<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: Option b
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Step-by-step explanation:
Linear equations have the following form:
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Where the exponents n, m, s and h are always 0 or 1
To know which equations are nonlinear, identify among the options given, those that have exponents other than 1 or 0
Note that in option b) the exponent of the variable x is
therefore the equation is nonlinear
Finally the answer is the option b
Answer:
I hope this helps:)
Step-by-step explanation:
Answer:
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Step-by-step explanation:

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You finish the leftover.