anonymous
  • anonymous
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..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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
anonymous
  • anonymous
Someone else should weigh in
anonymous
  • anonymous
How did you do the second problem? i don't get it.

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anonymous
  • anonymous
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.
anonymous
  • anonymous
Well my solution isn't a general solution. It does not really answer the general case.
anonymous
  • anonymous
Can you do the 2nd problem please?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
*Solving for d where (-d/2)^2+d(-d/2)+3(-d/2) = 0 will yield d = 0 and d = -6
anonymous
  • anonymous
The vertex for the second equation is not on the x axis. I think it is at 3.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 = \[+- \frac{ 2 }{ \sqrt{13} }\]
anonymous
  • anonymous
If d is what y is when x = 0 then d would equal 3.
anonymous
  • anonymous
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.
anonymous
  • anonymous
I think d is we are looking for when y = 0, not necessarily when x = 0.
anonymous
  • anonymous
I am unfamiliar with these types of questions but y never touches 0 in the second equation.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Is the second formula equal to f(x) = x^2 + 3?

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