Author Topic: How to do this. (Read 2284 times) Tweet Share

onlyfknhuman · « **on:** November 09, 2008, 11:09:15 am »

[IMG]http://img133.imageshack.us/img133/262/91156617ga5.jpg[/img]
[IMG]http://img133.imageshack.us/img133/91156617ga5.jpg/1/w640.png[/img]

It looks so simple but wth i cant do it, since its jsut tuching at 0 ' obviously x^2 ... intercept at c and b, so (x-c) and (x+b)

So why is it E. Not D.

Ken · « **Reply #1 on:** November 09, 2008, 11:10:26 am »

b is positive, it's not negative, therefore it's E

bec · « **Reply #2 on:** November 09, 2008, 11:13:02 am »

You mean, b is negative, not positive?

Yeah I always get these questions wrong, they seem to go against logic but then if you read the graph and the questions really carefully (or sub in values), it makes sense that the answer is E.

Ken · « **Reply #3 on:** November 09, 2008, 11:16:29 am »

no b is actually positive, if they put in "-b" (in the graph) then D would be correct but in this case "b" is positive so it must be E.

bturville · « **Reply #4 on:** November 09, 2008, 11:18:12 am »

The B "includes" the negative sign, if you want to think of it that way.

danieltennis · « **Reply #5 on:** November 09, 2008, 11:19:29 am »

All you do is re-arrange the negative sign.

$h(x)= -ax^2(x-b)(x-c)$
To $h(x)= ax^2(b-x)(x-c)$

onlyfknhuman · « **Reply #6 on:** November 09, 2008, 11:20:23 am »

oH the b includes teh neg sign

bec · « **Reply #7 on:** November 09, 2008, 11:22:07 am »

Yep we're saying the same thing I think, just in...opposite...ways haha

I meant: "b" is at the point where x=-ve

So, for example, if b is at x=-4, and c is at the point where x=2, the equation would be
$h(x) = -ax^2(x+4)(x-2)$
Substituting b and c back in..

$h(x) = -ax^2 (x-b)(x-c)$

onlyfknhuman · « **Reply #8 on:** November 09, 2008, 11:23:29 am »

Quote from: danieltennis on November 09, 2008, 11:19:29 am

All you do is re-arrange the negative sign.

$h(x)= -ax^2(x-b)(x-c)$
To $h(x)= ax^2(b-x)(x-c)$

that doesnt look like option E at all lol.

onlyfknhuman · « **Reply #9 on:** November 09, 2008, 11:24:41 am »

Quote from: bec on November 09, 2008, 11:22:07 am

Yep we're saying the same thing I think, just in...opposite...ways haha

I meant: "b" is at the point where x=-ve

So, for example, if b is at x=-4, and c is at the point where x=2, the equation would be
$h(x) = -ax^2(x+4)(x-2)$
Substituting b and c back in..

$h(x) = -ax^2 (x-b)(x-c)$

i see, man this question was shifty.

bec · « **Reply #10 on:** November 09, 2008, 11:25:39 am »

Quote from: onlyfknhuman on November 09, 2008, 11:23:29 am

Quote from: danieltennis on November 09, 2008, 11:19:29 am
All you do is re-arrange the negative sign.

$h(x)= -ax^2(x-b)(x-c)$
To $h(x)= ax^2(b-x)(x-c)$

that doesnt look like option E at all lol.

Yeah it's the same thing...just expressed differently

danieltennis · « **Reply #11 on:** November 09, 2008, 11:26:01 am »

Quote from: onlyfknhuman on November 09, 2008, 11:23:29 am

Quote from: danieltennis on November 09, 2008, 11:19:29 am
All you do is re-arrange the negative sign.

$h(x)= -ax^2(x-b)(x-c)$
To $h(x)= ax^2(b-x)(x-c)$

that doesnt look like option E at all lol.

All I'm sayin is you can rearrange the signs around to prove whether its correct or not.

Mao · « **Reply #12 on:** November 09, 2008, 11:31:22 am »

stating that $b=-|b|$ at the start would help.

$\implies x-b = x+|b|$

onlyfknhuman · « **Reply #13 on:** November 09, 2008, 11:41:21 am »

Thanks guys, what about this? fck.

$y = (x-3)^{2/3} + 1$
$x = (y-3)^{2/3} + 1$
$(x-1)^{3} = (y-3)^{6/3}$
$(x-1)^{3/2} = (y-3)$
$(x-1)^{3/2} + 3 = y$

Answer says its $y = 3 - (x-1)^{3/2}$

Mao · « **Reply #14 on:** November 09, 2008, 11:49:05 am »

was there a domain restriction?

otherwise, it should be $y^{-1}=\pm (x-1)^{3/2} + 3$ because you have taken the sqrt on both sides.

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Author Topic: How to do this. (Read 2284 times) Tweet Share

onlyfknhuman

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Ken

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bec

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Ken

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danieltennis

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onlyfknhuman

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onlyfknhuman

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onlyfknhuman

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bec

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onlyfknhuman

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