A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Someone else should weigh in

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

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Can you do the 2nd problem please?

anonymous
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Is the second formula equal to f(x) = x^2 + 3?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.