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rashley284
Group Title
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..
 one year ago
 one year ago
rashley284 Group Title
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..
 one year ago
 one year ago

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LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
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

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
Someone else should weigh in
 one year ago

rashley284 Group TitleBest ResponseYou've already chosen the best response.0
How did you do the second problem? i don't get it.
 one year ago

patdistheman Group TitleBest ResponseYou've already chosen the best response.0
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

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
Well my solution isn't a general solution. It does not really answer the general case.
 one year ago

rashley284 Group TitleBest ResponseYou've already chosen the best response.0
Can you do the 2nd problem please?
 one year ago

patdistheman Group TitleBest ResponseYou've already chosen the best response.0
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

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
Ok 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
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
*Solving for d where (d/2)^2+d(d/2)+3(d/2) = 0 will yield d = 0 and d = 6
 one year ago

patdistheman Group TitleBest ResponseYou've already chosen the best response.0
The vertex for the second equation is not on the x axis. I think it is at 3.
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
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

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
Likewise 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 = \[+ \frac{ 2 }{ \sqrt{13} }\]
 one year ago

patdistheman Group TitleBest ResponseYou've already chosen the best response.0
If d is what y is when x = 0 then d would equal 3.
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
I 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.
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
I think d is we are looking for when y = 0, not necessarily when x = 0.
 one year ago

patdistheman Group TitleBest ResponseYou've already chosen the best response.0
I am unfamiliar with these types of questions but y never touches 0 in the second equation.
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.1
If 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.
 one year ago

patdistheman Group TitleBest ResponseYou've already chosen the best response.0
Is the second formula equal to f(x) = x^2 + 3?
 one year ago
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