Asked by Tammie
Can someone show me how to do this? I got 5 more problems of the same kind. Thannk!
Consider the polynomial P(x), shown in both standard form and factored form.
P(x) = 1/2x^4 - 9/2x^3+21/2x^2+1/2x-15
= 1/2 (x+1)(x-2)(x-3)(x-5)
State the behavior at the ends: (up/down) at the left, and (up/down) at the right?
State the y-intercept
State the x-intercept
Create a graph depicting the polynomial above.
Consider the polynomial P(x), shown in both standard form and factored form.
P(x) = 1/2x^4 - 9/2x^3+21/2x^2+1/2x-15
= 1/2 (x+1)(x-2)(x-3)(x-5)
State the behavior at the ends: (up/down) at the left, and (up/down) at the right?
State the y-intercept
State the x-intercept
Create a graph depicting the polynomial above.
Answers
Answered by
Reiny
setting each of the factors to zero gives you the x-intercepts
so the graph cuts the x-axis at -1,2,3, and 5
(a fourth degree polynomials can cut the x-axis in at most 4 places, a cubic in at most 3 places, a 5th degree in at most 5 places, etc)
If the highest term is an even power and positive the graph rises up in the first and second quadrants, if an even power but negative coefficient, it will drop down into the third and fourth quadrants.
If the highest power has an odd exponent, but the term is positive, the graph will rise in the first quadrant and drop into the third quadrant.
the y-intercept is when x=0, easy to see it must be -15
Fourth degree functions tend to look like a "W"
Can you sketch the graph ?
so the graph cuts the x-axis at -1,2,3, and 5
(a fourth degree polynomials can cut the x-axis in at most 4 places, a cubic in at most 3 places, a 5th degree in at most 5 places, etc)
If the highest term is an even power and positive the graph rises up in the first and second quadrants, if an even power but negative coefficient, it will drop down into the third and fourth quadrants.
If the highest power has an odd exponent, but the term is positive, the graph will rise in the first quadrant and drop into the third quadrant.
the y-intercept is when x=0, easy to see it must be -15
Fourth degree functions tend to look like a "W"
Can you sketch the graph ?
Answered by
Tammie
I was trying to can you prvide me with a picture so i can see where I have gone wrong? Thanks!
Answered by
Tammie
I am going to say that based on what you described the behavior of this graph is
up at the left, and down) at the right? Is this correct?
up at the left, and down) at the right? Is this correct?
Answered by
MathMate
http://i263.photobucket.com/albums/ii157/mathmate/x4.png
Answered by
Reiny
I don't have a quick way to show the graph
I will describe how I would graph it.
As stated above, it will cross the x-axis at
-1,2,3, and 5
also (0,-15) is on the y-axis
using the factored version it is quite easy to see where the graph is at x=1 and x=4
I picked these values since they are roughly half-way between x-intercepts, and give you an idea how high or low the graph is between the x-intercepts.
I had (1,-8) and (4,-5)
so make your graph come down from the third quadrants, hit (-1,0), down to (0,-15) back up through (1,-8), then (2,0), rise up a bit and then down again to (3,0), downwards to roughly (4,-5), back up through (5,0) and then rising rapidly from there on.
I will describe how I would graph it.
As stated above, it will cross the x-axis at
-1,2,3, and 5
also (0,-15) is on the y-axis
using the factored version it is quite easy to see where the graph is at x=1 and x=4
I picked these values since they are roughly half-way between x-intercepts, and give you an idea how high or low the graph is between the x-intercepts.
I had (1,-8) and (4,-5)
so make your graph come down from the third quadrants, hit (-1,0), down to (0,-15) back up through (1,-8), then (2,0), rise up a bit and then down again to (3,0), downwards to roughly (4,-5), back up through (5,0) and then rising rapidly from there on.
Answered by
Tammie
I think i got it! Thanks!
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