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Mathematics 19 Online
OpenStudy (umulas):

Create your own quadratic equation and demonstrate how it would be solved by graphing, factoring, the quadratic formula, and by completing the square

OpenStudy (umulas):

Is there any way that a person can show the picture of the graph, that's my only problem, I've been doing this question for an hour now and still stuck.

OpenStudy (anonymous):

So did you already form an equation then?

OpenStudy (umulas):

x^2+7x+12 but the factor i did which is (x+3)(x+4), meaning x = -3, -4

OpenStudy (anonymous):

Ok. First, I assume that you know that this is in the form \(y = ax^{2} + bx + c ok?\) To find the vertex, use the formula \(-\frac{b}{2a}\) to get the x-value of the vertex. Then, you plug it back in for x. To graph it, pick numbers close to what you get when using \(-\frac{b}{2a}\) and plug it in for x. What you get is that point's y-value. Then just connect the points. If you need a demonstration, let me know :)

OpenStudy (anonymous):

That in the first line should be \(ax^{2} + bx + c\)

OpenStudy (umulas):

Hmm, I gotten -7/2, did I do that correct?

OpenStudy (anonymous):

Yes. Sorry I was so late :) That is the x-value of the vertex. Plug it back in to solve for y.

OpenStudy (umulas):

x^2+7x+12 y = (7/2)^2 + 7(7/2) +12 Crap, I'm getting more confused, can you guys give me a demonstration that's more simplier?

OpenStudy (anonymous):

y = x^2 + 2x + 1 -b/2a = -1 y = (-1)^2 + 2(-1) + 1 y = 1 - 2 + 1 Vertex is (-1, 0) Plug in these points in for x. x = -2, -3, -4, 0, 1, 2 y = ?

OpenStudy (umulas):

D:, why didn't I think of that! Thanks!

OpenStudy (anonymous):

np :)

OpenStudy (anonymous):

If you want, you can try changing your equation to mine because it's a lot easier.

OpenStudy (umulas):

Thanks: Quadractic functioning: -b sqrt +- b^2 - 4ac ------------------ 2a A: 1 B: 2 -2 sqrt +- 2^2 -4(1)(1) C: 1 -------------------- 2(1) -2 sqrt +- 4 - 4 = 0 -2 sqrt 0 -------- 2 so it's -1 right?

OpenStudy (anonymous):

Quadfratic Formula = \(\huge x = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a}\)

OpenStudy (anonymous):

Which equation are you using?

OpenStudy (umulas):

The quadractic formaula

OpenStudy (anonymous):

I know, but are you using my example or yours?

OpenStudy (umulas):

I am using your example, it's way more easire and better to understand.

OpenStudy (anonymous):

Alright. Yeah. You are correct. There is a double root by the way.

OpenStudy (umulas):

Alright just one more :P, the completing the square and I complete the question :P y = x^2 + 2x + 1 y - 1 = x^2 + 2x 2 /2 = 1^2 = 1 y - 1 + 1 = x^2 + 1 y = x^2 + 1 Is this correct "P

OpenStudy (anonymous):

No. \[y = x^{2} + 2x + 1\]\[y - 1 = x^{2} + 2x\]\[y - 1 + 1 = x^{2} + 2x + 1\]\[y = (x + 1)^{2}\]I didn't see that my example wsas a perfect square which is why the steps are a bit repetitive.

OpenStudy (umulas):

Oh I see the error I did, thanks man for everything you did :D

OpenStudy (anonymous):

np :)

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