Solving Linear Equations: Expanding Brackets, Subbing Values

[TheAkuma]

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.

 

[tiny-tim]

… 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!

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

What is the whole problem?

 

[TheAkuma]

Hi TheAkuma!

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

What is the whole problem?

I honestly don't know did you just ignore the 'x'?

 

[tiny-tim]

I honestly don't know 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?

 

[TheAkuma]

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

 

[tiny-tim]

What am I smoking? sorry about that.

he he

that's ok! ​

 

[TheAkuma]

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?

 

[tiny-tim]

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.

 

[TheAkuma]

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.

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 Now I can finish off my assignment