If f(1) = 160 and f(n + 1) = –2f(n), what is f(4)?
2 answers:
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Answer:
and as
Step-by-step explanation:
Given
-- Missing from the question
Required
The behavior of the function around its vertical asymptote at
Expand the numerator
Factorize
Factor out x + 1
We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)
We are only interested in the sign of the result
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As x approaches -2 implies that:
Say x = -3
We have a negative value (-12); This will be called negative infinity
This implies that as x approaches -2, p(x) approaches negative infinity
Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)
As x leaves -2 implies that:
Say x = -2.1
We have a negative value (-56.1); This will be called negative infinity
This implies that as x leaves -2, p(x) approaches negative infinity
So, the behavior is:
and as
Y=2-x^2
Vertex ( 0, 2 )
Focus : ( 0, 7/4 )
Axis of symmetry : x = 0
Directrix : y = 9/4
Vertex ( 0, 2 )
Focus : ( 0, 7/4 )
Axis of symmetry : x = 0
Directrix : y = 9/4