Tutor.com Quadratic Functions Session

Nov. 21, 2013

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Session Transcript - Live Math Tutoring - MAT/117, 10/11/2013 4:44PM - Tutor.comSession Date: 10/11/2013 4:44PM
Length: 20.3 minute(s)
Subject: Live Math Tutoring - MAT/117


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

RACHEL (Customer)
[00:00:00] find the minimum y value on the graph of y= f(x) f(x)= x^2+4x-1

Carolyn K (Tutor)
[00:00:07] Welcome to Tutor.com ! How are you today?
[00:00:18] Do you think that you know how to start this question?
[00:01:32] I'll only be able to wait for a few more moments. If I don't hear from you soon, I'll need to end this session so I can help other students. You can sign back in when you are ready to continue working.

RACHEL (Customer)
[00:03:14] hey
[00:03:15] sorry
[00:03:21] baby started crying
[00:03:31] i am lost here

Carolyn K (Tutor)
[00:03:47] Okay :) What is the shape of this function?

RACHEL (Customer)
[00:03:55] upward

Carolyn K (Tutor)
[00:04:20] Okay :) It's a parabola opening upward

RACHEL (Customer)
[00:04:23] yes

Carolyn K (Tutor)
[00:04:54] If you think about the graph, at what point is the y value a minimum?
[00:05:46] Right! What do we call this point?

RACHEL (Customer)
[00:06:00] vertex

Carolyn K (Tutor)
[00:06:12] Awesome!
[00:06:23] Do you think that you know how we can find the vertex for this function?

RACHEL (Customer)
[00:06:41] the loweset on the function above
[00:07:43] yes

Carolyn K (Tutor)
[00:07:45] Let's look at this :) There is a formula
[00:07:50] for finding the vertex

RACHEL (Customer)
[00:07:54] ok

Carolyn K (Tutor)
[00:08:04] that involves the a and b from the given function
[00:08:19] Once we have the x component of the vertex
[00:08:29] we can use it to find the minimum y value

RACHEL (Customer)
[00:08:29] ok

Carolyn K (Tutor)
[00:08:47] What do you think we can substitute for b and a in this formula?

RACHEL (Customer)
[00:09:04] 1 and 4
[00:09:07] maybe

Carolyn K (Tutor)
[00:09:34] Great! What will we get?

RACHEL (Customer)
[00:09:38] 5
[00:09:53] well 4

Carolyn K (Tutor)
[00:10:13] I'm not sure what point in the problem you are at....
[00:10:19] did you find this value?

RACHEL (Customer)
[00:10:23] doing the dunction
[00:10:26] fucntion

Carolyn K (Tutor)
[00:10:27] -4/2(1)

RACHEL (Customer)
[00:10:34] -2
[00:11:20] 4+-8
[00:11:23] -4

Carolyn K (Tutor)
[00:11:39] Okay :) Great! So far so good

RACHEL (Customer)
[00:11:42] ok

Carolyn K (Tutor)
[00:11:46] What comes next?

RACHEL (Customer)
[00:11:51] idrk
[00:12:08] -5

Carolyn K (Tutor)
[01:08:53] You got it!
[00:12:27] This is the minimum y value :)
[00:12:40] Did you have any other questions about this one?

RACHEL (Customer)
[00:12:49] -5 is

Carolyn K (Tutor)
[00:13:39] Are you still with me?

RACHEL (Customer)
[00:14:04] yes

Carolyn K (Tutor)
[00:14:21] Okay :) Did you have any other questions about this one?

RACHEL (Customer)
[00:14:30] no ty

Carolyn K (Tutor)
[00:14:37] You're welcome :D

RACHEL (Customer)
[00:14:42] can we do another

Carolyn K (Tutor)
[00:15:04] Yes :) We can work on one more.
[00:15:15] Here's a new board for the next problem.

RACHEL (Customer)
[00:15:39] okay
[00:16:16] it says to write the vertex form of the parabola shown in the graph. assume that a=+/- 1

Carolyn K (Tutor)
[00:16:46] Okay :) What does the graph look like?

RACHEL (Customer)
[00:17:38] there

Carolyn K (Tutor)
[00:17:52] ?

RACHEL (Customer)
[00:18:06] you cant see it

Carolyn K (Tutor)
[00:18:13] no :) sorry
[00:18:27] You might need to click away from it.

RACHEL (Customer)
[00:18:27] h darn
[00:18:34] i did
[00:18:45] I got it! Thanks!
[00:18:49] oops

Carolyn K (Tutor)
[00:18:51] This screen shot shares personal information.
[00:18:58] I will need to end the session.

RACHEL (Customer)
[00:19:09] what?

Carolyn K (Tutor)
[00:19:27] Please be sure to remove all personal information before sharing in a tutoring session.
[00:19:45] I'm not allowed to receive any personal information.
[00:20:04] Please log back in.