graphing polynomials

[Jeannie611]

f(x) = x⁴-2. Since the exponent is 4 I thought it is supposed to cross the x axis 4 times, but the answer I got only crosses it twice (around the 1.5 and -1.5). please help.

x	-3	-2	-1	0	1	2	3
y	78	18	-1	-2	-1	14	79

 

[mmm4444bot]

The degree of a polynomial tells us how many Complex roots the polynomial has.

Some Complex roots may have an imaginary part; other Complex Roots do not, and so they are Real roots.

Only Real roots show up as x-intercepts on the graph.

You had better show your work; your values are way off.

Also -- please follow the main points on THIS PAGE, when posting again.

Cheers :cool:

 

[mmm4444bot]

Actually, I don't know yet what the assigned exercise is because you did not post it.

Are you supposed to be using a graphing calculator to view the graph of y=x^4-2, followed by zooming in on x-intercepts?

Something else? What are you trying to do, exactly?

(This is why we have posting guidelines; to help prevent you from wasting your time.)

 

[JeffM]

The degree of a polynomial tells us how many complex roots (or zeroes) there are, but some of the roots may be duplicated. The number of distinct complex roots is less than or equal to the degree of the polynomial.

To rephrase it, a polynomial of degree n with real coefficients may have any number of distinct real zeroes from 0 through n.

\(\displaystyle f(x) = x^2 + 1.\) Obviously, f(x) is everywhere greater than or equal to 1 so no zeroes.

\(\displaystyle g(x) = x^2.\) Obviously, g(x) is greater than zero everywhere except x = 0, where g(x) = 0. So one zero.

\(\displaystyle h(x) = x^2 - 1.\)

If x < - 1, h(x) is positive. If x = - 1, h(x) = 0. If -1 < x < 1, h(x) is negative. If x = 1, h(x) = 0. If x > 1, h(x) is positive. So two zeroes.

 

[JeffM]

f(x) = x⁴-2.

x	-3	-2	-1		1	2	3
y	78	18	-1	-2	-1	14	79

Some errors

\(\displaystyle (-3)^4 - 2 = 81 - 2 = 79.\)

\(\displaystyle (-2)^4 - 2 = 16 - 2 = 14.\)

\(\displaystyle (-1)^4 - 2 = 1 - 2 = -1.\)

\(\displaystyle 0^4 - 2 = 0 - 2 = - 2.\)

\(\displaystyle 1^4 - 2 = 1 - 2 = - 1.\)

\(\displaystyle 2^4 - 2 = 16 - 2 = 14.\)

\(\displaystyle 3^4 - 2 = 81 - 2 = 79.\)

Now you should be able to see that if \(\displaystyle x > \sqrt[4]{2}\) the polynomial will be positive

\(\displaystyle x > \sqrt[4]{2} \implies x^4 > 2 \implies x^4 - 2 > 2 - 2 = 0.\) Does that make sense?


As mmm said, you did not tell us what the exercise requested you to do or how you were to do it so we really do not what further help we can give you unless you tell us.

 

[mmm4444bot]

What mmm meant to say is

I said what I meant to say. :cool:

A polynomial's degree is equal to the number of its Complex roots. (Said another way: the number of Complex roots is the same as the polynomial's degree.)

If the degree is four, then there are four roots.

My main point, however, is highlighted in red.

Cheers!

 

[mmm4444bot]

Hi Jeannie611:

It's a bad idea to edit your posts, after people have replied, because most people will not take the time to guess at how the original information has changed.

If you would like to alter or add to a post after people have replied, please spell-out your changes or additions in a new post to the thread.

You have yet to post the actual exercise! Were you told to graph the polynomial by hand or what? Did you read our guidelines at the link that I provided, or did you skip them?

 

[HallsofIvy]

f(x) = x⁴-2. Since the exponent is 4 I thought it is supposed to cross the x axis 4 times, but the answer I got only crosses it twice (around the 1.5 and -1.5). please help.

The graph of a polynomial of degree n crosses the x-axis at most n times- it may cross fewer. For example \(\displaystyle x^4+2\) is always larger than or equal to 2 (since \(\displaystyle x^4\) is never negative) and so its graph never crosses the x- axis.

Using "(a- b)(a+ b)= a²- b²", we can write \(\displaystyle x^4- 2= (x^2-\sqrt{2})(x^2+ \sqrt{2}\) and then using the same idea on \(\displaystyle x^2- \sqrt{2}\), write it as \(\displaystyle (x- \sqrt[4]{2})(x+ \sqrt[4]{2})(x^2+ \sqrt{2}\). That will be 0 when any of its factors is 0. \(\displaystyle x- \sqrt[4]{2}= 0\) when \(\displaystyle x= \sqrt[4]{2}\), \(\displaystyle x+ \sqrt[4]{2}= 0\) when \(\displaystyle x= -\sqrt[4]{2}\) but \(\displaystyle x^2+ \sqrt{2}\) is never 0 since it is never less than \(\displaystyle \sqrt{2}\). The graph of \(\displaystyle y= x^4- 2\) crosses the x-axis twice.

 

[mmm4444bot]

Hi Jeannie611:

I just noticed the image that you edited into your original post.

It's not working.

Please see the FAQ for information about uploading and displaying images on these boards.

Cheers :cool: