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find b and c

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3 Answers

Line of symmetry is where the vertex lies. 

X = -b/2a = -b/20

In this case x = 5 since the vertex has coordinates (5, -4)

5 = -b/20 

-b = 100

b = -100

To find c we input b and the coordinates of the vertex in the original equation.

-4 = 10(5)^2 - 100(5) + c

-4 = 250 - 500 + c

-4 = -250 + c

c = 246

Comments

Thank you Muhammad.  I learned something.

Comment

Standard form of parabola is y = ax2 + bx + c
Vertex form of parabola is y = a(x-h)2 + k where (h,k) is the vertex.
 
a is the same in each equation, which is given as 10, and if we also plug in the vertex given we get (in vertex form)
 
y = 10(x-(5))2 + (-4)
 
Now we just need to rewrite this in standard form to find b and c.
 
First, foil the (x-5)2
 
10(x-5)(x-5) - 4
 
10(x- 5x - 5x + 25) - 4
 
10(x2 - 10x + 25) - 4
 
Distribute the 10 
 
10x2 - 100x + 250 - 4
 
Finally,
10x2 - 100x + 246 shows b=-100 and c=246
 
 The formula to use is:
    
    Y = a ( X - b/ 2a )2  -  ( b-4ac ) / 4a2
 
    Y = 10( X - 5 ) 2  - 4 
 
 Y =10[ ( X - 10X + 25 ) - 4 ]
   
  Y = 10 X2 - 100X + 250 -40
  
 Y = 10 X -100X + 210
 
    another method:
 
 -b / 2a = 5          -b/20 = 5        b = -100 
 
 b- 4ac = 4         [ (-100) - 4 (10)(c) ] / 4 (10) 2 = 4
 
                          10000 - 40c / 4 ( 100) =4 
 
                          10000 -40c = 1600
                          8400 = 40 c
 
                          C = 210
 
 ax2 + bx+c 10X2 -100X + 210- 
 
 any quadratic in the form of aX2 + bX + c can be written equivalently as
a[( X - b/ 2a)2 - (b-4ac)/4a2