In Exercise,find the horizontal asymptote of the graph of the function. f(x) = 8x^3+2/2x^3+x
1 answer:
Answer:
Horizontal asymptote of the graph of the function f(x) = (8x^3+2)/(2x^3+x) is at y=4
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 at y=4. There is also a <em>vertical </em>asymptote at x=0
When denominator's degree (3) is the same as the nominator's degree (3) then the horizontal asymptote is at (numerator's leading coefficient (8) divided by denominator's lading coefficient (2))
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Answer:
Step-by-step explanation:
You need to pay very close attention to the triangle similarity statement. This says that triangle NML is similar to triangle NVU. But if you look at the way that triangle NVU is oriented in its appearance, it's laying on its side. We need to set it upright so that angle N is the vertex angle, angle V is the base angle on the left, and angle U is the base angle on the right. When we do that we see that sides NV and NM are corresponding and exist in a ratio to one another; likewise with sides VU and ML. Setting up the proportion:
Filling in:
Cross multiply to get
324 = 108x
and x = 3