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anonymous

  • one year ago

What is the equation of the axis of symmetry of the graph of y + 3x – 6 = –3(x – 2)2 + 4?

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  1. campbell_st
    • one year ago
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    well you need to distribute and collect like terms so the curve is in the form \[y = ax^2 + bx + c\] then the line of symmetry is \[x = \frac{-b}{2 \times a}\] hope it helps

  2. anonymous
    • one year ago
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    i know i need to simplify the equation, how do i do this?

  3. campbell_st
    • one year ago
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    well start with the perfect square \[(x -2)^2 =?\] what does that become

  4. anonymous
    • one year ago
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    x squared minus 4?

  5. campbell_st
    • one year ago
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    nope \[(x -2)^2 = (x -2) \times (x -2) \] which can be written as \[x(x -2) - 2(x-2)\] can you distribute this..?

  6. anonymous
    • one year ago
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    x^2-2x-2x+4?

  7. campbell_st
    • one year ago
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    great so its \[x^2 - 4x + 4\] so you now have \[y + 3x - 6 = -3(x^2 - 4x + 4) + 4\] so distribute the -3

  8. anonymous
    • one year ago
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    -3x^2+12x-12+4

  9. anonymous
    • one year ago
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    then subtract the 3x from 12x?

  10. campbell_st
    • one year ago
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    that's good so its \[y +3x - 6 = -3x^2 +12x -8\] next add 6 to both sides of the equation

  11. campbell_st
    • one year ago
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    then lastly subtract 3 from both sides of the equation

  12. campbell_st
    • one year ago
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    oops 3x

  13. campbell_st
    • one year ago
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    so you will then have the coefficients for a and b and can find the line of symmetry

  14. campbell_st
    • one year ago
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    an alternative method is to start by factoring the right side \[y + 3(x -2) = -3(x -2)^2 + 4\] subtract 3(x -2) from both sides \[y = -3(x -2)^2 -3(x -2) + 4\] so the coefficients are b = -3 and a = -3 and you are solving for (x -2) so the line of symmetry \[x -2 = \frac{-(-3)}{2 \times -3}\] which is \[x - 2 = \frac{-1}{2}\] now solve for x

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