1. ## Graphing Conics

4x^2 + 4y^2 +20x -16y + 37 =0
I completed the square and got:
(2x+10)^2 + (2y-8)^2 =127

x int ?
y int ?
Vertices ?
Center (-5, 4)
Domain ?
Range ?

I tried making y=0 to find the x intercepts but didn't work out. Same thing for y intercepts
Completely lost....

2. Hello, Honorable24!

$4x^2 + 4y^2 +20x -16y + 37 \:=\:0$

I completed the square and got: . $(2x+10)^2 + (2y-8)^2 \:=\:127$ .??

Find: . $x\text{-intercepts,}\quad y\text{-intercepts,}\quad \text{Vertices,}\quad \text{Center,}\quad \text{Domain,}\quad \text{Range}$

We have: . $4x^2 + 20x + 4y^2 - 16y \;=\;-37$

. . . . $4(x^2 + 5x \quad) + 4(y^2 - 4y \quad) \;=\;-37$

$4\left(x^2 + 5x + {\color{blue}\tfrac{25}{4}}\right)^2 + 4\left(y^2 - 4y + {\color{red}4}\right) \;=\;-37 + {\color{blue}25} + {\color{red}16}$

. . . . . . . . $4\left(x + \tfrac{5}{2}\right)^2 + 4(y-2)^2 \;=\;4$

. . . . . . . . . $\left(x + \tfrac{5}{2}\right)^2 + (y-2)^2 \;=\;1$

This is a circle: . $\text{Center: }\left(-\tfrac{5}{2},\:2\right),\;\text{ radius } 2$

There are no $x$-intercepts.

There are no $y$-intercepts.

A circle does not have vertices.

$\text{Center: }\:\left(-\tfrac{5}{2},\:2\right)$

$\text{Domain: }\;x \in \left[-\tfrac{7}{2},\:-\tfrac{3}{2}\right]$

$\text{Range: }\;y \in[1,\:3]$

3. Thanks

x^2-6x + y^2 +4x=3

How do you do completing the square for this
I don't know how to do these types becuase there is no single y term

4. Originally Posted by Honorable24
Thanks

x^2-6x + y^2 +4x=3

How do you do completing the square for this
I don't know how to do these types becuase there is no single y term
I recommend you go back and check the problem. I suspect it should be $x^2- 6x+ y^2+ 4y= 3$.

If $x^2- 6x+ y^2+ 4x= 3$ really is the problem, -6x+ 4x= -2x so this is the same as $x^2- 2x+ y^2= 3$. The fact that there is no "y to the first power" term is not a problem. It only means that y is already a "perfect square"" $y^2= (y- 0)^2$. Complete the square in x.

5. Originally Posted by Soroban
Hello, Honorable24!

We have: . $4x^2 + 20x + 4y^2 - 16y \;=\;-37$

. . . . $4(x^2 + 5x \quad) + 4(y^2 - 4y \quad) \;=\;-37$

$4\left(x^2 + 5x + {\color{blue}\tfrac{25}{4}}\right)^2 + 4\left(y^2 - 4y + {\color{red}4}\right) \;=\;-37 + {\color{blue}25} + {\color{red}16}$

. . . . . . . . $4\left(x + \tfrac{5}{2}\right)^2 + 4(y-2)^2 \;=\;4$

. . . . . . . . . $\left(x + \tfrac{5}{2}\right)^2 + (y-2)^2 \;=\;1$

This is a circle: . $\text{Center: }\left(-\tfrac{5}{2},\:2\right),\;\text{ radius } 1$ <<<<<<<< typo

There are no $x$-intercepts.

There are no $y$-intercepts.

A circle does not have vertices.

$\text{Center: }\:\left(-\tfrac{5}{2},\:2\right)$

$\text{Domain: }\;x \in \left[-\tfrac{7}{2},\:-\tfrac{3}{2}\right]$

$\text{Range: }\;y \in[1,\:3]$

... I don't want to pick at you but there is a tiny typo.