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rashley284

if the graph of the quadratic function f(x)=x^2+dx+3d has its vertex on the x-axis, what are the possible values of d? What if f(x)=x^2+3dx-d^2+1? NEED HELP PLEASE..

  • one year ago
  • one year ago

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  1. LogicalApple
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    If a parabola has a vertex on the axis, then that is the only point that touches the axis. This occurs whenever the constant term is 0. Or when 3d = 0 --> d = 0. For the other parabola, set -d^2 + 1 = 0

    • one year ago
  2. LogicalApple
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    Someone else should weigh in

    • one year ago
  3. rashley284
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    How did you do the second problem? i don't get it.

    • one year ago
  4. patdistheman
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    He didn't do the second problem. He only solved the first. Which I think is correct but I am not certain at this point.

    • one year ago
  5. LogicalApple
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    Well my solution isn't a general solution. It does not really answer the general case.

    • one year ago
  6. rashley284
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    Can you do the 2nd problem please?

    • one year ago
  7. patdistheman
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    The first one simplifies to f(x) = x^2 The second one simplifies to f(x) = x^2 + 3 However I am not certain what the answer is to your question at this point.

    • one year ago
  8. LogicalApple
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    Ok I think generally speaking, you would have to use the following reasoning: The x-coordinate of the vertex of a parabola is -b/2a. In the first case, b = d and a = 1 so -b/2a = -d/2 We know that the vertex is at the x-axis, which corresponds with a y value of 0. So f(-d/2) = 0. f(-d/2) = (-d/2)^2+d(-d/2)+3(-d/2) Solving for d here will yield d = 0 and d = -6

    • one year ago
  9. LogicalApple
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    *Solving for d where (-d/2)^2+d(-d/2)+3(-d/2) = 0 will yield d = 0 and d = -6

    • one year ago
  10. patdistheman
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    The vertex for the second equation is not on the x axis. I think it is at 3.

    • one year ago
  11. LogicalApple
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    I apologize I wrote down the wrong equation. The equation should be f(-d/2) = (-d/2)^2+d(-d/2) + 3 = 0 This yields d = 0 and d = 12 This is for the first question.

    • one year ago
  12. LogicalApple
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    Likewise if f(x) = x^2+3dx-d^2+1 Then the x-intercept of the vertex is -b/(2a) = -(3d) / 2 = -3d/2 Set f(-3d/2) = 0 to obtain (-3d/2)^2 + 3 * d * (-3d/2) - d^2 + 1 = 0 This yields d = \[+- \frac{ 2 }{ \sqrt{13} }\]

    • one year ago
  13. patdistheman
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    If d is what y is when x = 0 then d would equal 3.

    • one year ago
  14. LogicalApple
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    I pictured f(x)=x^2+3dx-d^2+1 as a family of functions. The restriction given was that there is a vertex at the x-axis. If there is a vertex at the x-axis. I suggested using the fact that the x-intercept of a parabola lies at x = -b/(2a). Then, knowing that the vertex is on the x-axis, that means f(-b/(2a)) = 0. I could have misinterpreted the question though. The way I have it is if you plug in the values of d obtained, then you will end up with a function that touches the x-axis on its vertex.

    • one year ago
  15. LogicalApple
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    I think d is we are looking for when y = 0, not necessarily when x = 0.

    • one year ago
  16. patdistheman
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    I am unfamiliar with these types of questions but y never touches 0 in the second equation.

    • one year ago
  17. LogicalApple
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    If the second equation is f(x)=x^2+3dx-d^2+1 Then if we let d = + 2/sqrt(13) then we have a function that looks like: f(x) = x^2 + 6/sqrt(13) * x + 9/13 This function does touch the y-axis at x = -3/sqrt(13). This is where its vertex is located.

    • one year ago
  18. patdistheman
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    Is the second formula equal to f(x) = x^2 + 3?

    • one year ago
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