Solving for Minimum Point of Curve C: x2 + 2x + 4

[discombobulated]

I'm a bit stuck on this question:
the curve C has the equation y=x² + 2x + 4
a) Express x² +2x + 4 in the form a(x+b)² + c and hence the coordinates of the minimum point C.
This is what I've done:
x² +2x +4 = a(x+b)²+c
a(x+b)(x-b) + c = x²+2x+4
a(x² + xb+ xb +b²) + c = x²+2x+4
ax² + 2abx + ab² + c = x² +2x+4
Therefore: a= 1, 2ab= 2, ab= 1, b= 1, ab² + c= 4
1+c = 4
c= 3
(x=0) y= 4
y= 1(x+1)²+ 2
..and that's all I've got so far. Please let me know if it's all wrong and how do i go about getting the minimum point from here.
Thanks!

 

[hotvette]

You are almost there. Take a look at your final equation ($y= 1(x+1)^2+ 2$). What value of x will make y the smallest it can possibly be?

 

[HallsofIvy]

I have a small problem with that. If y= (x+1)²+ 2 then
y= x²+ 2x+ 1+ 2= x²+ 2x+ 3 which doesn't appear to be what you started with!

 

[discombobulated]

sorry, that would be me unable to read my own scribbled notes! it's meant to be a 3.

 

[ivybond]

Standard (and easier way) of completing the square toget a(x+b)² + c is

1x² +2x +4 = 1 (x² +2 x + 1) + 3 =
(x + 1)² + 3.

Graphing (especially by hand) would also give a good idea of the minimum.

Looking into an algebra or precalculus textbook would work too.

 

[VietDao29]

Since you have:
A² ≥ 0.
So (x + 1)² ≥ 0.
Adding 3 to both sides gives:
(x + 1)² + 3 ≥ 3.
So what's the smallest value y can have, what x makes y smallest?
Viet Dao,

 

[discombobulated]

right, so..the minimum point is (-1,3)?

 

[HallsofIvy]

Yes, that's correct. If x=-1, y= 0²+ 3= 3. If x is any number other than -1, x+ 1 is not 0 so (x+1)² is positive and (x+1)²+ 3 is greater than 3.