find b and c so that y= 10x

^{2}+bx+c has vertex (5,-4)-
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find b and c so that y= 10x^{2}+bx+c has vertex (5,-4)

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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**

Standard form of parabola is y = ax^{2} + 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^{2 }- 5x - 5x + 25) - 4

10(x^{2} - 10x + 25) - 4

Distribute the 10

10x^{2 }- 100x + 250 - 4

Finally,

10x^{2} - 100x + 246 shows b=-100 and c=246

The formula to use is:

Y = a ( X - b/ 2a )^{2} - ( b^{2 }-4ac ) / 4a^{2}

Y = 10( X - 5 ) ^{2 }- 4

Y =10[ ( X^{2 } - 10X + 25 ) - 4 ]

Y = 10 X^{2} - 100X + 250 -40

another method:

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

b^{2 }- 4ac = 4 [ (-100)^{2 } - 4 (10)(c) ] / 4 (10) ^{2
}= 4

10000 - 40c / 4 ( 100) =4

10000 -40c = 1600

8400 = 40 c

C = **210**

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