In Exercise,find the horizontal asymptote of the graph of the function. f(x) = 16x/3+x^2
1 answer:
Answer:
Horizontal asymptote of the graph of the function f(x) = 16x/(3+x^2) is y=0
Step-by-step explanation:
I attached the graph of the function. Graphically it can be seen that the horizontal asymptote of the graph of the function is y=0.
When denominator's degree (2) is higher than nominator's degree (1) then the horizontal asymptote is y=0
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Answer:
a. V = 4/3
b. See attachment
Step-by-step explanation:
a.
Given
Z = x² + y²
V = ∫ ∫ (x² + y²) dxdy {0,1}{0,y} + ∫ ∫ (x² + y²) dxdy {0,1}{x,2-x}
V = ∫ ∫ (x² + y²) dxdy {0,1}{x,2-x}
Integrate with respect to y
V = ∫ x²y+ y³/3 dx {0,1}{x,2-x}
V = ∫ x²(2-x) + (2-x)³/3 - x²(2) - (2)³/3 dx {0,1}
V = ∫ 2x² -7x³/3 + (2-x)³/3 dx {0,1}
V = 2x³/3 - 7x⁴/12 + (2-x)⁴/12 {0,1}
V = (⅔ - 7/4 + 2/12) - (0-0+16/12)
V = 4/3
