# Thread: Completing the Square

1. ## Completing the Square

What is the minimum point of the graph of the equation

y = 2x^2 + 8x + 9?

I understand the minimum point is the vertex (h, k).

To complete the square, I set the function to zero first.

2x^2 + 8x + 9 = 0

I then subtracted 9 from both sides.

2x^2 + 8x = -9

I then took half of the second term and squared it.

2x^2 + 8x = 16 - 9

Where do I go from here?

2. Hello !

One way to answer your question is to write
y = 2x² + 8x + 9
y = 2 (x² + 4x + 9/2)
y = 2 ((x+2)²+1/2)

Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1

3. Originally Posted by magentarita
What is the minimum point of the graph of the equation

y = 2x^2 + 8x + 9?

I understand the minimum point is the vertex (h, k).

To complete the square, I set the function to zero first.

2x^2 + 8x + 9 = 0

I then subtracted 9 from both sides.

2x^2 + 8x = -9

I then took half of the second term and squared it.

2x^2 + 8x = 16 - 9

Where do I go from here?
===============================================
magentarita,
I see where your problem is.
2x^2 + 8x = -9 ------ ok so far.
Before you take half of the secod term and square,
Make sure the coefficient of suared term 2x^2 to be 1.
Therefore, 2x^2 + 8x = -9
becomes (x^2) + 4x = -9/2 -- simply didived by 2.
(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get
(x^2) + 4x + 2^2 = (-9/2 ) + 2^2
(x + 2)^2 = -1/2 now complete the square part is done at this point.
Rewrite original question to y = [(x + 2)^2 ]+ 1/2
Can you see y = a ( x-h )^2 + k --- standard form.
Vertex is at ( h, k ) = ( -2, 1/2) and a =1
Opens up vertically, therefor, mini. point is at vertex.

4. ## ok............

Originally Posted by 2976math
===============================================
magentarita,
I see where your problem is.
2x^2 + 8x = -9 ------ ok so far.
Before you take half of the secod term and square,
Make sure the coefficient of suared term 2x^2 to be 1.
Therefore, 2x^2 + 8x = -9
becomes (x^2) + 4x = -9/2 -- simply didived by 2.
(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get
(x^2) + 4x + 2^2 = (-9/2 ) + 2^2
(x + 2)^2 = -1/2 now complete the square part is done at this point.
Rewrite original question to y = [(x + 2)^2 ]+ 1/2
Can you see y = a ( x-h )^2 + k --- standard form.
Vertex is at ( h, k ) = ( -2, 1/2) and a =1
Opens up vertically, therefor, mini. point is at vertex.

ok but none the choices given for the minimum point is (-2, 1/2).

(2,33)

(2,17)

(-2,-15)

(-2,1)

5. ## ok....

Originally Posted by running-gag
Hello !

One way to answer your question is to write
y = 2x² + 8x + 9
y = 2 (x² + 4x + 9/2)
y = 2 ((x+2)²+1/2)

Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1
You got the right answer. But how did you get y = 1?

I got y = 1/2

6. Originally Posted by magentarita
You got the right answer. But how did you get y = 1?

I got y = 1/2
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1

7. Originally Posted by magentarita
ok but none the choices given for the minimum point is (-2, 1/2).

(2,33)

(2,17)

(-2,-15)

(-2,1)
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.

Reason: I forgot to divid y by 2.

Remember orginal question: y =( 2x^2 ) + 8x +9

(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side

(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side

y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side

y = a [ ( x-h)^2 ] + k

where a=2, h=-2, k=1 sorry

8. I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.

From $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.

$y=2x^2+8x+9$ Factor 2 out of first 2 terms

$y=2(x^2+4x)+9$ Next, complete the square in parentheses.

$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side

$y=2(x+2)^2+1$ Now, we're in vertex form.

$V(h, k)=V(-2, 1) = \ \ minimum \ \ point$

9. ## ok.........

Originally Posted by Moo
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1
By subbing -2 for x, I get y = 1.

I got it.

10. ## ok..........

Originally Posted by 2976math
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.

Reason: I forgot to divid y by 2.

Remember orginal question: y =( 2x^2 ) + 8x +9

(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side

(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side

y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side

y = a [ ( x-h)^2 ] + k

where a=2, h=-2, k=1 sorry

Be be sorry, be right. We all make mistakes, right?

11. ## ok..........

Originally Posted by masters
I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.

From $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.

$y=2x^2+8x+9$ Factor 2 out of first 2 terms

$y=2(x^2+4x)+9$ Next, complete the square in parentheses.

$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side

$y=2(x+2)^2+1$ Now, we're in vertex form.

$V(h, k)=V(-2, 1) = \ \ minimum \ \ point$
I enjoyed the steps as a guide. Well-done!