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anonymous

  • 5 years ago

y=2x^2-8x+7 i have to find the axis of symmetry the coordinates of the vertex and tell whether the vertex is a maximum or a minimum

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  1. amistre64
    • 5 years ago
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    -b/2a ia axis of symmetry --8/2.2 8/4 = 2

  2. amistre64
    • 5 years ago
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    ia is stupidity for "is" ...

  3. anonymous
    • 5 years ago
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    haha thank you

  4. anonymous
    • 5 years ago
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    \[y = 2x^2 - 8x + 7\] \[ = 2(x^2 - 4x) + 7\] \[= 2(x^2 -4x + 4) -8 + 7\] \[\implies y +1 = 2(x-2)^2\]

  5. anonymous
    • 5 years ago
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    So we have a parabola which is shifted 1 unit up in y and 2 units forward in x and grows twice as fast as the standard parabola.

  6. anonymous
    • 5 years ago
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    Err shifted 1 unit down in y, sorry.

  7. anonymous
    • 5 years ago
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    oh ok u got me confused 4 a sec

  8. anonymous
    • 5 years ago
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    what part of the problem are u helping with

  9. anonymous
    • 5 years ago
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    Finding the vertex, and the axis of symmetry.

  10. anonymous
    • 5 years ago
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    Actually this form of the function answers all of those questions.

  11. anonymous
    • 5 years ago
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    i did not learn it that way can u help me another way by chance

  12. anonymous
    • 5 years ago
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    Since the coefficient of x is positive, the graph is opening upward. Therefore the vertex will be a min. The location of the vertex can be found from the shift in x and y of the function from (0,0). In this case we shifted down 1 unit and 2 units to the right. \[y-y_0 = m(x-x_0)^2\] Is the standard form for a parabola. Where \(x_0,y_0\) are the x and y points of your vertex. And m is the scaling and directional coefficient. So we have \(y - (-1) = 2(x - 2)^2\) which tells us the vertex is at (2,-1) And since the graph is opening upward the line x = 2 will be the line of symmetry (the line going up/down that goes through the vertex)

  13. anonymous
    • 5 years ago
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    thank you soooooo much

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