anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@amistre64
anonymous
  • anonymous
You would need to graph both equations to solve. Are you able to do so?
anonymous
  • anonymous
set \[-(x+2)^2+3=-4\] and solve for \(x\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@satellite73 graphically no algebraically
anonymous
  • anonymous
oops typo there sorry \[-(x-2)^2+3=-4\] solve for \(x\) takes three steps
anonymous
  • anonymous
The only thing that is bothering me is the - before the "(" How would I foil it?
anonymous
  • anonymous
oh no!!!
anonymous
  • anonymous
don't foil nothing
1 Attachment
anonymous
  • anonymous
lets go through the steps, there aer only three of them
anonymous
  • anonymous
Lol. How would I do it then. I don't know how to do it graphically or algebraically
anonymous
  • anonymous
Are you able to graph y=-4?
anonymous
  • anonymous
\[-(x-2)^2+3=-4\] subtract 3 from both sides what do you get?
anonymous
  • anonymous
wouldnt you move four over?
anonymous
  • anonymous
no
anonymous
  • anonymous
-(x-2)^2+7
anonymous
  • anonymous
You need to isolate the variable.
anonymous
  • anonymous
subtract 3 from both sides of the equal sign
anonymous
  • anonymous
don't add 4 to both sides,you would be working backwards
anonymous
  • anonymous
@satellite73 He won't understand if you are just giving directions, you must explain why.
anonymous
  • anonymous
so... -(x-2)^2=-7
anonymous
  • anonymous
right now change the sing of both sides, i.e. get rid of both minus signs
anonymous
  • anonymous
*sign
anonymous
  • anonymous
(x-2)^2=7
anonymous
  • anonymous
explanation comes with the doing
anonymous
  • anonymous
right
anonymous
  • 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
  • anonymous
x=2\[\pm \sqrt{7}\]
anonymous
  • anonymous
that sort of came out right
anonymous
  • anonymous
bingo
anonymous
  • anonymous
\[x=2\pm\sqrt7\]
anonymous
  • 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
  • anonymous
x=4.6 or -.645
anonymous
  • anonymous
so then I plug it in to get Y.
anonymous
  • anonymous
notice no foil was harmed in solving this question
anonymous
  • anonymous
lol no you don't plug it in to get \(y\) you are told at the outset that \(y=-4\)
anonymous
  • anonymous
Oh yea, thats true
anonymous
  • anonymous
The solution to the system is (−1, −6) and (5, −6). thats the answer
anonymous
  • anonymous
@satellite73 would I round the answer to 5?
anonymous
  • anonymous
@satellite73 Anyway you can help me?
anonymous
  • anonymous
i am lost now
anonymous
  • anonymous
unless perhaps there is a typo in the question
anonymous
  • anonymous
@Beleaguer It would've been easier if you graphed the two.
anonymous
  • anonymous
could it have actually been \[-(x-2)^2-3=-4\]?
anonymous
  • anonymous
thats exactly how I am right now
anonymous
  • anonymous
Just graph the two and see where they meet.
anonymous
  • 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
  • anonymous
Nope, its +3 = -4. I pasted it from the book. Maybe the book is wrong? lol i have no idea
anonymous
  • anonymous
now that i see it, it is either a typo on their part of yours
anonymous
  • 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
  • anonymous
now i made a typo, first line should have been \[-(x-2)^2+3=-6\]
anonymous
  • anonymous
if they wrote \(y=-4\) they meant \(y=-6\) that is clear, because both answers have \(-6\) in the second coordinate
anonymous
  • anonymous
What about this y = −(x − 2)2 + 8 y = −9
anonymous
  • anonymous
I got x= 6.123
anonymous
  • anonymous
lorda mercy lets assume there is not typo here stop with the decimals already
anonymous
  • anonymous
thats what we are supposed to use
anonymous
  • anonymous
x= 2+rad17
anonymous
  • 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
  • anonymous
yeah you got it
anonymous
  • anonymous
thats what I got, but woah...theres more xD
anonymous
  • anonymous
this is the answer -.- The solution to the system is (−1, −1) and (5, −1).
anonymous
  • anonymous
I am going to look on the internet on how to solve it graphically..
anonymous
  • anonymous
the formula for vertex is -b/2a, right?
anonymous
  • anonymous
can you post a screen shot?
anonymous
  • anonymous
if \(y=-9\) it cannot be \(-1\) at the same time
anonymous
  • anonymous
that is the answer if you solve \[-(x-2)^2+8=-1\]
anonymous
  • anonymous
either there are massive typos here, of there is a problem in the translation
anonymous
  • 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
  • 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
  • anonymous
but that is \(y=-1\) NOT \(y=-9\)
anonymous
  • anonymous
I guess it is because we are solving it graphically can you help me that way @satellite73
anonymous
  • anonymous
it is the same graphically!!
anonymous
  • anonymous
solutions will be the same, makes no difference
anonymous
  • anonymous
Well, I literally pasted exactly what I saw
anonymous
  • 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
  • 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
  • anonymous
Ok, lets try something else.... maybe I will understand it. How about this problem. \[y=-(x-1)^{2}+11 y= x+4\]
anonymous
  • 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
  • anonymous
First step is set it equal to each other?
anonymous
  • anonymous
it is not a matter of you understanding there is a mistake in the question, take my word for it
anonymous
  • anonymous
Well this is from another source, so I am going to attempt it and see
anonymous
  • anonymous
I don't recall learning the other way
anonymous
  • 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
  • anonymous
but what about the one I just posted
anonymous
  • 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
  • anonymous
Where do I go from there
anonymous
  • 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
  • anonymous
So by moving the 4 over the - goes away?
anonymous
  • anonymous
then expand and combine like terms \[x^2+2x+1+x-7\\ x^2+3x+6=0\]
anonymous
  • anonymous
i added \((x+1)^2\) and subtracted \(11\) from both sides easier to work with \((x+1)^2\) than \(-(x+1)^2\)
anonymous
  • anonymous
any chance i can convince you to post a screen shot of the original two questions?
anonymous
  • 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
  • anonymous
hope your teacher doesn't make the same mistake on the test!
anonymous
  • 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
  • anonymous
let me make sure i have the question correct is it \[y=-(x+1)^2+11\\ y=x+4\]?
anonymous
  • anonymous
I keep thinking I would distribute the - into the x and -1 therefore making it (-x+1)^2
anonymous
  • anonymous
hell no!!
anonymous
  • anonymous
Yes
anonymous
  • 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
  • anonymous
Oh yea. So what is the other option
anonymous
  • anonymous
\[(x+1)^2=x^2+2x+1\] so \[-(x+1)^2=-x^2-2x-1\]
anonymous
  • anonymous
but it is easier to work with \((x+1)^2\) than \(-(x+1)^2\)
anonymous
  • 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
  • anonymous
OHHHHHHHHHH. Wow, I can't believe I didn't see that
anonymous
  • anonymous
Alright, so I combined -11+4 and got -7... next step? what do I do with the x
anonymous
  • anonymous
then you solve the quadratic equation by expanding and combining like terms still waiting for that screen shot
anonymous
  • anonymous
this one you have to expand
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
what do you mean by expanding
anonymous
  • anonymous
wait, let me give it a try. I think I got it :) your honestly the best
anonymous
  • anonymous
\[(x+1)^2+x-7=x^2+2x+1+x-7=x^2+3x-6\]
anonymous
  • anonymous
i made some typo there last line is right however
anonymous
  • anonymous
Exactly what I got. Would I then complete the sq?
anonymous
  • anonymous
i wouldn't because \(3\) is odd i would use the quadratic formula
anonymous
  • anonymous
Is that something I should follow? If the B is odd, use quadratic formula?
anonymous
  • 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
  • anonymous
I got -3 plus or minus rad 33 over 2. Sorry, couldn't get the equations to work
anonymous
  • anonymous
yes me too
anonymous
  • 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
  • anonymous
hmmm... the answers are (3,7)(-2,2) which I didnt get

Looking for something else?

Not the answer you are looking for? Search for more explanations.