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# Help

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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\$.

Feb 21, 2018

#1
+18057
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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

Feb 21, 2018
edited by ElectricPavlov  Feb 21, 2018
edited by ElectricPavlov  Feb 21, 2018

#1
+18057
+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
edited by ElectricPavlov  Feb 21, 2018
edited by ElectricPavlov  Feb 21, 2018