I really need help
1 answer:
Answer:
See step by step
Step-by-step explanation:
To find if a function intervals are possible, make sure the points are above the x axis.
The graph is getting positve around x=-3 but it isn't positive yet.
The graph is positve until x=-1
The graph becomes positive again again at x=2 and is positive infinitely forever.
So let set up our interval
We are going to start at -3 since that where it start getting positive, we then going to use the greater than sign since it get positve as we go to the right as of right now. We use the greater than becuase we are not going to include -3 as our solutions.
We then use a less than sign since the answer must be less than -1 becuase around-1 it start getting negative.
It becomes positive again at x=2 so since we going to left to right use the greater than sign becuase we are not including 2.
It infinitely becomes positive so our interval is x must be cause greater than w becuase it can be any interval greater than two.
and
The roots or zeroes are the x intercept of the graph,(when y=0, what is the x value.)
The x intercepts are -3 and 2.
You might be interested in
Answer:
Step-by-step explanation:
If we have 5 terms to the right of the -1 outside the box thing, that means that our original polynomial is a 4th degree (the degree is one lower than the number of terms in order to allow for the constant which has no variable). Set up the synthetic division as shown in the problem. The rule is to bring down the first term, the 1 to the right of the box, multiply it by the number outside, -1, and put that product up under the next number in line inside the box. Like this:
-1 | 1 7 15 9 7
<u> -1 </u>
1
Now add straight down, multiply the sum by -1 and put that product up under the next term in line, the 15:
-1 | 1 7 15 9 7
<u> -1 -6 </u>
1 6
Then add straight down again, multiply the sum by -1 and put that product up under the next term in line, the 9:
-1 | 1 7 15 9 7
<u> -1 -6 -9 </u>
1 6 9
Then do the same thing. Add straight down, multiply the sum by -1 and put the product up under the next term in line, the 7:
-1 | 1 7 15 9 7
<u> -1 -6 -9 0</u>
1 6 9 0 7
This is the final result. The last number, the 7 is the remainder, but the rest of it makes up what we call the depressed polynomial and is a polynomial that is one degree less than the degree we started with. So this depressed polynomial is a 3rd degree. The numbers are the coefficients on the x terms:
or more simply stated: