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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 $a$. 

Guest Feb 21, 2018

Best Answer 

 #1
avatar+12735 
+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
 #1
avatar+12735 
+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
edited by ElectricPavlov  Feb 21, 2018
edited by ElectricPavlov  Feb 21, 2018

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