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There exist constants  a,h,k  and  such that 3x^2+12x+4=a(x-h)^2+k
for all real numbers x  Enter the ordered triple (a,h,k)

 

Find the vertex of the graph of the equation x-y^2+8y=13

 

 

find the vertex of the graph of the equation y=-2x^2+8x-15

 Jun 30, 2020
 #1
avatar+21953 
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1)  Rewrite  3x2 + 12x + 4  into the form  a(x - h)2 + k

 

     3x2 + 12x + 4

Separate the x-terms from the constant:  (3x2 + 12x) + 4

Factor out the 3:                                        3(x2 + 4x) + 4

Complete the square:                                3(x2 + 4x + 4) - 12 + 4 * (see below)

Factor:                                                       3(x + 2)2 - 8

Pick out a, h, and K:      a = 3     h = -2     k = -8

 

*Completing the square:  divide the coefficient of the x-term (4x) by 2 and square the result:

               4 / 2  =  2     --->     22 = 4

  place this number within the parentheses

  because you have entered a 4 in the parentheses and the parentheses have a '3' before it,

     you have added a '12', so you'll need to subtract a 12.

 

2)  x - y2 + 8y  =  13

 

     Rearrange:                             - y2 + 8y  =  13 - x

     Multiply everything by -1:          y2 - 8y  =  x - 13

     Complete the squre:         y2 - 8y + 16  =  x - 13 + 16

     Factor:                                      (y - 4)2  =  x + 3

     Vertex:       x = -3     y = 4

 

3)  Like number 2.

 Jun 30, 2020

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