I am not smart enough to answer your question!
I do apologize
2007-03-07 02:49:37
·
answer #1
·
answered by Mr realistic...believer in truth 6
·
0⤊
0⤋
I haven't done equations like this one in years, but I did find a website that tells you what to do, step by step using a similar problem. Hope it helps.
Some don't answer when a question takes any chunk of time to answer and explain. It is sad to see a teacher not willing to help you, though.
Scroll down to where he states the problem:
Find the vertex of y = 3x2 + 2x – 1
Use the steps he's listed on the left side.
Purple math seems like a nice website to learn from. Link below.
2007-03-07 11:13:03
·
answer #2
·
answered by ? 6
·
0⤊
0⤋
I even checked Mathematica, their answer:
Integrals That Can and Cannot Be Done
Evaluating integrals is much more difficult than evaluating derivatives. For derivatives, there is a systematic procedure based on the chain rule that effectively allows any derivative to be worked out. But for integrals, there is no such systematic procedure.
One of the main problems is that it is difficult to know what kinds of functions will be needed to evaluate a particular integral. When you work out a derivative, you always end up with functions that are of the same kind or simpler than the ones you started with. But when you work out integrals, you often end up needing to use functions that are much more complicated than the ones you started with.
This integral can be evaluated using the same kind of functions that appeared in the input.
In[1]:= Integrate[Log[x]^2, x]
Out[1]= 2 x - 2x Log [x] + x Log [x]sq'd
Humm?
2007-03-07 11:12:25
·
answer #3
·
answered by matrix5.3neo 1
·
0⤊
0⤋
I don't know if this is right, as it has been a while since I tackled this type of problem. But your equation can be simplified down to y=(x+2)(x+4), which means that your two vertex are (-2,0) and (-4,0)...
As I said, I'm not sure! Good luck.
2007-03-07 10:53:43
·
answer #4
·
answered by Belle E 1
·
0⤊
0⤋
Sorry you feel ignored and I would gladly answer your question if I knew how. I don't believe the people on here (at least not most of the) are ignoring you. Maybe they, like me , just don't know the answer. One thing you can do is go to your couselor. She can at least get your teacher to give you some help.
2007-03-07 10:53:06
·
answer #5
·
answered by sarge 6
·
0⤊
0⤋
I found a site that explains this very well. Please click on the link as it has several pages of explanation.
I do not have the math knowledge to personally make sense of this, but perhaps with this help you can understand it better.
If this problem came from a textbook, you can get help at http://hotmath.com/.
2007-03-07 11:03:38
·
answer #6
·
answered by Deb 4
·
0⤊
0⤋
Can't help you as I'm an English teacher, not a math teacher, but sorry that everyone is ignoring you :-)
2007-03-07 10:46:50
·
answer #7
·
answered by ♥♥♥ Mommy to Two ♥♥♥ 5
·
0⤊
0⤋
*= times (multiply)
y=(x+3)*2-1
y=(x+3)*1
y=1x+3
2007-03-07 13:23:17
·
answer #8
·
answered by Arielle K 2
·
0⤊
0⤋
x^2+6x+9-1
x^2+6x+8
(x+2)(x+4)
x=-2,-4
2007-03-07 10:53:42
·
answer #9
·
answered by Dave aka Spider Monkey 7
·
0⤊
0⤋
i'm answering you cuz i feel bad for you, having no answers to your question... but i can't help you a bit!! sorry!
2007-03-07 10:51:15
·
answer #10
·
answered by Anonymous
·
0⤊
0⤋