Create your own quadratic equation and demonstrate how it would be solved by graphing, factoring, the quadratic formula, and by completing the square
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.
So did you already form an equation then?
x^2+7x+12 but the factor i did which is (x+3)(x+4), meaning x = -3, -4
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 :)
That in the first line should be \(ax^{2} + bx + c\)
Hmm, I gotten -7/2, did I do that correct?
Yes. Sorry I was so late :) That is the x-value of the vertex. Plug it back in to solve for y.
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?
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 = ?
D:, why didn't I think of that! Thanks!
np :)
If you want, you can try changing your equation to mine because it's a lot easier.
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?
Quadfratic Formula = \(\huge x = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a}\)
Which equation are you using?
The quadractic formaula
I know, but are you using my example or yours?
I am using your example, it's way more easire and better to understand.
Alright. Yeah. You are correct. There is a double root by the way.
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
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.
Oh I see the error I did, thanks man for everything you did :D
np :)
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