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clara1223

  • one year ago

How to graph f(x) = ax^2 + bx + c when a<0, b^2-4ac=0?

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  1. anonymous
    • one year ago
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    it looks something like this|dw:1438599255017:dw|

  2. clara1223
    • one year ago
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    @saseal Could you please explain?

  3. anonymous
    • one year ago
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    this is a quadratic equation, which makes it a parabola. since a for ax^2 is greater than 0 the parabola opens upward. if b^2-4ac=0, the curve theres only 1 x-intercept, instead of 2 roots

  4. anonymous
    • one year ago
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    c is always your y-intercept

  5. anonymous
    • one year ago
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    you can understand that?

  6. clara1223
    • one year ago
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    @saseal from b^2-4ac how did you find that there is only one root and where c is?

  7. anonymous
    • one year ago
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    |dw:1438600059555:dw||dw:1438600096383:dw||dw:1438600146075:dw|

  8. anonymous
    • one year ago
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    if you remember the y=mx+c equation, c is the y-intercept. thats how we know. c is always the constant number

  9. clara1223
    • one year ago
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    how did you find that the equation moves to the left?

  10. anonymous
    • one year ago
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    this is something you need to figure it out by plugging the numbers in after you solve the equation

  11. clara1223
    • one year ago
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    There are no numbers provided in the question.

  12. anonymous
    • one year ago
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    then x could be anywhere on the line, but your curve should only touch the x-axis once

  13. anonymous
    • one year ago
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    the curve can be anywhere along the x-axis |dw:1438600611313:dw|

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