polynomial

[Guest]

1. Not all cubic functions have zeros. Is this statement true or false? Explain

2.r1=1,r2=1,r3=0,r4=-2:y-intercpt=0determine the equation.

 

[stapel]

1) What do you know about "end behavior" and the general shapes of all odd-degree polynomials?

2) What do the "r_i" stand for?

When you reply, please show the progress you have made on each exercise.

Thank you.

Eliz.
_____________
Edit: Ne'mind; solution posted below.

 

[pka]

We will assume that by ‘cubic function’ you mean a polynomial of degree three with real coefficients. Such a function has three roots. However these roots need not be real numbers. But complex roots occur in conjugate pairs. How does that tell you the there must be at least one real zero?

 

[soroban]

Hello, batticaloa!

Not all cubic functions have zeros. \(\displaystyle \;\)Is this statement true or false? \(\displaystyle \;\) Explain

False . . .

A function has a zero if its graph crosses the x-axis.

The graph of a cubic function: \(\displaystyle \,y\;=\;ax^3\,+\,bx^2\,+\,cx\,+\,d\) looks like this (basically):

          ***                                                   ***
       *       *                 *         *                 *       *
     *           *                                         *           *
    *             *             *     or    *             *             *
                   *           *             *           *
   *                 *       *                 *       *                 *
                        ***                       ***

Since the graph is continuous and goes to both positive infinity and negative infinity,
\(\displaystyle \;\;\)it will cross the x-axis somewhere.

It will have an x-intercept and therefore a zero.

\(\displaystyle 2.\;r_1\,=\,1,\:r_2\,=\,1,\:r_3\,=\,0,\:r_4\,=\,-2,\;\sout{y\text{-intercept }= 0}\;\) ??
Determine the equation.

This problem is poorly written . . . is there a typo?
Since one of the roots is \(\displaystyle x\,=\,0\), the curve passes through the origin.
The statement "y-intercept = 0" is redundant . . . and adds nothing to the information.

Since we know the four roots of the equation,
\(\displaystyle \;\;\)we have: \(\displaystyle \,a(x\,-\,1)(x\,-\,1)(x\,-\,0)(x\,+\,2)\;=\;0\)

The equation is: \(\displaystyle \,a(x^4\,-\,3x^2\,+\,2x)\;=\;0\;\) . . . for any nonzero constant \(\displaystyle a\).

 

[Guest]

Thanks for your help.