1. Find the center of the circle with equation $9x^2-18x+9y^2+36y+44=0$

2.The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers. Find $a$.

Guest Feb 21, 2018

#1**+2 **

1. 9x^2-18x +9y^2 + 36y = -44 divide thru by 9

x^2 - 2x + y^2 +4 = -44/9 'complete the square' for x and y

( x-1)^2 + (y+2)^2 = -44/9 + 1 + 4

(x-1)^2 + (y+2)^2 = 1/9 Now this is standard form for a circle (x-h)^2 + (y-k)^2 = r^2

where center is h,k and r = radius

center = 1,-2 radius = 1/3

2. Let's set this up as vetex form for a parabola h, k = vertex = 1,3 (from graph)

so a(x-1)^2 + 3 =y is the equation for this parabola...now let's find 'a'

a convineint point which must satisfy this equation (on the graph) is 0,1 ...substitute this

a (0-1)^2 + 3 = 1

a+3 = 1

a= -2 so the equation becomes y= -2 (x-1)^2 +3

ElectricPavlov
Feb 21, 2018

#1**+2 **

Best Answer

1. 9x^2-18x +9y^2 + 36y = -44 divide thru by 9

x^2 - 2x + y^2 +4 = -44/9 'complete the square' for x and y

( x-1)^2 + (y+2)^2 = -44/9 + 1 + 4

(x-1)^2 + (y+2)^2 = 1/9 Now this is standard form for a circle (x-h)^2 + (y-k)^2 = r^2

where center is h,k and r = radius

center = 1,-2 radius = 1/3

2. Let's set this up as vetex form for a parabola h, k = vertex = 1,3 (from graph)

so a(x-1)^2 + 3 =y is the equation for this parabola...now let's find 'a'

a convineint point which must satisfy this equation (on the graph) is 0,1 ...substitute this

a (0-1)^2 + 3 = 1

a+3 = 1

a= -2 so the equation becomes y= -2 (x-1)^2 +3

ElectricPavlov
Feb 21, 2018