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Forums - General Discussion - I have a math question

dtewi said:
So the vertex is the point where the parabola becomes symmetrical over the x-axis?

The vertex is the minimum point (if the parabole is concave to the upside, or "happy"), or the maximum point (if the parabole is concave to the downside, or "sad"). In your example, 0,0




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dtewi said:
So the vertex is the point where the parabola becomes symmetrical over the x-axis?

 

Yes, but only in the parabola case. Every geometrical figure has vertixes.



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zexen_lowe said:
dtewi said:
So the vertex is the point where the parabola becomes symmetrical over the x-axis?

The vertex is the minimum point (if the parabole is concave to the upside, or "happy"), or the maximum point (if the parabole is concave to the downside, or "sad"). In your example, 0,0

 

Happy and sad, LOL. What do we do, in order to memoryze it

 

remember, if it is "happy", a>0. If "sad", a<0.



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Okay then. Thank you.



Kimi wa ne tashika ni ano toki watashi no soba ni ita

Itsudatte itsudatte itsudatte

Sugu yoko de waratteita

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No prob. Anything else, just ask me or Mr. Niko Lowe (Zexen_lowe)



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Okay then. Another question para usted.

I have trouble factoring quadratic equations wanna help?

For example, how would I factor 2x^2 - 16x + 4 = 0?



Kimi wa ne tashika ni ano toki watashi no soba ni ita

Itsudatte itsudatte itsudatte

Sugu yoko de waratteita

Nakushitemo torimodosu kimi wo

I will never leave you

dtewi said:
Okay then. Another question para usted.

I have trouble factoring quadratic equations wanna help?

For example, how would I factor 2x^2 - 16x + 4 = 0?

Use the almighty quadratic formula: (-b +/- (b^2 - 4ac) ^ (1/2)) / 2a

That is

a=2
b=-16
c=4

Then it's: (16+/- (256-32)^ (1/2)) / 4 = (16 +/- 14.96)/4 = approx 31/4 and approx 1/4

So it results= 2 (x-31/4) (x-1/4)




Okay then. Using the quadratic equation is a nice way to factor.

But what if I want to use the shorter way instead of all that radicaling and squaring and dividing. Factoring it straight into a binomial.



Kimi wa ne tashika ni ano toki watashi no soba ni ita

Itsudatte itsudatte itsudatte

Sugu yoko de waratteita

Nakushitemo torimodosu kimi wo

I will never leave you

I don't think Product & sum method would work for factoring that...
So quad might be best option



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dtewi said:
Okay then. Using the quadratic equation is a nice way to factor.

But what if I want to use the shorter way instead of all that radicaling and squaring and dividing. Factoring it straight into a binomial.

Well, other option is to use Gauss' theorem (in a polynomial of integer coefficients, if there's a rational root, then the numerator divides the independent coefficient (the last one) and the denominator divides the principal coefficient (the first one)) so if you find a quick root, you can divide the polynomial by x-root and quickly factorize it. But I think that takes more time, and it's not guaranteed to get you an answer (if there are no rational roots, Gauss' theorem doesn't do jack). The utility of it is when the polynomial is of grade 3 or higher, since in grade 2 the quadratic equation always work.

Of course, there are more rules, but they only work on some polynomials, like a perfect trinomial (x^2+2x+1) can be factored easily into (x+1)^2, and some I don't remmeber. But when you grow up and learn more math, you only keep what's really useful, therefore the only factorization methods I remember are the quadratic equation and Gauss' theorem