Alegra help!!

[cwells0014]

I would really appreciate if someone could help me solve these problems and the steps I need to take to solve them. I have looked at several examples of both but I still do not understand. I am studying functions and here are the two homework problems that have me stumped...


1) Graph P(x)= (5x-7)(x-4)²(x+6)² and label all real zeros exactly!


2) Find all real zeros exactly for P(x)=2x³+6x²-7x+4

Thank you so much![/tex]

 

[stapel]

1) This is already factored, so finding the zeroes is simple enough. With which part of the graphing are you having difficulty? Please be specific.

2) For this one, you'll have to factor first. The list of possible zeroes (from the Rational Roots Test) is pretty small. Are you checking the possible zeroes with synthetic division or evaluation?

Thank you.

Eliz.

 

[cwells0014]

Im not really sure where to start. For the zeros, I put each one equal to zero to find x values but what about the ones that are squared...What do I do with those? The as far as graphing goes, after I plot x values how do I know exactly how the lines behave over the graphs?

As far as your second question, I don't even knoe what the roots test is? I know you have to look for factors that go into the whole thing and then continue to pull stuff out until it is completely factored but I just don't get it. I also am not very good with using the calculator to find factors.

Thank you for taking time with me.

 

[stapel]

...what about the ones that are squared...What do I do with those?

Squaring just means the factor occurs twice. So set the factor equal to zero... twice.

...how do I know exactly how the lines behave over the graphs?

Graph enough points that you have a pretty good idea. (You can't graph "every" point, so "enough to get a good idea" is all they're looking for anyway.) So find the zeroes, the y-intercept, and a few points between the zeroes (enough that you can "see" about where the line goes). Then use your knowledge of end behavior to finish the graph.

I don't even knoe what the roots test is?

:shock:

If they haven't covered the Rational Roots Test, then I don't see how they're expecting you to solve this. What method does the book use? What method did the instructor use in class?

Thank you.

Eliz.

 

[soroban]

Hello, cwells0014!

Permit me to baby-talk through this one . . .

Graph: \(\displaystyle P(x)\:=\5x\,-\,7)(x\,-\,4)^2(x\,+\,6)^2\) and label all real zeros exactly!

You found the zero, right? . . . \(\displaystyle x\,=\,\frac{7}{5},\,4,\,-6\)

Here's the rule.

If a factor has an odd exponent, the graph passes through that x-intercept.

If a factor has an even exponent, the graph is tangent to the x-axis.


The zero \(\displaystyle x = \frac{7}{5}\) came from \(\displaystyle (5x\,-\,7)^1.\)
. . The factor had exponent 1 ("multiplicity one").
. . Hence, the graph goes through \(\displaystyle (\frac{7}{5},\,0)\)

The zero \(\displaystyle x = 4\) came from \(\displaystyle (x\,-\,4)^2.\)
. . This factor has "multiplicity two".
. . Hence, the graph is tangent to the x-axis at \(\displaystyle (4,\,0).\)

Similarly, \(\displaystyle x=-6\) came from \(\displaystyle (x\,+\,6)^2\): multiplicity two.
. . Hence, the graph is tangent to the x-axis at \(\displaystyle (-6,0).\)

Plot the zeros on a graph.

                    |
      - - * - - - - + - - * - - - * - - -
         -6         |    7/5      4

Now choose any value of \(\displaystyle x\), say, \(\displaystyle x=0.\)

\(\displaystyle P(0)\:=\0-7)(0-4)^2(0+6)^2\:=\-7)(16)(36)\:=\:\)negative

. . When \(\displaystyle x=0,\,y\) is some negative number, -\(\displaystyle b.\) . Plot that point \(\displaystyle (0,-b).\)

                    |
                    |           
      - - * - - - - + - - * - - - * - - -
         -6         |    7/5     4
                    |
                    *(-b,0)
                    |

There is a zero at \(\displaystyle (\frac{7}{5},\,0)\)
. . and we know the graph goes <u>through</u> that point.

                    |
                    |        *
                    |      *
      - - * - - - - + - - * - - - * - - -
         -6         |    *        4
                    |  *
                    *(-b,0)
                    |

We know there is a zero at \(\displaystyle (4,\,0).\)
. . The graph must come back down
. . and we know that it is <u>tangent</u> to the x-axis there.

                    |
                    |        * *      *
                    |      *    *   *   
      - - * - - - - + - - * - - - * - - -
         -6         |    *        4
                    |  *
                    *
                    |

We know there is a zero at \(\displaystyle (0,\,-6)\) at the left.
. . The graph must go up that point
. . and we know it is <u>tangent</u> to the x-axis there.

                    |
                    |        * *      *
         -6         |      *    *   *   
      - - * - - - - + - - * - - - * - - -
       *     *      |    *        4
      *         *   |  *
                    *
                    |

See? . . . We have a good idea of what the graph looks like
. . without plotting dozen (hundreds?) of points.