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anonymous

  • one year ago

Using the completing-the-square method, find the vertex of the function f(x) = 2x2 − 8x + 6 and indicate whether it is a minimum or a maximum and at what point. Maximum at (2, –2) Minimum at (2, –2) Maximum at (2, 6) Minimum at (2, 6)

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  1. Mehek14
    • one year ago
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    since the first term is positive = \(2x^2\) that means it is opening upwards so the vertex would be the minimum that eliminates A and C

  2. Mehek14
    • one year ago
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    to find the vertex, use \(\dfrac{-b}{2a}\) in your equation, you have b = -8 a = 2 \(\dfrac{-(-8)}{2*2}=\dfrac{8}{4}=2\) so x = 2

  3. Mehek14
    • one year ago
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    plug x = 2 into the equation

  4. Mehek14
    • one year ago
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    @boots_2000 can you do that?

  5. anonymous
    • one year ago
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    Yeah so is it maximum 2,6?

  6. anonymous
    • one year ago
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    @Mehek14

  7. IrishBoy123
    • one year ago
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    it says "Using the completing-the-square method"

  8. Mehek14
    • one year ago
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    no did you plug in x = 2 in the equation

  9. anonymous
    • one year ago
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    yeah

  10. welshfella
    • one year ago
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    using completing the square:- = 2(x^2 - 4x + 3) = 2[(x - 2)^2 - c + 3] can you tell me the value of c - can you remember from the last post?

  11. Mehek14
    • one year ago
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    \(f(2)=2*2^2-8*2+6\\2^2=4\\2*4=8\\8*2=16\\8-16=-8\) add \(-8+6\)

  12. IrishBoy123
    • one year ago
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    +1 @welshfella

  13. anonymous
    • one year ago
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    so its minimum? lol

  14. Mehek14
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Mehek14 since the first term is positive = \(2x^2\) that means it is opening upwards so the vertex would be the minimum that eliminates A and C \(\color{#0cbb34}{\text{End of Quote}}\)

  15. welshfella
    • one year ago
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    its a minimum but what are the coordinates at the minimum?

  16. anonymous
    • one year ago
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    2,6

  17. Mehek14
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Mehek14 \(f(2)=2*2^2-8*2+6\\2^2=4\\2*4=8\\8*2=16\\8-16=-8\) add \(-8+6\) \(\color{#0cbb34}{\text{End of Quote}}\)

  18. welshfella
    • one year ago
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    No mehek has worked the y coordinate for you you can also get from the completing the square method

  19. anonymous
    • one year ago
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    so 2,-2

  20. Mehek14
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    could you guys come and help me quick when you're all done here?

  22. welshfella
    • one year ago
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    I'll carry from where i left off 2[(x - 2)^2 - c + 3] c = 4 because -2^2 = + 4 = 2(x - 2)^2 - 1) = 2(x - 2)^2 - 2 so the vertex is at ( 2,-2) you get this by comparing your expression with the genarl form for the vertex

  23. anonymous
    • one year ago
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    please

  24. welshfella
    • one year ago
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    |dw:1440256991772:dw|

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