<span>The y-intercept of is .
Of course, it is 3 less than , the y-intercept of .
Subtracting 3 does not change either the regions where the graph is increasing and decreasing, or the end behavior. It just translates the graph 3 units down.
It does not matter is the function is odd or even.
is the mirror image of stretched along the y-direction.
The y-intercept, the value of for , is</span><span>which is times the y-intercept of .</span><span>Because of the negative factor/mirror-like graph, the intervals where increases are the intervals where decreases, and vice versa.
The end behavior is similarly reversed.
If then .
If then .
If then .
The same goes for the other end, as tends to .
All of the above applies equally to any function, polynomial or not, odd, even, or neither odd not even.
Of course, if polynomial functions are understood to have a non-zero degree, never happens for a polynomial function.</span><span> </span>
Answer:
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Answer:
The area of the shaded region is 21.45cm².
Answer:
90
Step-by-step explanation:
so first do 120 divided by 4 is 30. Then multiply it by 3 to get 90.
Answer: the tuition in 2020 is $502300
Step-by-step explanation:
The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.
The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = $20500
The fee in 2018 is the 19th term of the sequence. Therefore,
T19 = $45,4120
n = 19
Therefore,
454120 = 20500 + (19 - 1) d
454120 - 20500 = 19d
18d = 433620
d = 24090
Therefore, an
equation that can be used to find the tuition y for x years after 2000 is
y = 20500 + 24090(x - 1)
Therefore, at 2020,
n = 21
y = 20500 + 24090(21 - 1)
y = 20500 + 481800
y = $502300