Question

y=1/2x^2-6x-3 what is the vertex/how to find.., what is the axis of symmetry, range and exact value for x intercepts?

Answer 1

= 1/2(x^2 - 12 x - 6)
= 1/2[x - 6)^2 - 36 - 6)]
= 1/2((x-6)^3 - 42)

the vertex is at x = -6 , y = -42* 1/2 = 21 = ( -6,21)

Answer 2

?? i dont get what u did..
oh u comple ted the square?

Answer 3

the x - intercepts occur when y = 0
x^2 - 12x - 6 = 0

Answer 4

how did u get -6,21?

Answer 5

yes - sorry - i should have explained - i did
complete the square
this changed the function to the vertex form - the x cood. of the vertex is the -6 in the parentheses

Answer 6

1/2 * 42

Answer 7

3rd line should be (x - 6)^2 not (x-6)^3

Answer 8

completing the square is stumping me.. it's not adding up to yours for spme reason..

Answer 9

shouldnt it be +36 since -12 halved is -6 squared is positive 36?

Answer 10

go we have
x^2 - 12x - 6
divide -12 by 2 to give -6
= (x - 6)^2 - 6
the expansion of x-6 gives x^2 - 12 + 36 so we need -36
and the -6 gives us the -42

Answer 11

ok i understand vertex.. how do i get the axis of symmetry and range?

Answer 12

the range of the function is >= the lowest value which is -21
range ix [-21,+infinity]

Answer 13

y>-21

Answer 14

the axis of symmetry is the vertical line thru the vertex which is x = 6

i'm sorry - its early in morning and i made a mistake with vertex
(x-6)^3 give x = +6 as the vertex
vertex is at (6,-21) NOT (-6,21)