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

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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
 4 years ago
Best ResponseYou've already chosen the best response.0Someone else should weigh in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you do the second problem? i don't get it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0He 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
 4 years ago
Best ResponseYou've already chosen the best response.0Well my solution isn't a general solution. It does not really answer the general case.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you do the 2nd problem please?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ok I think generally speaking, you would have to use the following reasoning: The xcoordinate 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 xaxis, 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
 4 years ago
Best ResponseYou've already chosen the best response.0*Solving for d where (d/2)^2+d(d/2)+3(d/2) = 0 will yield d = 0 and d = 6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The vertex for the second equation is not on the x axis. I think it is at 3.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0Likewise if f(x) = x^2+3dxd^2+1 Then the xintercept 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
 4 years ago
Best ResponseYou've already chosen the best response.0If d is what y is when x = 0 then d would equal 3.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I pictured f(x)=x^2+3dxd^2+1 as a family of functions. The restriction given was that there is a vertex at the xaxis. If there is a vertex at the xaxis. I suggested using the fact that the xintercept of a parabola lies at x = b/(2a). Then, knowing that the vertex is on the xaxis, 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 xaxis on its vertex.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think d is we are looking for when y = 0, not necessarily when x = 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am unfamiliar with these types of questions but y never touches 0 in the second equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If the second equation is f(x)=x^2+3dxd^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 yaxis at x = 3/sqrt(13). This is where its vertex is located.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is the second formula equal to f(x) = x^2 + 3?
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