Question

math help

Answer 1

http://prntscr.com/jdsgqe

Answer 2

alright so we have to complete the square on both quadratic parts (the x^2 part and the y^2 part)

let's start with the x terms

x^2 - 4x

to convert this to a quadratic equation, we take the coefficient of the b-term (-4), divide it by 2, and square it, then add it to the final equation to get [ (-4)/2 ] ^ 2 = 4

so your equation becomes

x^2 - 4x + 4

can you try repeating this process with the y terms?

Answer 3

y^2 + 2y + 1?

Answer 4

awesome so we have:

x^2 - 4x + 4

y^2 + 2y + 1

we can factor these to get

(x-2)^2 + (y+1)^2

now, there's a -11 on the left side of the equation that we didn't include yet, so we have:

(x-2)^2 + (y+1)^2 - 11 = 0

we added 4 and 1 (from our previous calculations) so we must also add 4 + 1 to the other side

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

adding 11 to both sides of the equation gives us

(x-2)^2 + (y+1)^2 = 16

^ do you know how to get the center and radius from this?

Answer 5

(x – h)2 + (y – k)2 = r2

Answer 6

i take the h and k points

Answer 7

good, so if we compare that equation to our equation we get

(x-2)^2 = (x-h)^2
(y-k)^2 = (y+1)^2
16 = r^2

what are your r, h, k values?

Answer 8

for the first one we dont have somehting on the h?

Answer 9

if it helps you can take the square root of all sides to eliminate the exponents

(x-2) = (x-h)

h = ?

Answer 10

(x-2) = (x-h)

subtract x from both sides

-2 = -h

h = ?

Answer 11

sorry i was confused about somehing else but
h=2

Answer 12

good

(y-k) = (y+1)

k = ?

Answer 13

k=-1

Answer 14

awesome

and r^2 = 16 so r = 4

so your center = (2,-1) and your radius = 4 which is what they are asking for

be sure to show the calculations we have done in addition to the final answer

Answer 15

will do , thank you for your patience!!

Answer 16

yeah np

i have to go now but will be on later if you have any additional questions

Answer 17

thank you lol