# Zeros of cubic functions

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• Nov 16th 2006, 08:17 AM
alexia1huff
Zeros of cubic functions
>Cubic function f(x)=2x^3 + 6x^2 - 4.5x -13.5.
>1) Find the roots and confirm them by remainder theorem.
>2) Taking two roots at a time, find the equations of the tangent
>lines to the average of two of the three roots?
>3) Find where the tangent lines at the average of the two roots
>intersect the curve again.
>4) State a conjecture concerning the roots of the cubic and tangent
>lines at the average value of the roots. Proove it and
>investigate: one root, two roots and, one real and two complex roots.

I found the roots from the graph, (1.5,0), (-3,0) and (-1.5,0) but could some one tell me in steps how to factorise it to find the roots in the form
a(x-b)(x-c)(x-d). Or any other way of finding the roots.
When I'm supposed to proove it by remainder thorem, do I divide the function by one at a time, roots?
Then when I'm suppossed to find the tangent and I guess you take
((x-b)+(x-c))/2 for the average of roots, and i think you're suppposed to use differentiation to find it. Differentiate the function and then put the average of the roots in it, and if so what do you do with the number you get from this calculation? In what form should the tangent be written?
I'm also suppossed to see some pattern in my calculations (of 3. i guess)
For 3. am i supposed to look at each pair of roots, because I guess that it won't be all the pair that will intersect the curve again.
• Nov 16th 2006, 12:15 PM
Soroban
Hello, Alexia!

Here's part (1) . . .

Quote:

Cubic function: .$\displaystyle f(x)\:=\:2x^3 + 6x^2 - \frac{9}{2}x -\frac{27}{2}$

1) Find the roots and confirm them by remainder theorem.

We want to solve: .$\displaystyle 2x^3 + 6x^2 - \frac{9}{2}x - \frac{27}{2}\;=\;0$

It is easier to factor without those fractions.

Factor out $\displaystyle \frac{1}{2}\!:\;\;\frac{1}{2}\left[4x^3 + 12x^2 - 9x - 27\right] \;= \;0$

Factor "by grouping": .$\displaystyle \frac{1}{2}\left[4x^2(x + 3) - 9(x + 3)\right] \;= \;0$

Take out the common factor: .$\displaystyle \frac{1}{2}(x + 3)(4x^2 - 9) \;= \;0$

Difference of squares: .$\displaystyle \frac{1}{2}(x + 3)(2x - 3)(2x + 3) \;= \;0$

Set each factor equal to zero and solve:

. . $\displaystyle \begin{array}{ccccc}x + 3\:=\:0 \\ \\ 2x-3\:=\:0 \\ \\ 2x+3\:=\:0\end{array} \begin{array}{ccc}\Rightarrow \\ \\ \Rightarrow \\ \\ \Rightarrow\end{array} \boxed{\begin{array}{ccc} x\,=\,\text{-}3 \\ \\ x\,=\,\frac{3}{2} \\ \\ x\,=\,\text{-}\frac{3}{2}\end{array}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

If $\displaystyle a$ is a root of $\displaystyle f(x)$, then $\displaystyle f(a) = 0.$

$\displaystyle x = -3:\;\;f(-3) \;= \;2(-3)^3 + 6(-3)^2 - \frac{9}{2}(-3) - \frac{27}{2}(-3)$

. . . . . . . . . . . .$\displaystyle = \;-54 + 54 + \frac{27}{2} - \frac{27}{2} \;= \;0$ . . . check!

$\displaystyle x = \frac{3}{2}:\;\;f\left(\frac{3}{2}\right) \;= \;2\left(\frac{3}{2}\right)^3 + 6\left(\frac{3}{2}\right)^2 - \frac{9}{2}\left(\frac{3}{2}\right) - \frac{27}{2}$

. . . . . . . . . . . .$\displaystyle = \;\frac{27}{4} + \frac{27}{2} - \frac{27}{4} - \frac{27}{2} \;= \;0$ . . . check!

$\displaystyle x = \text{-}\frac{3}{2}:\;\;f\left(\text{-}\frac{3}{2}\right)\;=\;2\left(\text{-}\frac{3}{2}\right)^3 + 6\left(\text{-}\frac{3}{2}\right)^2 - \frac{9}{2}\left(\text{-}\frac{3}{2}\right) - \frac{27}{2}$

. . . . . . . . . . . . $\displaystyle = \;-\frac{27}{4} + \frac{27}{2} + \frac{27}{4} - \frac{27}{2} \;= \;0$ . . . check!

• Nov 16th 2006, 01:16 PM
alexia1huff
thanks so far
Thanks for the so far answer, it was great and very well detailed, exactly what i needed and what i didn't understand.
However the other parts is still hard to understand for me,
what do you use to find the tangents, the intersecting point and showing the roots,
i think that even just the formulae topic or brief explanation would be sufficient, at least to get me going. But if someone can give me the whole picture i would be really thankful.
• Nov 16th 2006, 11:39 PM
topsquark
Quote:

Originally Posted by alexia1huff
2) Taking two roots at a time, find the equations of the tangent lines to the average of two of the three roots?

My problem with this question is I can't make any sense out of it!

Surely you have an example that is in your book or from your class notes?

-Dan
• Nov 17th 2006, 12:13 AM
alexia1huff
There isn't any example, but what they mean with the average of roots is that you take two of them and divide by 2, so a+b/2. But i don't know if for the tangent i need to check for all each average of roots. What is the formula for the tangent in this case and how do you use it?
• Nov 17th 2006, 12:31 AM
topsquark
Quote:

Originally Posted by alexia1huff
There isn't any example, but what they mean with the average of roots is that you take two of them and divide by 2, so a+b/2. But i don't know if for the tangent i need to check for all each average of roots. What is the formula for the tangent in this case and how do you use it?

I considered for simplicity the roots $\displaystyle x = \pm 3/2$. The average of the roots is 0. Then I found the line tangent to the cubic function at x = 0. The graph below is the result. Perhaps what they want you to show is that the tangent line to the curve passes through one of the roots?

-Dan
• Nov 17th 2006, 12:57 AM
alexia1huff
well, i still don't know what is the equation of the tangent or how to show were it intersects the curve again.
• Nov 17th 2006, 04:33 AM
earboth
Quote:

Originally Posted by alexia1huff
well, i still don't know what is the equation of the tangent or how to show were it intersects the curve again.

Hello, Alexia,

If (and only if) I understand your problem right, thenn you should do the following stepes:

1. Calculate the roots (see post of Soroban). Result: x1 = -3;x2 = -3/2; x3 = 3/2

2. Calculate the average values between each two roots. Result:
$\displaystyle x_A=\frac{-3-\frac{3}{2}}{2}=-\frac{9}{4}$
$\displaystyle x_B=\frac{\frac{3}{2}-\frac{3}{2}}{2}=0$

3. Calculate the y-value corresponding with xA or xB to get the points A or B, where the tangents touch the curve:
$\displaystyle f(x_A)=\frac{135}{32}$
$\displaystyle f(x_B)=-\frac{27}{2}$

4. To calculate the equation of the two tangents you need the slope of the line. The slope is equal to the gradient of f in xA or xB. First you need the first derivative of f:
$\displaystyle \frac{df}{dx}=f'(x)=6x^2+12x-\frac{9}{2}$

Now calculate the slope sA or sB of the tangents:
$\displaystyle s_A=f'(x_A)=-\frac{9}{8}$
$\displaystyle s_B=f'(x_B)=-\frac{9}{2}$

5. You now have two points and the corresponding slopes. Use the point-slope-formula of a straight line to get the equations of the tangents tA or tB:

$\displaystyle t_A:\frac{y-\frac{135}{32}}{x+\frac{9}{4}}=-\frac{9}{8}$
$\displaystyle t_B:\frac{y+\frac{27}{2}}{x-0}=-\frac{9}{2}$

$\displaystyle t_A:y=-\frac{9}{8}x+\frac{27}{16}$
$\displaystyle t_B:y=-\frac{9}{2}x-\frac{27}{2}$

6. I've attached a diagram to demonstrate what I've calculated so far. Pay attetion please: The scales on the axes are not equal!

7. You can see that the tangents intercept the curve again, that means you have to calculate the intercepts:
Intercept between tB and f is C;
intercept between tA and f is F;

8. To calculate an intercept you set equal the terms of f and the tangent:
$\displaystyle 2x^3+6x^2-\frac{9}{2}x-\frac{27}{2}=-\frac{9}{2}x-\frac{27}{2}$
$\displaystyle 2x^3+6x^2 = 0\ \Longleftrightarrow\ x=-3 \ \vee \ x=0$
the intercept at x = 0 is the point where the tangent touches the curve. Thus the additional intercept is C(-3,0)

$\displaystyle 2x^3+6x^2-\frac{9}{2}x-\frac{27}{2}=-\frac{9}{8}x+\frac{27}{16}$
$\displaystyle 2x^3+6x^2-\frac{27}{8}x-\frac{243}{32}=0$

I don't know if you know polynomial division(?). This equation has 2 known solutions: x = -9/4 or x = -9/4. Thus you can divide the LHS of the equation by $\displaystyle \left(x+\frac{9}{4} \right)^2$. The remainder contains the missing solution. You'll get: x = 3/2. Thus the additional intercept is F(3/2,0)

EB
• Nov 17th 2006, 05:28 AM
topsquark
This is actually a kinda cool theorem. I had never known this before. :)

-Dan
• Nov 17th 2006, 08:25 AM
earboth
Quote:

Originally Posted by topsquark
This is actually a kinda cool theorem. I had never known this before. :)

-Dan

Hello topsquark,

thanks for this kind remark.

Obviously I havn't expressed the following property very clearly:

If you have a curve and a straight line there are possible the following situations:
1. The curve and the straight line have no point in common so the line is a passante. If you try to calculate the intercept you will not get a real solution.

2. The curve and the straight line have two points in common so the line is a sekante. If you calculate the intercept you will get 2 real solutions.

3. The curve and the straight line have exactly one point in common so the line is a tangent. This is a special case of nr.2 so if you calculate the intercept=touching point(?) you will get 2 real solutions but they are equal.

By the way: If the tangent is the x-axis you have found a critical point if you get a "double zero".

EB
• Nov 17th 2006, 09:04 AM
alexia1huff
This is really great!!!! :D
But :p ;) , In the last post where you show the roots, i don't exactly understand what you mean by the last one.
Anyway I'm really impressed and thanful for your help.
• Nov 18th 2006, 10:58 AM
alexia1huff
Thanks to you I got very far, but i'm stuck on the last part about investigating roots, how can i investigate them, is there a formula?
• Nov 18th 2006, 01:00 PM
topsquark
Quote:

Originally Posted by alexia1huff
Thanks to you I got very far, but i'm stuck on the last part about investigating roots, how can i investigate them, is there a formula?

Well, now that you see that tangent at the correct point on the curve will intersect the remaining real root of a cubic. So what you need to do now is generalize this. Make up some cubics with real and imaginary roots and apply the method. (Notice that the average value of two imaginary roots of a cubic will be a real number since the two imaginary roots have to be complex conjugates.)

If you've forgotten how to construct a cubic equation with known roots, recall that
$\displaystyle y = (x - r_1)(x - r_2)(x - r_c)$
is the desired cubic with zeros at $\displaystyle r_1, r_2, r_3$ and if you have a root $\displaystyle a+ib$ then you must include another root $\displaystyle a-ib$.

-Dan
• Nov 19th 2006, 10:10 AM
alexia1huff
I still don't really know how am i to investigate the different roots.

For one root i took (x-1)[cubed] (as you said) and then i found the average root, which is 1 and then the tangent is 0!!! So I don't exactly understand what it means. I guess i'm suppossed to see if my conjecture is true here as well.

i can't find a two roots one. :(

For one real and two complex i took x^3 - 4x^2 + x + 8
but the root that i get from it is not a whole number and i don't really know how to find the complex roots. :confused:

apart from that i have another question, my conjecture is: in cubics the tangent at the average of two roots intersects the curve at the third root. What similar cubics can i use to prove it and they should not be exactly the same, for other proofs you usually have to try functions with different powers, e.g. fractions, negative, but here you can't really do it, so do you have any ideas.
Can you think of any algebric way to prove my conjecture?
• Nov 19th 2006, 10:30 AM
putnam120