Question

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

Answer 1

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.

Answer 2

So did you already form an equation then?

Answer 3

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

Answer 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 :)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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?

Answer 9

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 = ?

Answer 10

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

Answer 11

np :)

Answer 12

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

Answer 13

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?

Answer 14

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

Answer 15

Which equation are you using?

Answer 16

The quadractic formaula

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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.

Answer 22

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

Answer 23

np :)