Question

4(x-5)^2+4y^2=16 find the center(h,k) and the radius

Answer 1

Eyeball problem.

Center: (5,0) -- You should see this from a quick glance.
Radius is a little trickier. Divide by 4 and that should be another quick glance away.

Answer 2

so I don't use (x-h)^2+(x-k)^2=r^2

Answer 3

Yes you do. Divide by 4 and it will be in that form.

Answer 4

4x^2-40+4y^2=16 add 40 to both sides
4x^2+4y^2=56 divide by 4 both sides
x^2+y^2=sqrt of 14
now im a little lost

Answer 5

What are you doing? What happened to that nice, convenient form.

You have: \(4(x-5)^2 + 4y^{2} = 16\)
Divide by 4: \((x-5)^2 + y^{2} = 4 = 2^{2}\)

You are now done with the problem.
Center: (5,0)
Radius: 2

Answer 6

I told you I was lost...guess I confused myself
How do I find the intercepts?

Answer 7

That's why we invent "forms". They contain useful information. When presented with and important form, don't mess it up!

Intercepts look like this: ALWAYS

x-intercepts: (r,0)
y-intercepts: (0,s)

In words, substitute y = 0 into the equation to find x-intercepts. And, substitute x = 0 into the equation to find y-intercepts.

Answer 8

is the x (2,-6)

Answer 9

y=(0,14)

Answer 10

You must pay closer attention. There MUST be a zero in there, somewhere or it is NOT an intercept. Give it another go.

Answer 11

when I plug in I get 14 for both x and y @tkhunny

Answer 12

ok so x=(2,0) and y=(0,14)

Answer 13

You have \((x-5)^{2} + y^{2} = 4\)

To find the y-intercept, substitute x = 0

\((0-5)^{2} + y^{2} = 4\)

Solve for y

\((5)^{2} + y^{2} = 4\)

\(25 + y^{2} = 4\)

\(y^{2} = 4 - 25\)

\(y^{2} = -21\)

Since \(y^{2}\) cannot be a negative number, there are no y-intercepts.

You find the x-intercepts, if they exist.

Answer 14

I added 25 and got 29

Answer 15

Why would you add 25? It's positive over there on the left. You must subtract it. Don't let your desire for a satisfying solution cloud your judgment.

Answer 16

ok and I dropped my sign on the neg 5. careless mistakes

Answer 17

will my graph look like this

Answer 18

or close to it

Answer 19

That's no good. The radius is WAY too big. There are NO y-intercepts. Use your x-intercepts to guide you. What are they?

Answer 20

x=(2,0)

Answer 21

do I point my radius first

Answer 22

You must find your x-intercepts!

You have You have \((x−5)^{2} +y^{2} =4 \)

To find the x-intercepts, substitute y = 0

\((x−5)^{2} +0^{2} = 4 \)

\((x−5)^{2} +0 = 4 \)

\((x−5)^{2} = 4 \)

x - 5 = 2 or x - 5 = -2
x = 7 or x = 3

There are the x-intercepts. (7,0) and (3,0)

You have to do the work. It never will be magic.