3x² + 5x - 2
f(-2) = 3(-2)² + 5(-2) - 2
f(-2) = 3(4) - 10 - 2
f(-2) = 12 - 12
f(-2) = 0
(x, f(x)) = (-2, 0)
Vertical Asymptotes: ¹/₃ or -2
Horizontal Asymptotes: 0
Oblique Asymptote: No Asymptotes
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Hi there!
We can calculate dy/dx using implicit differentiation:
xy + y² = 6
Differentiate both sides. Remember to use the Product Rule for the "xy" term:
(1)y + x(dy/dx) + 2y(dy/dx) = 0
Move y to the opposite side:
x(dy/dx) + 2y(dy/dx) = -y
Factor out dy/dx:
dy/dx(x + 2y) = -y
Divide both sides by x + 2y:
dy/dx = -y/x + 2y
We need both x and y to find dy/dx, so plug in the given value of x into the original equation:
-1(y) + y² = 6
-y + y² = 6
y² - y - 6 = 0
(y - 3)(y + 2) = 0
Thus, y = -2 and 3.
We can calculate dy/dx at each point:
At y = -2: dy/dx = -(-2) / -1+ 2(-2) = -2/5.
At y = 3: dy/dx = -(3) / -1 + 2(3) = -3/5.