## rashley284 Group Title 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

1. LogicalApple Group Title

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

2. LogicalApple Group Title

Someone else should weigh in

3. rashley284 Group Title

How did you do the second problem? i don't get it.

4. patdistheman Group Title

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.

5. LogicalApple Group Title

Well my solution isn't a general solution. It does not really answer the general case.

6. rashley284 Group Title

Can you do the 2nd problem please?

7. patdistheman Group Title

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.

8. LogicalApple Group Title

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

9. LogicalApple Group Title

*Solving for d where (-d/2)^2+d(-d/2)+3(-d/2) = 0 will yield d = 0 and d = -6

10. patdistheman Group Title

The vertex for the second equation is not on the x axis. I think it is at 3.

11. LogicalApple Group Title

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.

12. LogicalApple Group Title

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 = $+- \frac{ 2 }{ \sqrt{13} }$

13. patdistheman Group Title

If d is what y is when x = 0 then d would equal 3.

14. LogicalApple Group Title

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.

15. LogicalApple Group Title

I think d is we are looking for when y = 0, not necessarily when x = 0.

16. patdistheman Group Title

I am unfamiliar with these types of questions but y never touches 0 in the second equation.

17. LogicalApple Group Title

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.

18. patdistheman Group Title

Is the second formula equal to f(x) = x^2 + 3?