Question

Can someone help me for a medal?

Answer 1

Determine which of the following statements is true concerning the values described in column #1 and column #2.


Column #1 Column #2
The x-coordinate of the vertex of the graph of
y = −2x2 − 4x + 12 The x-coordinate of the vertex of the graph of
y = x2 − 4x + 3

Answer 2

Options are:

A: The value found in column #1 is greater than the value found in column #2.

B: The value found in column #1 is less than the value found in column #2.

C: The value found in column #1 is equivalent to the value found in column #2.

D: The relationship between column #1 and column #2 cannot be determined by the information given.

Answer 3

A

Answer 4

Am I right?

Answer 5

Lol, so sorry for this delay. I'm not sure if you are correct or not, thats why I'm asking for help.

Answer 6

Should be A

Answer 7

How do you know?

Answer 8

With you way you posted it, it's really difficult to determine what is in column 1 and what is in column 2.

Answer 9

Hannah, why you sideways! >.<
Pick yerself up! :D

Answer 10

simplify everything

Answer 11

Oh, sorry. I'll take a screenshot and post it on here. I' not gonna lie, what you said was actually pretty funny :D

Answer 12

I'm not sure how to simplify them. Can you explain how you simplified?

Answer 13

Oh :) So in order to identify the vertex points, you need these equations in vertex form.
That usually requires you to `complete the square`.
Does that sound familiar?

Answer 14

Yeah, it does. It's just that when I was going over the lesson, I found it very difficult to understand.

Answer 15

y = -2x^2 - 4x + 12 = -2(x^2 + 2x) + 12 = -2(x^2 + 2x + 1) + 12 + 2 = -2(x + 1)^2 + 14

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

The vertex is (-1, 14) in the first; (2, -1) in the second.

ANSWER: The value found in column #1 is less than the value found in column #2.

Answer 16

zepdrix can you confirm if his answer is correct or not?

Answer 17

It seems correct to me, but you seem to be the smarter then me

Answer 18

Whut? 0_o
I dunno, I'm busy doing a problem, and eating yogurt. It's SO good.

Answer 19

I love yogurt :D. Also, its fine if your doing another problem. I can wait

Answer 20

you're*

grumble grumble <.<

Answer 21

Should I just go with Damneny's answer?

Answer 22

Oops, my bad.

Answer 23

ya his work looks correct c:

Answer 24

So you having trouble with this whole `complete the square` business? :o

Answer 25

Yeah, I do sadly. And thanks for confirming that. I'll give you a fan and him a medal

Answer 26

So HannahBanana,
what you do is like uhhhh...
ok ok let's look at the first equation,\[\Large\rm -2x^2+4x+12\]Err no no, let's deal with that in a sec, lemme try to be more general first.

We would LIKE to have an equation in this form:\[\Large\rm y=x^2+bx\]To complete the square,
we take `half` of the b coefficient,
and `square` it.
\[\Large\rm \left(\frac{b}{2}\right)^2\]This is the value that completes the square for us.
So we'll add it in.\[\Large\rm y=x^2+bx+\left(\frac{b}{2}\right)^2\]But holdddd up!
We can't just add something willy nilly like that.
We have a couple of options.
The one you're probably most familiar with is "adding it to both sides".
That's a way to keep it balanced, yah?

But we want to try another trick instead, we'll add and subtract the quantity at the same time. That's another way to balance things.\[\Large\rm y=\color{orangered}{x^2+bx+\left(\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]I colored this in orange because these first three give us our perfect square trinomial.
They will condense down like this:\[\Large\rm y=\color{orangered}{\left(x+\frac{b}{2}\right)^2}-\left(\frac{b}{2}\right)^2\]

Answer 27

Ok ok, but maybe that's too general, lemme just explain how to do it on your first equation then maybe you'll get the hang of it.

Answer 28

\[\Large\rm y=-2x^2+4x+12\]There is too much stuff here, we gotta try to get that nice form that I talked about before.
We don't care about this +12, let's ignore him for now.\[\Large\rm y=-2x^2+4x\qquad\qquad\qquad+12\]We also don't want anything on our x^2 term.
But we need to keep the x's grouped together.
So we'll factor a -2 out of each x term.\[\Large\rm y=-2(x^2-2x)\qquad\qquad\qquad+12\]Ok with those steps so far? >.<

Answer 29

Bananz where you at? D:
I see how it isssss