How to interpret the function: (1/9)x^2 + y^2 = 1

I need an explanation that explains how the x coefficient and term plus y term factor into creating the graph to better understand it all

For the record, this is NOT a function. It is the equation of an ellipse.
You can tell a lot just by looking at the equation.

Put your equation into standard ellipse form by moving the coefficients to the denominators.
x²/3² + y²/1² = 1
The denominators determine the orientation of the ellipse (horizontal or vertical) and its dimensions.

Compare the denominators.
x² has the larger denominator, so the ellipse is HORIZONTAL.
(If y² had the larger denominator, the ellipse would be vertical. If the denominators were the same, the ellipse would be a circle.)

The general equation of a horizontal ellipse is
     (x−h)²/a² + (y−k)²/b² = 1
where
     center is (h, k)
     a ≥ b > 0
     vertices are (h±a, k)
     length of transverse (major) axis = 2a
     co-vertices are (h, k±b)
     length of conjugate (minor) axis = 2b
     foci (h±c, k) where c² = a² − b²

Apply your equation.
h = k = 0, so the ellipse is centered over (0, 0)
a = 3
b = 1
vertices (h±a, k) = (±3, 0)
length of transverse (major) axis = 2a = 6
co-vertices (h, k±b) = (0, ±1)
length of conjugate (minor) axis = 2b
 
c² = 3² - 1² = 3
c = √3 ≅ 2.8
foci (h±c, k) = (±2.8, 0)

http://www.flickr.com/photos/dwread/6406…

(x - h) ^2/a^2 +(y - k) ^2/b^2 = 1 is the equation of an ellipse in standard form
a = Semi-major axis is 3 (horizontal)
b = Semi-minor axis is 1
(h, k) = (0, 0) is the center.

(1/9)x^2 + y^2 = 1

x^2/3^2 + y^2 = 1
(x^2 + (3^2)(y^2) ) / 3^2 = 1
9y^2 = -x^2 +9
9y^2 / 9 = (-x^2 + 9) / 9
y^2 = (-x^2+3^2)/9
y^2 = (3^2-x^2)/9
y^2= ((3-x)(3+x))/9

y= ( (-x+3)^(1/2) (x+3)^(1/2) ) /3
y= - ( (-x+3)^(1/2) (x+3)^(1/2) ) /3

If you are under the grade of 8th grade and in regular math (not advanced) you have some problems. this is so easy!

This is the equation of an ellipse.