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
Slope of the given line = 4
slope of the req. line = 4
equation of the line
y - y1 = m(x - x1)
y + 8 = 4(x - 3)
y + 8 = 4x - 12
4x - y - 12 - 8 = 0
4x - y - 20 = 0
Answer:
42 feet
Step-by-step explanation: