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Math Help - Completing the Square

  1. #1
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    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?
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  2. #2
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    Hello !

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

    Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1
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  3. #3
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    Quote Originally Posted by magentarita View Post
    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.


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

    Quote Originally Posted by 2976math View Post
    ===============================================
    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)
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  5. #5
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    ok....

    Quote Originally Posted by running-gag View Post
    Hello !

    One way to answer your question is to write
    y = 2x + 8x + 9
    y = 2 (x + 4x + 9/2)
    y = 2 ((x+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
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  6. #6
    Moo
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    Quote Originally Posted by magentarita View Post
    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) + 1/2 ] = 2 [ 0+1/2 ] = 1
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  7. #7
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    Quote Originally Posted by magentarita View Post
    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




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  8. #8
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    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
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  9. #9
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    ok.........

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

    I got it.
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  10. #10
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    ok..........

    Quote Originally Posted by 2976math View Post
    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?
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  11. #11
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    ok..........

    Quote Originally Posted by masters View Post
    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!
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