Question

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

Answer 1

@amistre64

Answer 2

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

Answer 3

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

Answer 4

@satellite73 graphically no algebraically

Answer 5

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

Answer 6

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

Answer 7

oh no!!!

Answer 8

don't foil nothing

Answer 9

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

Answer 10

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

Answer 11

Are you able to graph y=-4?

Answer 12

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

Answer 13

wouldnt you move four over?

Answer 14

no

Answer 15

-(x-2)^2+7

Answer 16

You need to isolate the variable.

Answer 17

subtract 3 from both sides of the equal sign

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

*sign

Answer 23

(x-2)^2=7

Answer 24

explanation comes with the doing

Answer 25

right

Answer 26

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

Answer 27

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

Answer 28

that sort of came out right

Answer 29

bingo

Answer 30

\[x=2\pm\sqrt7\]

Answer 31

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

Answer 32

x=4.6 or -.645

Answer 33

so then I plug it in to get Y.

Answer 34

notice no foil was harmed in solving this question

Answer 35

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

Answer 36

Oh yea, thats true

Answer 37

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

thats the answer

Answer 38

@satellite73 would I round the answer to 5?

Answer 39

@satellite73 Anyway you can help me?

Answer 40

i am lost now

Answer 41

unless perhaps there is a typo in the question

Answer 42

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

Answer 43

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

Answer 44

thats exactly how I am right now

Answer 45

Just graph the two and see where they meet.

Answer 46

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 \[y=-6\]?

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

I got x= 6.123

Answer 54

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

Answer 55

thats what we are supposed to use

Answer 56

x= 2+rad17

Answer 57

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

Answer 58

yeah you got it

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

can you post a screen shot?

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

it is the same graphically!!

Answer 72

solutions will be the same, makes no difference

Answer 73

Well, I literally pasted exactly what I saw

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

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 \[-(x-2)^2=-9\]

Answer 78

First step is set it equal to each other?

Answer 79

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

Answer 80

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

Answer 81

I don't recall learning the other way

Answer 82

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

Answer 83

but what about the one I just posted

Answer 84

\[-(x+1)^2=11\\
y=x+4\] 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

Answer 85

Where do I go from there

Answer 86

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

Answer 87

So by moving the 4 over the - goes away?

Answer 88

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

Answer 89

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

Answer 90

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

Answer 91

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

Answer 92

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

Answer 93

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

Answer 94

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

Answer 95

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

Answer 96

hell no!!

Answer 97

Yes