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1. Find the sum of the constants a, h, k and  such that
\(2x^2 - 8x + 7 = a(x - h)^2 + k\)
for all real numbers x.

 

2. Find the vertex of the graph of the equation \(x -y^2 + 6y =8\).

 

3. Find the area of the region enclosed by the graph of the equation \(x^2 + y^2 = 4x + 6y+13\).

Guest Jan 6, 2018
 #1
avatar+7336 
+1

1.  2x2 - 8x + 7

                                          Factor  2  out of the first two terms.

=   2(x2 - 4x) + 7

                                          Add and subtract  (4/2)2 , which is  4 .

=   2(x2 - 4x + 4 - 4) + 7

                                          Factor  x2 - 4x + 4  as a perfect square trinomial.

=   2( (x - 2)2 - 4 ) + 7

                                          Now distribute the  2  to the two terms in parenthesees.

=   2(x - 2)2 - 8 + 7

 

=   2(x - 2)2 - 1

 

So...   a = 2  ,   h = 2  ,   k = -1     ...and...     2  +  2  +  -1   =   3

hectictar  Jan 6, 2018
 #2
avatar+7336 
+1

2.

 

x - y2 + 6y   =   8

                                         Add  y2  to both sides, subtract  6y  from both sides.

x   =   y2 - 6y + 8

                                         Complete the square on the right side.

x   =   y2 - 6y + 9 - 9 + 8

 

x   =   (y - 3)2 - 1

 

The vertex is at  (-1, 3)

hectictar  Jan 6, 2018
 #3
avatar+7336 
+1

3.

 

x2 + y2   =   4x + 6y + 13     This is the equation of a circle. Let's get it in standard form...

 

x2 - 4x        +  y2 - 6y         =   13       Add  4  and  9  to both sides.

 

x2 - 4x + 4  +  y2 - 6y + 9   =   13 + 4 + 9

 

(x - 2)2  +  (y - 3)2   =   26

 

Let the radius of the circle be  r  ,  now we can see that  r2  =  26  .

 

area of the circle   =   pi * r2   =   pi * 26   =  26pi    sq. units

hectictar  Jan 6, 2018

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