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Bookworm14

  • one year ago

I am so confused, please help me! (I know HOW to write the equations, but just not when the info is given to me like this :'( )

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  1. Bookworm14
    • one year ago
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  2. anonymous
    • one year ago
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    Vertex form: \(y=a(x-h)^2+k\), where (h, k) is the vertex. Looking at the info they gave you, can you tell what either coordinate for the vertex is?

  3. Bookworm14
    • one year ago
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    (-2, ? ) im sorry this is a new concept for me

  4. anonymous
    • one year ago
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    no problem. The axis of symmetry gives the x-coordinate of the vertex. The maximum (or minimum when you get a problem like that) height gives the y-coordinate. So the vertex is (2, 2)

  5. anonymous
    • one year ago
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    Put that into the equation to get \(y = a(x -2)^2+2\)

  6. anonymous
    • one year ago
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    make sense so far?

  7. Bookworm14
    • one year ago
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    because it is maximum, wont that make it a negative cooridinate or does that not apply here?

  8. anonymous
    • one year ago
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    no. The negative doesn't come into play here. Just take the number as they give it to you. The negative comes in as a check. Since it's a maximum \(a\) is supposed to be negative. But you have to plug in the point (3, 0) to find it.

  9. Bookworm14
    • one year ago
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    okay, i dont know how to plug in the points in runs through, im not really sure about that part

  10. Bookworm14
    • one year ago
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    and yes, i was looking at my chart wrong, i see now that the a is whats supposed to be negative

  11. anonymous
    • one year ago
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    You have \(y=a(x-2)^2+2\) You have to plug in 3 for x and 0 for y, then solve for a. \[0=a(3-2)^2+2\]

  12. Bookworm14
    • one year ago
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    a is our maximum value which is 2, but its negative? so -2?

  13. anonymous
    • one year ago
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    a is not the maximum value. The y-coordinate of the vertex is the maximum value.

  14. Bookworm14
    • one year ago
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    ohhh ok, my bad

  15. anonymous
    • one year ago
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    |dw:1443061442308:dw|

  16. Bookworm14
    • one year ago
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    so the final answer is \[y=2(x-2)^2-2\]

  17. Bookworm14
    • one year ago
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    so how does a=2 ?

  18. anonymous
    • one year ago
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    no, it's not 2 \(0=a(3−2)^2+2\) \(0=a + 2\) \(-2=a\)

  19. Bookworm14
    • one year ago
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    oh ok, i see what you did there. do you think you could help me with one more? (its similar, but worded differently, i will definitely contribute more this time, because i think i understand slightly better now...)

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