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
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?
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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.
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
I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.
From , you want to convert to , where represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.
Factor 2 out of first 2 terms
Next, complete the square in parentheses.
Notice that we added 2 times 4, so we subtract 8 to balance the side
Now, we're in vertex form.