put function y=x2-6x+8 in the form y=a(x-h)2+k?

Answer 1

y= (1)(x- 3)2+(-1)

x2 - 3x + 9 -1

a= 1
h= 3
k= -1

Answer 2

We first set the two values of y =.
x^2 - 6x + 8 = a(x-h)^2 + k
we then expand the binomial square
1x^2 - 6x + 8 = a(x^2 - 2hx + h^2) + k
we distribute the a through the trinomial
1x^2 - 6x + 8 = ax^2 - 2ahx +ah^2 + k
we next group coefficients for the terms of the trinomials
(1)x^2 - (6)x + (8) = a(x^2) - (2ah)x +(ah^2 + k) ==>
if these two equations are equal the coefficients of each term must be equal so we set:
1 = a, 6 = 2ah, and 8 = ah^2 + k which now gives us a
substituting a= 1 in the second we get 6 = 1(2h) ==> 3 = h
substituting a and h in the third we get 8 = 1(3^2) + k ==>
8 = 9 + k ==> -1 = k using the three values we get the second equation as:
a(x-h)^2 + k = 1(x - 3)^2 + (-1)

Answer 3

hi man to answer this kind of maths problems sympol (a) represents the coefficient of (X2) in the given equatin and here it equals 1 the h represents half of the coefficient of x which equals 6/2=3 and about the k it is deduced by comparing ur absolute coefficient with the 8 if u solved the squared bracket u will get h2=9 so the k has to be equals -1 to make the 9 get back to 8 anyway ur answer will b y=(x-3)2 -1

Answer 4

to transform to that variety, you ought to complete the sq. for the x element: x^2 -6x +8 turns into (x-3)^2 +a million =y a. the line of symmetry is in the process the vertex of the parabola or the x-coordinate of the vertex (3,a million) so x=3 b. the graph is shifted over by using 3 gadgets to the properly suited and one unit up c. properly i could graph it, yet purely plug it intoyour TI 89

Answer 5

y= (x-3)^2-1

Answer 6

y=(x-3)^2-1

Answer 7

yet another homework i guess:
y = (x-3)^2 -9