Solve By Factoring, and Equation of Cubic Function whose graph passes thru

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Homework Help Overview

The discussion revolves around two pre-calculus problems: one involves solving a cubic equation by factoring, and the other requires finding a cubic function that passes through specific points and is tangent to the x-axis at the origin.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the factoring of the cubic equation by rearranging terms. There is also a focus on determining the coefficients of a cubic function based on given points and tangency conditions.

Discussion Status

Some participants have provided guidance on how to approach the problems, including rearranging the equation and setting up conditions for the cubic function. Multiple interpretations of the second problem are being discussed, particularly regarding the implications of the points provided.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the type of assistance they can provide. There is an emphasis on understanding the relationships between the points and the behavior of the cubic function.

rought
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Homework Statement




I am stuck on these two pre-cal problems... can anyone help?

Solve By Factoring: 2x^3 + 2x^2 = 4x + 4

This next one I have no idea how to do

Find an equation of the cubic function whose graph passes through the points (3,0) and (1,4) and is tangent to the x-axis at the origin...
 
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rought said:

Homework Statement



Solve By Factoring: 2x^3 + 2x^2 = 4x + 4

Try putting all the terms on one side so that they all equal 0.
 
alrite I think I am ok with the first problem, but I'm still stuck on the second one...
 
What can you tell from the point (3,0)?

EDIT: And what can you say since the line passes through the origin?
 
You need to find a, b, c, and d of y= ax3+ bx2+ cx+ d so that
a) when x= 3, y= 0
b) when x= 1, y= 4
c) when x= 0, y= 0 and y'= 3ax2+ 2b2+ c= 0.

Putting those values in the equation gives you four linear equations for a,b,c, and d.
 
Alright, I think i have it...

It is x^2 because there is a tangent on the x-axis at (0,0), and the other zero is (x-3)

Through a bit of trial and error I have come to: -2x^2(x-3) and I am pretty sure that works just fine.


Thanks again everyone who helped you guys are great! :smile:
 

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