Solving Linear Equations: Expanding Brackets, Subbing Values

  • Context: High School 
  • Thread starter Thread starter TheAkuma
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    Linear Linear equations
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Discussion Overview

The discussion revolves around the process of expanding brackets in a cubic equation and substituting values into a polynomial form. Participants explore the implications of their substitutions and the resulting graphing issues related to the equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expanded the expression (x+4)(x-4)(x-4) to x³ - 4x² - 16x + 64 and sought confirmation on substituting these values into the equation y = ax³ + bx² + cx - d.
  • Another participant questioned whether the coefficients a, b, c, and d should be defined as constants without 'x' in them, suggesting a = 1, b = -4, c = -16, d = 64.
  • A participant expressed confusion about the inclusion of 'x' in the coefficients and whether they misunderstood the question.
  • One participant later clarified that the question was already solved, confirming the correct form of the equation as y = x³ - 4x² - 16x - 64.
  • Concerns were raised about graphing the cubic equation, with one participant noting that their graphing calculator displayed a blank graph and their manual graphing efforts resulted in the curve being positioned incorrectly.
  • Another participant described the behavior of the cubic function, noting its intersections with the x-axis and suggesting that the graph might be off due to scaling issues.

Areas of Agreement / Disagreement

Participants generally agree on the form of the expanded equation, but there is uncertainty regarding the graphing of the function and the appropriate scaling for visualization. The discussion remains unresolved regarding the specific graphing issues encountered.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the coefficients and the graphing scale, which may affect the interpretation of the cubic function's behavior.

TheAkuma
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Okay, so I expanded the brackets and simplified the equation from (x+4)(x-4)(x-4) to equal
x3-4x2-16x+64 and I needed to sub those values into this equation y=ax3+bx2+cx-d where a=x3 b=-4x2 c=-16 d=64 which were the values I got from the first equation. I need confirmation If it would be y=x6[/SUP]-4x4-16x2-64
I dunno, it just doesn't seem right.
 
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TheAkuma said:
… x3-4x2-16x+64 and I needed to sub those values into this equation y=ax3+bx2+cx-d where a=x3 b=-4x2 c=-16 d=64 which were the values I got from the first equation.

Hi TheAkuma! :smile:

Are you sure you don't mean a = 1, b = -4, c = -16, d = 64? :confused:

What is the whole problem?
 
tiny-tim said:
Hi TheAkuma! :smile:

Are you sure you don't mean a = 1, b = -4, c = -16, d = 64? :confused:

What is the whole problem?

I honestly don't know:confused: did you just ignore the 'x'?
 
TheAkuma said:
I honestly don't know:confused: did you just ignore the 'x'?

Well, if x3-4x2-16x+64 is the same as y=ax3+bx2+cx-d, then a b c and d can't have any x in them.

Or am I misunderstanding the question?
 
tiny-tim said:
Well, if x3-4x2-16x+64 is the same as y=ax3+bx2+cx-d, then a b c and d can't have any x in them.

Or am I misunderstanding the question?

What am I smoking? sorry about that. The question is already solved. I just had to expand the brackets into the form of y=ax3+bx2+cx-d which when solved is y=x3-4x2-16x-64 where 'a' in that equation would be 1 and 'b' would be '-4' and so on. Sorry about that
 
TheAkuma said:
What am I smoking? sorry about that.

he he :biggrin:

that's ok! :smile:​
 
uh oh. When I try to graph it on my calculator it comes up blank. Even though I spent one whole day trying to fix my graphics calculator. And when I try to do it manually, the parabola or whatever the hell its called is waay at the bottom of my graph. What am I doing wrong?
 
TheAkuma said:
uh oh. When I try to graph it on my calculator it comes up blank. Even though I spent one whole day trying to fix my graphics calculator. And when I try to do it manually, the parabola or whatever the hell its called is waay at the bottom of my graph. What am I doing wrong?

y = (x+4)(x-4)(x-4) is a cubic …

it goes up, down, and up …

it cuts the x-axis at x = -4, goes up to about 100, comes down and touches the x-axis at x = 4, and goes up again …

is that why it's off the graph?

Try dividing everything by 10. :smile:
 
tiny-tim said:
y = (x+4)(x-4)(x-4) is a cubic …

it goes up, down, and up …

it cuts the x-axis at x = -4, goes up to about 100, comes down and touches the x-axis at x = 4, and goes up again …

is that why it's off the graph?

Try dividing everything by 10. :smile:

Yeah, my calculator is being a retard so i made the scale the same as my mate's one which is the same model, only newer. Thanks for all your help tiny tim:cool: Now I can finish off my assignment:smile:
 

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