Solve the given linear-quadratic system graphically.
y = −(x − 2)2 + 3
y = −4
Please lists steps... I need to know how to do it

- anonymous

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- anonymous

@amistre64

- anonymous

You would need to graph both equations to solve. Are you able to do so?

- anonymous

set \[-(x+2)^2+3=-4\] and solve for \(x\)

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## More answers

- anonymous

@satellite73 graphically no algebraically

- anonymous

oops typo there
sorry
\[-(x-2)^2+3=-4\] solve for \(x\) takes three steps

- anonymous

The only thing that is bothering me is the - before the "("
How would I foil it?

- anonymous

oh no!!!

- anonymous

don't foil nothing

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- anonymous

lets go through the steps, there aer only three of them

- anonymous

Lol. How would I do it then. I don't know how to do it graphically or algebraically

- anonymous

Are you able to graph y=-4?

- anonymous

\[-(x-2)^2+3=-4\] subtract 3 from both sides
what do you get?

- anonymous

wouldnt you move four over?

- anonymous

no

- anonymous

-(x-2)^2+7

- anonymous

You need to isolate the variable.

- anonymous

subtract 3 from both sides of the equal sign

- anonymous

don't add 4 to both sides,you would be working backwards

- anonymous

@satellite73 He won't understand if you are just giving directions, you must explain why.

- anonymous

so... -(x-2)^2=-7

- anonymous

right
now change the sing of both sides, i.e. get rid of both minus signs

- anonymous

*sign

- anonymous

(x-2)^2=7

- anonymous

explanation comes with the doing

- anonymous

right

- anonymous

now take the square root of both sides, i.e. get rid of the square on the left
don't forget the \(\pm\)

- anonymous

x=2\[\pm \sqrt{7}\]

- anonymous

that sort of came out right

- anonymous

bingo

- anonymous

\[x=2\pm\sqrt7\]

- anonymous

here is the graph if you really need it
http://www.wolframalpha.com/input/?i=y%3D-%28x-2%29^2%2B3%2Cy%3D-4

- anonymous

x=4.6 or -.645

- anonymous

so then I plug it in to get Y.

- anonymous

notice no foil was harmed in solving this question

- anonymous

lol no you don't plug it in to get \(y\) you are told at the outset that \(y=-4\)

- anonymous

Oh yea, thats true

- anonymous

The solution to the system is (−1, −6) and (5, −6).
thats the answer

- anonymous

@satellite73 would I round the answer to 5?

- anonymous

@satellite73 Anyway you can help me?

- anonymous

i am lost now

- anonymous

unless perhaps there is a typo in the question

- anonymous

@Beleaguer It would've been easier if you graphed the two.

- anonymous

could it have actually been \[-(x-2)^2-3=-4\]?

- anonymous

thats exactly how I am right now

- anonymous

Just graph the two and see where they meet.

- anonymous

actually it makes no sense anyway
how can the answer be (−1, −6) and (5, −6). when you are told \(y=-4\)
was it really \[y=-6\]?

- anonymous

Nope, its +3 = -4. I pasted it from the book. Maybe the book is wrong? lol i have no idea

- anonymous

now that i see it, it is either a typo on their part of yours

- anonymous

\[-(x-2)^2+3=-9\\
-(x-2)^2=-9\\
(x-2)^2=-9\\
x-2=3, x-2=-3\\
x=-1,x=-5\]

- anonymous

now i made a typo, first line should have been \[-(x-2)^2+3=-6\]

- anonymous

if they wrote \(y=-4\) they meant \(y=-6\)
that is clear, because both answers have \(-6\) in the second coordinate

- anonymous

What about this y = −(x − 2)2 + 8
y = −9

- anonymous

I got x= 6.123

- anonymous

lorda mercy lets assume there is not typo here
stop with the decimals already

- anonymous

thats what we are supposed to use

- anonymous

x= 2+rad17

- anonymous

\[--(x-2)^2+8=-9\\
-(x-2)^2=-17\\
(x-2)^2=17\\
x-2=\pm\sqrt{17}\\
x=2\pm\sqrt{17}\]

- anonymous

yeah you got it

- anonymous

thats what I got, but woah...theres more xD

- anonymous

this is the answer -.- The solution to the system is (−1, −1) and (5, −1).

- anonymous

I am going to look on the internet on how to solve it graphically..

- anonymous

the formula for vertex is -b/2a, right?

- anonymous

can you post a screen shot?

- anonymous

if \(y=-9\) it cannot be \(-1\) at the same time

- anonymous

that is the answer if you solve \[-(x-2)^2+8=-1\]

- anonymous

either there are massive typos here, of there is a problem in the translation

- anonymous

you cannot have \(y=-9\) and have a solution that has anything other than \(-9\) in the second coordinate
maybe your math teacher was on the sauce when he/she wrote these, or copied and pasted incorrectly

- anonymous

\[-(x-2)^2+8=-1\\
-(x-2)^2-9\\
(x-2)^2=9\\
x-2=3,x-2=-3\\
x=5,x=-1\] solutions are \((5,-1),(-1,-1)\)

- anonymous

but that is \(y=-1\) NOT \(y=-9\)

- anonymous

I guess it is because we are solving it graphically
can you help me that way @satellite73

- anonymous

it is the same graphically!!

- anonymous

solutions will be the same, makes no difference

- anonymous

Well, I literally pasted exactly what I saw

- anonymous

there is the graphical solution to \[y=-(x-2)^2+8\\
y=-1\]
notice that it is \((5,-1),(-1,-1)\)
http://www.wolframalpha.com/input/?i=y%3D-%28x-2%29^2%2B8%2Cy%3D-1

- anonymous

i know you are confused because of the mistake in the question, but try to ignore that
if \(y=-1\) then \(y=-1\) and if \(y=-9\) then \(y=-9\) it cannot be otherwise

- anonymous

Ok, lets try something else.... maybe I will understand it. How about this problem. \[y=-(x-1)^{2}+11 y= x+4\]

- anonymous

perhaps the teacher made a mistake and did the first step mentally and therefore wrote \[-(x-2)^2+8=-9\] instead of \(-(x-2)^2+8=-1\) because the first step in solving \[-(x-2)^2+8=-1\] would be to turn it in to \[-(x-2)^2=-9\]

- anonymous

First step is set it equal to each other?

- anonymous

it is not a matter of you understanding
there is a mistake in the question, take my word for it

- anonymous

Well this is from another source, so I am going to attempt it and see

- anonymous

I don't recall learning the other way

- anonymous

look if you went to solve a simple one like \[y=2x+1,y=9\] you would set \[2x+1=9\\2x=8\\
x=4\] and the solution would be \((4,9)\) not some other number in the second slot

- anonymous

but what about the one I just posted

- anonymous

\[-(x+1)^2=11\\
y=x+4\] is a different kind of question because of the \(x+4\)
in this case you would have to solve \[-(x+1)^2+11=x+4\]
here you don't know what \(y\) is, so you would have to find it

- anonymous

Where do I go from there

- anonymous

for this one, you need to put it all on one side of the equal sign and solve \[(x+1)^2+x+4-11=0\]

- anonymous

So by moving the 4 over the - goes away?

- anonymous

then expand and combine like terms \[x^2+2x+1+x-7\\
x^2+3x+6=0\]

- anonymous

i added \((x+1)^2\) and subtracted \(11\) from both sides
easier to work with \((x+1)^2\) than \(-(x+1)^2\)

- anonymous

any chance i can convince you to post a screen shot of the original two questions?

- anonymous

Yea, I will. I just want to get this problem straight because I know this will be on my test tomorrow 100%, im not sure about the other ones

- anonymous

hope your teacher doesn't make the same mistake on the test!

- anonymous

So, do you mind going over the process of eliminating the -? I am lost on that part, sorry for being so picky and ty for sticking through

- anonymous

let me make sure i have the question correct
is it \[y=-(x+1)^2+11\\
y=x+4\]?

- anonymous

I keep thinking I would distribute the - into the x and -1 therefore making it (-x+1)^2

- anonymous

hell no!!

- anonymous

Yes

- anonymous

ok you cannot distribute the minus sign
think of the order of operations
you would have to square first, then distribute the minus sign

- anonymous

Oh yea. So what is the other option

- anonymous

\[(x+1)^2=x^2+2x+1\] so \[-(x+1)^2=-x^2-2x-1\]

- anonymous

but it is easier to work with \((x+1)^2\) than \(-(x+1)^2\)

- anonymous

so if i had to solve \[-(x+1)^2+11=x+4\] i would subract \(11\) and add \((x+1)^2\) to get \[(x+1)^2-11+x+4=0\]

- anonymous

OHHHHHHHHHH. Wow, I can't believe I didn't see that

- anonymous

Alright, so I combined -11+4
and got -7... next step?
what do I do with the x

- anonymous

then you solve the quadratic equation by expanding and combining like terms
still waiting for that screen shot

- anonymous

this one you have to expand

- anonymous

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- anonymous

what do you mean by expanding

- anonymous

wait, let me give it a try. I think I got it :)
your honestly the best

- anonymous

\[(x+1)^2+x-7=x^2+2x+1+x-7=x^2+3x-6\]

- anonymous

i made some typo there
last line is right however

- anonymous

Exactly what I got. Would I then complete the sq?

- anonymous

i wouldn't because \(3\) is odd
i would use the quadratic formula

- anonymous

Is that something I should follow? If the B is odd, use quadratic formula?

- anonymous

completing the square and the quadratic formula are the same, one is a process and the other is that process condensed
completing the square requires taking half of the coefficient and squaring it
if it is even, then it is easy to take half and to square, you will have no denominator
if it is odd, you will get a fraction and have to square that
might as well use the formula

- anonymous

I got -3 plus or minus rad 33 over 2. Sorry, couldn't get the equations to work

- anonymous

yes me too

- anonymous

NOW this time if you want to find \(y\) you have to add 4 to those numbers, because you have \(y=x+4\)

- anonymous

hmmm... the answers are (3,7)(-2,2) which I didnt get

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