Question

Check:
is the vertex form of -3x^2-6x+3.....this:(x+1)^2+6

Answer 1

@radar

Answer 2

@zepdrix

Answer 3

Hmm close :d

Answer 4

Your vertex form should look like this:\[\large\rm y=\color{orangered}{a}(x-h)^2+k\]Where your vertex is located at (h,k).
Yes, you have the correct vertex,
but you forgot about your \(\large\rm \color{orangered}{a}\).

Answer 5

yeah, how do i find a

Answer 6

\[\Large\rm \color{blue}{a}x^2+\color{red}{b}x+\color{green}{c}=\color{blue}{-3}x^2+\color{red}{-6}x+\color{green}{3}\]

Answer 7

Oh I should have colored the a in orange again,
not blue :) lol

Answer 8

Anyway, it's the same a that you started with.

Answer 9

The same a that you used to find the x-coordinate of your vertex in the -(b/2a) formula

Answer 10

-3?

Answer 11

yup! :)\[\large\rm y=\color{orangered}{-3}(x+1)^2+6\]

Answer 12

yasss. one more question how do you find the roots to this problem

Answer 13

Oh I was waiting for you to post a problem. haha,
roots to the same problem? XD

Answer 14

haha do you want me to open another question

Answer 15

`Roots` is another fancy word that we use.
Remember earlier I was calling them `zeros` or `x-intercepts`.
All three of those terms mean the same thing.

So recall, to find roots/zeros/solutions, we let y=0.

Answer 16

\[\large\rm 0=-3(x+1)^2+6\]And solve for x.

Answer 17

Understand how to do that? :O

Answer 18

i think , i am trying to work it out right now

Answer 19

wait can't i just set -3x^2-6x+3

Answer 20

equal to zero

Answer 21

Yes, and use the `quadratic formula` from there.
That is your other option.
:)

Answer 22

It's a little bit easier to solve the vertex form
than it is to simplify the quadratic formula.

Both are quite fine though.
Whichever you're more familiar with.

Answer 23

fabulous, i got something weird like \[1\pm \sqrt{17}/-1\]

Answer 24

Hmm that square root doesn't look quite right +_+

Answer 25

Which method did you use?

Answer 26

i did the quadratic formula of -3x^2-6x+3

Answer 27

\[\large\rm x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Then plugging in the numbers,\[\large\rm x=\frac{6\pm\sqrt{(-6)^2-4(-3)(3)}}{2(-3)}\]it should look like that, ya?

Answer 28

sí

Answer 29

i'll do the rest

Answer 30

\[\left( 6\pm \sqrt{102} \right)/-6\]

Answer 31

\[\large\rm x=\frac{6\pm\sqrt{(-6)^2-4(-3)(3)}}{2(-3)}\]

\[\large\rm x=\frac{6\pm\sqrt{36+36}}{-6}\]
\[\large\rm x=\frac{6\pm\sqrt{2\cdot36}}{-6}\]Hmm 106? :o

Answer 32

omg no 72

Answer 33

We would like to simplify the root,
so it's actually preferable to write it this way :)\[\large\rm \sqrt{36+36}\quad=\quad\sqrt{2\cdot36}\quad=\quad\sqrt{2}\cdot\sqrt{36}\]

Answer 34

\[\large\rm x=\frac{6\pm\sqrt{2}\cdot\sqrt{36}}{-6}\]Do you see how it will simplify further?

Answer 35

yes eventually it will be \[1\pm \sqrt{12}/-1\]
right?

Answer 36

12? :o hmm

Answer 37

i divide 72 by 6

Answer 38

And if you're going to write it like that,
make sure you place brackets around your numerator,\[(1\pm \sqrt{12})/-1\]But no not 12 :(

Answer 39

Divide 72 by 36, not 6.

Answer 40

If you divide it by 6,\[\large\rm \sqrt{72}\quad=\quad \sqrt{12\cdot6}\]Then there is no way to get the 6 out of the square root.

Answer 41

2

Answer 42

\[\large\rm x=\frac{6\pm\sqrt{2}\cdot\sqrt{36}}{-6}\]
\[\large\rm x=\frac{6\pm\sqrt{2}\cdot6}{-6}\]Dividing everything by 6,\[\large\rm x=\frac{1\pm\sqrt{2}}{-1}\quad=\quad -1\pm\sqrt{2}\]Ah yes, 2 looks better! :)

Answer 43

That was a lot of nasty algebra to get there.
Let's remember what the question was asking for.
Roots, yes?

So we have two of them:
One corresponding to the positive sqrt(2)
and the other for the negative sqrt(2).\[\large\rm x=-1+\sqrt2\qquad\qquad and\qquad\qquad x=-1-\sqrt2\]

Answer 44

fab, so we are done?

Answer 45

yes :)

Answer 46

now go eat a strawberry :U
even a blueberry will do.

just not the watermelon, i beg of you!! >.< lol

Answer 47

hahaha fine ill eat watermelon tomorrow :)
thanks for all the help!!!