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
You might be interested in
Answer:
Tamara's example is in fact an example that represents a linear functional relationship.
- This is because the cost of baby-sitting is linearly related to the amount of hours the nany spend with the child: the more hours the nany spends with the child, the higher the cost of baby-sitting, and this relation is constant: for every extra hour the cost increases at a constant rate of $6.5.
- If we want to represent the total cost of baby-sitting in a graph, taking the variable "y" as the total cost of baby-sitting and the variable "x" as the amount of hours the nany remains with the baby, y=5+6.5x (see the graph attached).
- The relation is linear because the cost increases proportionally with the amount of hours ($6.5 per hour).
- See table attached, were you can see the increses in total cost of baby sitting (y) when the amount of hours (x) increases.
