# Find The Equation Of A Cubic Relationship Given This Data

1. Apr 17, 2005

### drunkenfool

1. Find the equation of a cubic relationship given that the graph has only two x intercepts, one at point (3,0) and the other at (-1,0). It is also known that the graph passes through the point (4,-10) and has a y-intercept of (0,-18)

2. Write down the exact co-ordinates of the two turning points of the equation found in question 1.

2. Apr 17, 2005

### Hurkyl

Staff Emeritus
So, what thoughts have you had on this problem?

3. Apr 17, 2005

### leon1127

first y = k(x-3)(x+1)(x-q)
q has to be 3 or -1 since the graph only touches one of these points.
and k*(-3)*1*(-q)=-18
so k = -6/q
subsitute into first equation and become
y = -6/q(x-3)(x+1)(x-q)
and it pass through the point 4, -10
then it becomes
-10 = -6/q(4-3)(4+1)(4-q)
solve for q
q=3
k = -2

2. turning point?? means inflection point or extrema? but since it said 2 turning points... i guess it means extrema. first, take first derivative and then find zero. that is your extrema.(turning point)?
x @ 1/3, 3 ???

Last edited: Apr 17, 2005
4. Apr 17, 2005

### drunkenfool

my thoughts were: y=a(x-3)(x+1)(x+k) and i was kinda stuck as..
thanks for d speedy help and uve made it understandable thanks.

so the answer to (a) is: y= -2(x-3)(x-3)(x+1)
i get it now great..

but then B? use graphical calc

Last edited: Apr 18, 2005
5. Apr 17, 2005

### drunkenfool

am i right on part b?

Last edited: Apr 17, 2005
6. Apr 17, 2005

### OlderDan

My calcultor gives the same result. If you are taking calculus, you can approach this problem by taking the dirivative of the function you found in part a), setting it equal to zero, and solving the quadratic equation for x. If you are not taking calculus, that will not mean anything to you and the calculator solution is probably what is expected of you.

7. Apr 17, 2005

### drunkenfool

wow im so good

Last edited: Apr 18, 2005
8. Apr 17, 2005

### LilSciWizGirl

Find The Equation Of A Cubic Relationship Given This Data?

Does this math Have to do with Point-Slope fomulas. Slope=Rise/Run.
Ifso,then I have the anser of how to solve it, for you. If, you want to know the anser , then just let me know.

I'm Hear to help in any way possable. "THANK YOU"

With many Smiles,
LilSciWizGirl.

9. Apr 17, 2005

### dextercioby

Nope,no point slopes here.Just some well done factoring and simple algebra.

Daniel.

10. Apr 17, 2005

### leon1127

The easier way to solve slope problem is calculus.....
but i believe he is not in calculus, and point-slope formula means nothing if we dont know the actually point and slope.

11. Apr 18, 2005

### drunkenfool

yea this isjust normal yr 11 maths..acually i think i got part b right aswell so yer thnx older dan. and thanks for all the other help esp leon1127. anyway the assignments handed in and over. thanks

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