cubic function

[sajoshi21]

Hi, I have a homework problem I'm totally stuck on.

It says to sketch a cubic function (third degree polynomial function) y=p(x) where p(x)>0 on the intervals (-infinity, 3) and (5,8) then determine a formula for the function

I can't find anywhere in our book that explains this and the way I tried to work it ended up with 3 lines instead of a parabola? Any help is appreciated!

 

[stapel]

...sketch a cubic function (third degree polynomial function) y=p(x) where p(x)>0 on the intervals (-infinity, 3) and (5,8) then determine a formula for the function

If the polynomial is greater than zero on the indicated intervals, where is it equal to zero? This gives you a good lead on the factors of polynomial. Multiply them together, perhaps with an appropriate "-1" thrown in somewhere, to get the "formula".

 

[JeffM]

Hi, I have a homework problem I'm totally stuck on.

It says to sketch a cubic function (third degree polynomial function) y=p(x) where p(x)>0 on the intervals (-infinity, 3) and (5,8) then determine a formula for the function

I can't find anywhere in our book that explains this and the way I tried to work it ended up with 3 lines instead of a parabola? Any help is appreciated!

OK This is a hard problem for elementary algebra. We'd best start by pointing out that the graph of a cubic function is NOT a parabola.

Start by sketching the very simple cubic y = x^3. Now try y = - x^3. They do not look anything like parabolas, do they?

Now, unfortunately, not all cubics are that simple. At a very large scale they look generally like one of the two that I asked you to sketch in the previous paragraph. But there is a small (relatively small compared to plus and minus infinity) region where the sign of the slope changes.
You can see an example of such a local region at this site: http://en.wikipedia.org/wiki/Cubic_function

You may remember that a quadratic with real coefficients function has zero, one, or two distinct real roots. A cubic with real coefficients has one, two, or three distinct real roots.

OK Enough background to help get you started. Question time.

Where is p(x) NOT positive? Where is p(x) negative? So where is p(x) zero?

Do you know the relationship between the zero product property and polynomials?

 

[mmm4444bot]

the way I tried to work it ended up with 3 lines instead of a parabola

Please show what you tried, so that we see whether you're on the right track... :cool:

(I'm not sure what "3 lines" means, above.)

 

[HallsofIvy]

Hi, I have a homework problem I'm totally stuck on.

It says to sketch a cubic function (third degree polynomial function) y=p(x) where p(x)>0 on the intervals (-infinity, 3) and (5,8) then determine a formula for the function

I can't find anywhere in our book that explains this and the way I tried to work it ended up with 3 lines instead of a parabola? Any help is appreciated!

It sounds to me like you are over-thinking this problem and making it too hard. The problem is to sketch a cubic function and I think you are trying to get a specific cubic. You can't do that, you don't have enough information.

Mark points (3, 0), (5, 0), and (8, 0) (not "lines", just those three points). If I read this correctly the function value is positive for x< 3, then negative for x between 3 and 5, positive for x between 5 and 8 and, finally, negative for x greater than 8. So draw freehand "hump" below the x axis from (3, 0) to (5, 0) and then continue that to a "hump" above the x-axis from (5, 0) to (8, 0). Continue the graph down toward negative infinity to the right of x= 8 and upward toward infinity to the left of x= 3.

 

[JeffM]

It sounds to me like you are over-thinking this problem and making it too hard. The problem is to sketch a cubic function and I think you are trying to get a specific cubic. You can't do that, you don't have enough information.

Mark points (3, 0), (5, 0), and (8, 0) (not "lines", just those three points). If I read this correctly the function value is positive for x< 3, then negative for x between 3 and 5, positive for x between 5 and 8 and, finally, negative for x greater than 8. So draw freehand "hump" below the x axis from (3, 0) to (5, 0) and then continue that to a "hump" above the x-axis from (5, 0) to (8, 0). Continue the graph down toward negative infinity to the right of x= 8 and upward toward infinity to the left of x= 3.

Halls of Ivy is correct that, with the information provided in your question, you are unable to specify a unique cubic that includes those three points. But you can specify any one of an infinite number of cubics that include those three points. "The" formula does not exist, but a formula does, and as you have written the problem, it asks for a formula, not "the" formula.

I now see what may have confused you about parabolas. Any three distinct points determine a parabola uniquely. Those same three points are included in an infinite number of cubics, quartics, quintics, etc. To determine a cubic uniquely requires four distinct points.

 

[sajoshi21]

Ok so I have p(x)=(x+3)(x-5)(x-8) then when combining I get x^3-10x^2+120 as my formula? When I try and graph this it is huge, out to 400 on the y axis, is this right? we are supposed to draw the function and submit it online so I feel like that's really big. Thanks for all the help!!

 

[HallsofIvy]

Ok so I have p(x)=(x+3)(x-5)(x-8) then when combining I get x^3-10x^2+120

No. The polynomial zeros are at 3, 5, and 8 so the polynomial is (x- 3)(x- 5)(x- 8)= x^3- 16x^2+ 21 x- 120.

as my formula? When I try and graph this it is huge, out to 400 on the y axis, is this right? we are supposed to draw the function and submit it online so I feel like that's really big. Thanks for all the help!!

If you don't like that, then multiply by something like, say, 1/400. p(x)= (1/400)(x^3- 16x+ 21x- 120) will have exactly the same zeros as x^3- 16x+ 21x- 120.

 

[lookagain]

No. The required polynomial zeros are at 3, 5, and 8 so the polynomial is (some negative coefficient multiplied by) (x- 3)(x- 5)(x- 8)= x^3 - 16x^2 + 79x - 120.
If you don't like that, then multiply by something like, say, negative 1/400. \(\displaystyle \ \ \ \) p(x)= (-1/400)(x^3 - 16x^2 + 79x - 120) will have exactly the same zeros as x^3 - 16x^2 + 79x - 120 \(\displaystyle \ \ \) or \(\displaystyle \ \ \ \) - (x^3 - 16x^2 + 79x - 120), for example.

Don't forget, a desired cubic function for the original poster necessarily has a negative lead coefficient.