So I need to understand how to solve this type of problem...
y=(x+2)^2-6
y-2+x
I think by...substitution? And I know you probably have to foil at some point but I have tried this problem so many times and I can't figure it out.
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OpenStudy (anonymous):
do you mean y=2+x for the second function?
OpenStudy (anonymous):
And what are you trying to solve for?
OpenStudy (supernova_sonntag):
my bad I meant y-2=x
silly shift key :)
the directions only said solve...I am assuming it meant solve the system of equations
OpenStudy (xapproachesinfinity):
So we have \(y=(x+2)^2-6 \\ y-2=x\)
OpenStudy (xapproachesinfinity):
is that what yu have?
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OpenStudy (anonymous):
I'm gonna let Mr. Mathlete take it from here
OpenStudy (supernova_sonntag):
Yes
OpenStudy (xapproachesinfinity):
let us substitute x for y-2 in equation #1
OpenStudy (xapproachesinfinity):
so to get \(y=(y-2+2)^2-6\)
OpenStudy (xapproachesinfinity):
you can see that i replaced x by y-2 right?
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OpenStudy (supernova_sonntag):
Yeah
OpenStudy (xapproachesinfinity):
ok now just simplify that more
\(y=y^2-6\)
OpenStudy (xapproachesinfinity):
-2 +2 cancel out
OpenStudy (xapproachesinfinity):
now it need more clean up
\(y^2-y-6=0\)
OpenStudy (xapproachesinfinity):
i subtract y from both sides to get that
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OpenStudy (xapproachesinfinity):
what kind of equation is that?
OpenStudy (supernova_sonntag):
Okay I see what you're doing, i'm not sure what you mean though, by what kind of equation?
OpenStudy (xapproachesinfinity):
i meant that's a quadratic equation, you should have faced it various times?
OpenStudy (supernova_sonntag):
Yes, but honestly I didn't recognize that; in my math class we don't really do much with the vocabulary for things, unfortunately.
OpenStudy (xapproachesinfinity):
vocabulary is really part of mathematics
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OpenStudy (xapproachesinfinity):
we need a language to cross our ideas!
OpenStudy (supernova_sonntag):
Haha very true.
So, is there anything more that needs to be done to the problem? Or, is that the type of answer they are looking for?
OpenStudy (xapproachesinfinity):
so back to our mean thing.
\(y^2-y-6=0 \Longrightarrow (y-3)(y+2)=0 \)
OpenStudy (xapproachesinfinity):
that is called factoring
and we are not done until we figure out what x is and y
OpenStudy (xapproachesinfinity):
now we pretty much close in finding y
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OpenStudy (supernova_sonntag):
ooooh
OpenStudy (xapproachesinfinity):
any idea what to do with \((y-3)(y+2)=0\)
OpenStudy (supernova_sonntag):
Um to solve for y you set them both equal to zero?
OpenStudy (xapproachesinfinity):
yes
OpenStudy (supernova_sonntag):
okay so y=3 and y=-2 right?
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OpenStudy (xapproachesinfinity):
yep!
OpenStudy (xapproachesinfinity):
so we got two values for y
we take one value at a time and look for x
OpenStudy (xapproachesinfinity):
to find x we go back the equations we started with
OpenStudy (xapproachesinfinity):
what equation do you think is easier to spit out an answer for us
OpenStudy (supernova_sonntag):
the second one
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OpenStudy (xapproachesinfinity):
yes
OpenStudy (xapproachesinfinity):
now pick one y value you found and plug it to that equation and find x
OpenStudy (supernova_sonntag):
so x=1 and -4?
OpenStudy (xapproachesinfinity):
oh you already tried both values of y
OpenStudy (xapproachesinfinity):
yes that good answers
(1, 3) and (-4, -2)
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OpenStudy (xapproachesinfinity):
the answer is two points
OpenStudy (supernova_sonntag):
Okay.
Thank you so much, I understood this a while ago, but I completely forgot. You jogged my memory!!
OpenStudy (xapproachesinfinity):
geometrically this mean line y-2=x intersects the parabola y=(x+2)^2-6
in two points
OpenStudy (xapproachesinfinity):
no problem :)
OpenStudy (xapproachesinfinity):
let me show you what is happening Geometrically
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