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Question 1177496: Write an equation for a polynomial function that satisfies each set of characteristics.
extending from quadrant 2 to quadrant 4, two turning points, y-intercept of 1
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
According to the context, the plot of the polynomial should be located in the 2nd and 4th quadrants
with the y-intercept at y= 1,
with no parts in the 1st and/or 3rd quadrant.
A polynomial satisfying such conditions DOES NOT EXIST.
If you mean something different, then the problem should be formulated accordingly.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
It doesn't say "with no parts in the 1st and/or 3rd quadrant", it just
says "extending from quadrant 2 to quadrant 4". It can extend from quadrant
2 to quadrant 4 by passing through quadrant 1 on its way to quadrant 4.
It might very well look like the graph below, with two turning points,
and having zeros at, arbitrarily, -3 and 4, and going up on the extreme
right. So we'll make sure it has a positive leading term.
To have 2 turning points, it must have degree 3 or more. So let's try to
make it have 3 real zeros.
To have a zero at -3 it must have factor (x+3)
To have a zero at 4 it must have factor (x-4)
It must have another zero between -3 and 4. Let's suppose that third zero
has value a. So the polynomial must also have a factor (x-a)
It would have degree 3 if it were of the form
and hopefully "a" will turn out to be between -3 and 4.
In order for the polynomial to have y-intercept 1, it must go through (0,1).
So we substitute x=0 and p(0)=1
 <--that is between -3 and 4
So such a polynomial would be
If you multiply that out, you get:
That's what the graph above is of.
Edwin
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