ksaimouli
find intersection in terms of y of y^2-4x=4and 4x-y=16
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ksaimouli
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\[y^2-4x=4 and 4x-y=16\]
ksaimouli
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@ZeHanz
Hero
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Hint:
y = 4x - 16
ksaimouli
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i did but i dont know whre i was wrong
ksaimouli
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i know dude to solve for x and set it eual and then solve for x
ksaimouli
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|dw:1357491125167:dw|
ksaimouli
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|dw:1357491191134:dw|
Hero
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Don't make it complicated bro
ksaimouli
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|dw:1357491240591:dw|
ksaimouli
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solve for y am i right this is to find area between curve sooo i must to with respect y
Hero
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Actually, that's a good approach after all.
ksaimouli
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but something is wrong in solving hmm i am not getting the answer
ksaimouli
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multiply both sides by 4
Hero
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What did you get for the answer?
ksaimouli
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.08 and -12.08
Hero
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That's what you got for y values?
ksaimouli
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nope
ksaimouli
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i gote it 5 and -4
Hero
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Okay, so what's the problem?
ksaimouli
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arithmetic mistake lol
ksaimouli
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now i have to integrate thats it
ksaimouli
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|dw:1357491558149:dw|
ksaimouli
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but not sure which one goes top and which goes bottom with respect to y
Hero
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Here's another way to do it to avoid confusion:
\[\frac{y^2 - 4}{4} - \frac{16 + y}{4} = 0\]
ksaimouli
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ya i tried that worked well
Hero
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Bro, a is always the lowest value. b is always the highest value.
ksaimouli
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i mean the function not the points (curves)
ksaimouli
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\[x=\frac{ y^2-4 }{ 4 }\]
Hero
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Are you sure you're supposed to be using y values for a and b?
ksaimouli
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or\[x=\frac{ 16+y }{ 4 }\]
ksaimouli
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ya i am damn sure
Hero
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I remember these. They can be tricky. What you do is just turn your graph sideways. And it will be the same as like when you're working with x-axis.
ksaimouli
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if \[\sqrt{y}\]
ksaimouli
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|dw:1357491948129:dw|
ksaimouli
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?
ksaimouli
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alright thx
Hero
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Bro, I graphed it as
\[\frac{16 + x}{4}\] and \[\frac{x^2 - 4}{4}\]
You get the same area when integrating from a = -4 to b = -5
The only difference is you're using variable y instead of variable x
Hero
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@ksaimouli
ksaimouli
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ya i noticed it thx