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For the function y = x^2 - 6x + 8, how do I put in the form y = a(x - h)^2 + k? Please show me how it should look and then explain why you did what you did please?

2007-03-29 03:36:57 · 3 answers · asked by Timberwolf 3 in Science & Mathematics Mathematics

Also, can you help me find out what the equation for the line of symmetry is for the graph of this function?

2007-03-29 03:39:01 · update #1

3 answers

First multiply the equation out

a(x - h)^2 + k

= a(x-h)(x-h) +k

= a(x^2 - 2hx + h^2) + k

= ax^2 - 2ahx +ah^2 + k = x^2 - 6x + 8

Therefore a = 1 (coefficient of x^2 must match)

and 2ah = 6 (coefficients of x must match)

2h=6, h= 3

ah^2 + k = 8 (the number must balance)

1*3*3 +k =8

9 + k =8

k =-1

The line of symmetry is x = h or x = 3 in this case.

2007-03-29 03:44:15 · answer #1 · answered by The exclamation mark 6 · 1 0

This process hinges on the coefficient of x, which is -6 in this case. If the coefficient of x is b, and the coefficient of x^2 is 1, which it is in this case, then h = -b/2. So h = -(-6)/2 = 3. In addition, a is just the coefficient of x^2, which is 1 as already observed. So y = (x - 3)^2 + k. Since (-3)^2 = 9 and the constant term of the polynomial is 8, we must have k = -1. The final result is y = (x - 3)^2 - 1. The line of symmetry is x = h, or x = 3 in this case.

2007-03-29 10:42:14 · answer #2 · answered by DavidK93 7 · 0 0

x^2 -6x + 8 by completing the square
General form
ax^2 - bx + c = x^2 - (b/a)x + (b/2a)^2 - (b/2a)^2 +c

x^2 -6x + 8 = x^2 - 6x + (6/2)^2 - (6/2)^2 + 8
factoring x^2 - 6x + 3^2 = (x - 3)^2
(x - 3)^2 - (3^2) + 8 = (x - 3)^2 - 9 + 8
(x - 3)^2 -1 comparising a(x - h)^2 + k
a = 1, h = 3, k = -1

2007-03-29 10:46:51 · answer #3 · answered by Azanuddin m 2 · 0 0

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