express x^2 - 6x + 10 in the form (x+a)^2+c and find minimun value of x^2-6x+10
2) http://i25.tinypic.com/263ezus.jpg
tyty
Originally posted by yiha093:express x^2 - 6x + 10 in the form (x+a)^2+c and find minimun value of x^2-6x+10
2) http://i25.tinypic.com/263ezus.jpg
tyty
(x-3)^2 + 1 i and min value should be 3 i think
2i) Not sure whether any circle properties can be used if not using sine rule,
r/sin30 = (r+5)/1
2r=r+5
r=5
ii) XOV = 60, XOY = 120 so OXY = 30
iii) XZY = 60 = XTY (angles in same segment)
iv) VYT = 90 - 40 -30 = 20 = VZT (angles in same segment)
v) ZVY = 30 (angles in same segment), VRY = 180 - 30 - 20 = 130
Originally posted by dkcx:(x-3)^2 + 1 i and min value should be 3 i think
2i) Not sure whether any circle properties can be used if not using sine rule,
r/sin30 = (r+5)/1
2r=r+5
r=5ii) XOV = 60, XOY = 120 so OXY = 30
iii) XZY = 60 = XTY (angles in same segment)
iv) VYT = 90 - 40 -30 = 20 = VZT (angles in same segment)
v) ZVY = 30 (angles in same segment), VRY = 180 - 30 - 20 = 130
(x-3)^2 + 1 i and min value should be 3 i think
can teach this 1? ?
Hi,
When we express x^2 - 6x + 10 in the form (x + a)^2 + c, we are completing the square of a quadratic expression.
In general, x^2 + px + q can be written as (x + p/2)^2 + q - (p/2)^2, and this fact should be committed to memory.
By completing the square of a quadratic equation (i.e. y = x^2 + px + q), we will be able to find its minimum y-value, which will be q - (p/2)^2. The x-coordinate of this minimum turning point is x = -p/2.
For Q1,
x^2 - 6x + 10 = (x - 3)^2 + 10 - 9
= (x - 3)^2 + 1,
so that its minimum value is 1, and this occurs at x = 3.
Hope these help to clarify! Thanks!
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
When we express x^2 - 6x + 10 in the form (x + a)^2 + c, we are completing the square of a quadratic expression.
In general, x^2 + px + q can be written as (x + p/2)^2 + q - (p/2)^2, and this fact should be committed to memory.
By completing the square of a quadratic equation (i.e. y = x^2 + px + q), we will be able to find its minimum y-value, which will be q - (p/2)^2. The x-coordinate of this minimum turning point is x = -p/2.
For Q1,
x^2 - 6x + 10 = (x - 3)^2 + 10 - 9
= (x - 3)^2 + 1,
so that its minimum value is 1, and this occurs at x = 3.
Hope these help to clarify! Thanks!
Cheers,
Wen Shih
eh. my fren also taught me another way
is to expand the (x+a)^2
into x^2+ 2ax+a^2 +c or sth like tat
then we compare(x^2-6x+10) with that eqn
?
Hi,
That is another alternative :)
Cheers,
Wen Shih
Originally posted by yiha093:eh. my fren also taught me another way
is to expand the (x+a)^2
into x^2+ 2ax+a^2 +c or sth like tat
then we compare(x^2-6x+10) with that eqn
?
this is a shortcut my teacher just told me this week lol!
all this time i have been using the formula for the completing the square... haha!
okok, thx guys~
xD
Originally posted by ItchyArmpit:this is a shortcut my teacher just told me this week lol!
all this time i have been using the formula for the completing the square... haha!
but actually using formulae is much quicker and less error prone than using expansion of terms.
