# Minimum Point

1. Oct 17, 2005

### 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!

2. Oct 17, 2005

### 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?

Last edited: Oct 17, 2005
3. Oct 17, 2005

### HallsofIvy

Staff Emeritus
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!

4. Oct 17, 2005

### discombobulated

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

5. Oct 17, 2005

### 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.

Last edited: Oct 17, 2005
6. Oct 18, 2005

### 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,

7. Oct 18, 2005

### discombobulated

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

8. Oct 18, 2005

### HallsofIvy

Staff Emeritus
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.