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Algebra II

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Mon May 24, 2010 4:48 pm
Loller65 says...



Honors Algebra 2. Graphing parabolas. How do I find equations in vertex form?
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Mon May 24, 2010 10:04 pm
Snoink says...



Basically? You have to change:

y = ax^2 + bx + c

into the form:

y = a(x - h)^2 + k

To make this slightly more manageable, let's do a little bit of factoring!

y = a(x^2 + (b/a)x) + c

Still a little freaky... but it's looking better!

Now, let's pretend that our equation only is x^2 + (b/a)x. We have to turn this to (x - h)^2. So let's set it equal!

x^2 + (b/a)x = (x - h)^2

Now, (x-h)^2 in its expanded form is x^2 - 2hx + h^2. So:

x^2 + (b/a)x = x^2 - 2hx + h^2

Looking good! So now, let's subtract the x^2...

(b/a)x = -2hx + h^2

Not good! This show us that we need another term for the other side in order for this to be equivalent! Otherwise, we can't get a constant term! So we're going to add something to this later. Now, let's ignore the h^2 term. We'll figure it out later.

(b/a)x = -2hx

Solve for h!

h = - b/(2a)

So you now now your h!

Now.

y = a(x^2 + (b/a)x + h^2) + c - ah^2

Solve! You will get:

a(x - h)^2 + (c - ah^2)

This is in vertex form. :)

(Side note: When you learn calculus, this whole process will be much easier.)
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Tue May 25, 2010 1:48 am
Loller65 says...



Uh....right. Still confused. I'm a regular mathlexic....
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Tue May 25, 2010 4:13 am
Snoink says...



Hahaha... yeah, it can get a bit confusing, especially with all the letters in there!

Do you have a specific example I can use? Maybe that will help. :)
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Sun Aug 15, 2010 2:24 am
seeminglymeaningless says...



You're crazy, Snoink. But you look like you know what you're doing :P

Could I ask for some help? Algebra, first year university, second semester unit.

The last question of the assignment is ridiculous, as we haven't even gone over it in lectures, so I think it was meant to be a research question or something.

"Use Newton's Method to find the solutions of the equation xlnx = 6, correct to six decimal places.

After a bit of wikipedia-ing, I got as far as:

xlnx = 6
f(x) = xlnx - 6
f'(x) = 1/x

and substituted a range of numbers (5 and 1 amongst them) into the formula from the wiki page (below), and kept on getting really varied answers that swung from -5 --> 8 -- > -104 and 7 --> -46 --> 7616. Just too varied and weird to keep on trying.

x1 = x0 - [f(x0)/f'(x0)] = X1
x2 = x0 - [f(x1)/f'(x1)] = X2
x3 = x0 - [f(x2)/f'(x2)] = X4
etc etc

Am hoping you know how to tackle something like this, but if you don't, that's fine :)




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Sun Aug 15, 2010 6:01 am
Snoink says...



XD

Don't you love it when teachers assign you calculus in an algebra class? XD

Also... attached! Download the PDF. Hopefully, it won't be confusing. If you have any questions, either PM me or post something on this thread. But make sure to say my name.
Attachments
seemingmath.pdf
Teh file of DOOM.
(111.77 KiB) Downloaded 16 times
Ubi caritas est vera, Deus ibi est.

"The mark of your ignorance is the depth of your belief in injustice and tragedy. What the caterpillar calls the end of the world, the Master calls the butterfly." ~ Richard Bach

Moth and Myth <- My comic! :D




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Sun Aug 15, 2010 6:18 am
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seeminglymeaningless says...



Hey Snoink! You're fantastic :)

:/ Sorry, it's Algebra & Calculus MAB120, I didn't mean to sound as if I didn't know what I was doing :P I understood how to use Newton's Method, I just supplied it there for anyone who wanted to give it a shot.

Basically, that was the last question, so fair enough it'd be hard.

I know you used a calculator to differentiate xlnx - 6, but... When I differentiate that in my head, the first x turns into 1, the ln(x) becomes 1/x (as per log rules), and the 6 becomes nothing.

So:

f(x) = xln(x) - 6
f'(x) = 1/x

I'm so confused, you got: 1 + ln(x)...

Waaaaaaaaaait...

*slaps forehead*

Product rule, right?

u = x
u' = 1
v = ln(x)
v' = 1/x

dy/dx = u'v + v'u
= 1ln(x) + x(1/x)
= ln(x) + 1

*slaps forehead again*

I really need to remember that x, and ln or e^x are all separate functions, not just one when in a mash like 2xe^2x.

Aaaaanyway, yes, I can see that the rest of what you did is perfectly correct.

Thank you so much :) Did you do math at uni or something?

- Jai




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Sun Aug 15, 2010 6:29 am
Snoink says...



Yes! That's the correct way of doing it. :)

I'm a biochemical engineer, so I had to do calculus, differential equations (ordinary and partial), and linear algebra. So I like math!
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Sun Aug 15, 2010 6:52 am
Lava says...



seeminglymeaningless wrote:
*slaps forehead*

Product rule, right?


:D That was one of the things I always forgot, when we first did differentials.
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Sun Aug 15, 2010 7:00 am
seeminglymeaningless says...



Wow, a biochemical engineer sounds like hard work - where do you find the time to be a very active moderator on YWS?

Also, there was another question that I couldn't quite do, if you were bored and felt like some more math. No need for you to do it though, just thought I'd share.

"Find the equations to the tangents to the curve whose equation is y^3 + yx^2 + x^2 - 3y^2 = 0, at the points where y = 1."

As I saw no other way, I used implicit differentiation and wound up with a huge mess and an answer at the end I don't know what to do with :3

0 = y^3 + yx^2 + x^2 - 3y^2
0 = 3y^2 + d/dx(yx^2) + 2x - 6y

u = y
u' = 1
v = x^2
v' = 2x

0 = 3y^2 + (x^2 + 2yx)dy/dx + 2x - 6y
dy/dx = -3y^2 - x^2 - 2yx - 2x + 6y

y = 1
= -3(1)^2 - x^2 - 2(1)x - 2x + 6(1)
= -x^2 - 4x + 3

AND OF COURSE... !!!! *RAGES*

DUH *slaps self again*

Looking over it again, now I have to factorize...

= (-x - 1)(x + 3)

So x = -1, x = -3.

Right? *hopes I'm right* :P

- Jai

(Edit for own failing at simple math)

EDIT: Yeah Lava :P It's not a hard rule, it's just that when xlnx is there for me to see, I see it as one function, instead of the two that it is *shakes fist at person who created product rule*




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Sun Aug 15, 2010 7:23 am
Lava says...



Seem, yup. That's how you do it. Though, since the question is just "Find the equations..." you could stop with

y= -x^2 - 4x + 3

Though of course, The x values correspond to the point of the tangent.
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Sun Aug 15, 2010 8:04 am
seeminglymeaningless says...



Hey Lava, it says "find the equations" does that mean I should have more than the final one I ended up with? o.O

Because if so, I have no idea what other equations I can get out of that.

Unless I differentiate again? Or substitute? Oh, right, omg, why do I only figure things out when I ask for help?

y= -x^2 - 4x + 3
dy/dx = -2x -4
x = -1
y = -2(-1) -4
= -1

y= -x^2 - 4x + 3
dy/dx = -2x -4
x = -3.
y = -2(-3) -4
= -2

Now what I am I meant to do with these values? o.O

That doesn't even make sense, isn't y supposed to be 1?




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Tue Aug 17, 2010 9:26 pm
Kyllorac says...



You only want the solutions where y = 1. I assume by equation(s), they want f(x) = -x - 1 and f(x) = x + 3.
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