1. ## function

1. Find the minimum value of the quadratic y = 2x2 - 8x + 10 by completing the square. Graph the resulting parabola.

Is this correct?

y= 2(x^2-4x)+5
y= 2(x^2-4x+4-4)+5
y= 2(x-2)^2+1
Vertex being (2,1)

If correct, how do I graph this? and which value is the minimum?

2. $\displaystyle y=2x^2-8x+10$
$\displaystyle y=2(x^2-4x+5)$
$\displaystyle y=2(x^2-4x+4)+(10-8)$
$\displaystyle y=2(x^2-4x+4)+2$
$\displaystyle y=2(x-2)^2+2$

So the vertex is (2,2)
Where does the mini occur on a parabolic function?
You can graph this by plotting points with (2,2) as the vertex.

3. Originally Posted by Linnus
$\displaystyle y=2x^2-8x+10$
$\displaystyle y=2(x^2-4x+5)$
$\displaystyle y=2(x^2-4x+4)+(10-8)$
$\displaystyle y=2(x^2-4x+4)+2$
$\displaystyle y=2(x-2)^2+2$

So the vertex is (2,2)
Where does the mini occur on a parabolic function?
You can graph this by plotting points with (2,2) as the vertex.
So the min is 2?
why is it 4 inside the bracket and not 5? I don't understand how you did that.

4. Originally Posted by william
So the min is 2?
why is it 4 inside the bracket and not 5? I don't understand how you did that.
Oh haha sorry. It is a bit confusing.
Here is a less confusing work

$\displaystyle y=2x^2-8x+10$
$\displaystyle y=2(x^2-4x)+10$
$\displaystyle y=2(x^2-4x+4)+(10-8)$
$\displaystyle y=2(x^2-4x+4)+2$
$\displaystyle y=2(x-2)^2+2$

5. Originally Posted by Linnus
Oh haha sorry. It is a bit confusing.
Here is a less confusing work

$\displaystyle y=2x^2-8x+10$
$\displaystyle y=2(x^2-4x)+10$
$\displaystyle y=2(x^2-4x+4)+(10-8)$
$\displaystyle y=2(x^2-4x+4)+2$
$\displaystyle y=2(x-2)^2+2$
Thanks.
So the min is 2, correct?

6. Originally Posted by william
Thanks.
So the min is 2, correct?
Yes, the min is 2.