# Graphing Conics

#### Honorable24

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....

#### Soroban

MHF Hall of Honor
Hello, Honorable24!

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

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

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

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

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

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

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

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

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

There are no $$\displaystyle x$$-intercepts.

There are no $$\displaystyle y$$-intercepts.

A circle does not have vertices.

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

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

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

#### 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

#### HallsofIvy

MHF Helper
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 $$\displaystyle x^2- 6x+ y^2+ 4y= 3$$.

If $$\displaystyle x^2- 6x+ y^2+ 4x= 3$$ really is the problem, -6x+ 4x= -2x so this is the same as $$\displaystyle x^2- 2x+ y^2= 3$$. 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""$$\displaystyle y^2= (y- 0)^2$$. Complete the square in x.

• earboth

#### earboth

MHF Hall of Honor
Hello, Honorable24!

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

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

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

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

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

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

There are no $$\displaystyle x$$-intercepts.

There are no $$\displaystyle y$$-intercepts.

A circle does not have vertices.

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

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

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

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