find b and c

Question

find b and c so that y= 10x²+bx+c has vertex (5,-4)

Answer 1

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

Answer 2

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(x² - 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

Answer 3

 

 The formula to use is:

    

    Y = a ( X - b/ 2a )²  -  ( b² -4ac ) / 4a²

 

    Y = 10( X - 5 ) ²  - 4 

 

 Y =10[ ( X²  - 10X + 25 ) - 4 ]

   

  Y = 10 X² - 100X + 250 -40

  

 Y = 10 X²  -100X + 210

 

    another method:

 

 -b / 2a = 5          -b/20 = 5        b = -100 

 

 b² - 4ac = 4         [ (-100)²  - 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 - (b² -4ac)/4a²]