A polynomial function has a root of -5 with multiplicity 3, a root of 1 with multiplicity of 2, and a root of 3 with multiplicit
y 7.
If the function has a negative leading coefficient and is of even degree which statement about the graph is true?
a) function is positive on (-∞,-5)
b) fuction is negative on (-5,3)
c) function is positive on (-∞,1)
d) function is negative on (3,∞)
If the function has a negative leading coefficient and is of even degree which statement about the graph is true?
a) function is positive on (-∞,-5)
b) fuction is negative on (-5,3)
c) function is positive on (-∞,1)
d) function is negative on (3,∞)
2 answers:
Answer:
d) function is negative on (3,∞)
Step-by-step explanation:
The even degree and negative leading coefficient tell you that the function is negative as x ⇒ ±∞. (Selections A and C cannot be correct.)
The odd multiplicity tells you the function crosses the x-axis at x=-5 and x=3, so will be <em>non-negative</em> between those values. (Selection B cannot be correct.)
The function is negative on (3, ∞).
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Answer:
Let the to consecutive odd integers be 2n-1 and 2n+1
their average is
=>
=
=
The square of the average is =
One less than the square of their average = -----------------------------(1)
Now the product of the two consecutive odd numbers
(2n-1)( 2n+1) = =
------------------------------------(2)
From (1) and (2) ,
We can say that the product of two consecutive odd integers is always one less than the square of their average is true
For consecutive even integers
The product of two consecutive integers
(2n)(2n+2) = ----------------------------(3)
Whereas, the one less than the square of their average is
=
=
=
= 2n--------------------------------------------(4)
From (3) and (4) it is clear that product of two consecutive even integers is not one less than the square of their average.