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

 

Find the vertex of the graph of the equation $y = -2x^2 + 8x - 15.$

 

Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6$

 

The graph of $y = ax^2 + bx + c$ has an axis of symmetry of $x = 4.$ Find $\frac{b}{a}$.

 

The graph of $y = ax^2 + bx + c$ is shown below. Find $a \cdot b \cdot c$.  (The distance between the grid lines is one unit.)        

 


 

 May 28, 2021
 #2
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1. The vertex of x - y^2 + 8y = 13 is (-8,4).

 

2. The vertex of y = -2x^2 + 8x - 15 is (4,-15).

 May 28, 2021
 #3
avatar+32758 
+2

1.)  x = y^2 -8y + 13

       re-arrange to vertex form

              x = (y-4)^2  -3        vertex is   -3,4

 May 28, 2021
 #4
avatar+32758 
+2

2 )   -2x^2+8x-15     re-arrange to vertex form

       -2 (x^2 - 4x) -15

         -2 ( x-2)^2  - 7         vertex = 2, -7

 May 28, 2021
 #5
avatar+32758 
+2

3)  x^2 -2x    + y^2  + 6y   =  6     Arrange into standard circle equation form

 

     (x-1)^2      +      ( y+3)^2 = 6 +1 + 9           

                                                6+ 1 + 9 = 16  = r^2

 

     Area of a circle = pi r^2 = pi (16)  = 16 pi  units2

 May 28, 2021
 #6
avatar+27 
0

Can you answer the last two questions? Thank you so much for the other answers.... all correct!

 Jun 5, 2021

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