Question

Vertex form of: x^2+3x-(1/2)+y=0

The (1/2) is one half.

Answer 1

Will be willing to metal someone.

Answer 2

y=a(x-h)^2+k is vertex form

Answer 3

I mean't the equation, not the formula or whatever that is :p

Answer 4

\[y=-x^2-3x+\frac{1}{2}\]

Answer 5

So we are going to complete the square so we can put it in the form y=a(x-h)^2+k

Answer 6

did you divide by -1 to do that?

Answer 7

unless you are saying you don't want to put it in vertex form now...

Answer 8

I do, I just wasn't taught that today.

Answer 9

I subtracted everything on both sides except the y

Answer 10

yeah you subtracted the -y to the other side, and then divided by -1, to keep the y positive?

Answer 11

Well I subtracted everything on both sides EXCEPT the y

Answer 12

ohhh.

Answer 13

But you could subtract y on both sides then multiply both sides by -1

Answer 14

but you'll still get the same outcome

Answer 15

alright so would i complete the square or quadratic formula?

Answer 16

Well you want the vertex form which is y=a(x-h)^2+k so we need to complete the square

Answer 17

Can you teach me how to complete the square? I know how to do it, but I forgot how to do it with odd numbers. :'p

Answer 18

\[y=(-x^2-3x)+\frac{1}{2} \\ y=-(x^2+3x)+\frac{1}{2} \\ y=-(x^2+3x+?)+\frac{1}{2}+?\]

Answer 19

:o, i thought completing the square was half of three times that by 2, add it to both sides?

Answer 20

\[x^2+bx+(\frac{b}{2})^2=(x+\frac{b}{2})^2\]

Answer 21

Jaja, sorry but my teacher hasn't taught us anything like that. x'D

Answer 22

That is how you complete the square...

Answer 23

oh i reread the post, i get it.

Answer 24

i thought the steps were different with odd numbers

Answer 25

So your b was 3
and you said let's add inside that ( ) (3/2)^2 right?

Answer 26

right

Answer 27

\[x^2+3x+(\frac{3}{2})^2=(x+\frac{3}{2})^2 \]

Answer 28

wait i am confused, what did you do with the (-) -> -x^2-3x

Answer 29

\[y=(-x^2-3x)+\frac{1}{2} \\ y=-(x^2+3x)+\frac{1}{2} \\ y=-(x^2+3x+?)+\frac{1}{2}+? \\ \text{ the ? is } (3/2)^2 \\ y=-(x^2+3x+(\frac{3}{2})^2)+\frac{1}{2}+(\frac{3}{2})^2\]

Answer 30

did you do this -(x^2-3x)

Answer 31

well no i did this: -(x^2+3x)

Answer 32

yeah typo in the ( ) x'D

Answer 33

3/2 = 1.5 * 2 = 3

Answer 34

Why are you doing 3/2*2?

Answer 35

i meant 3 divided by 2, the outcome times by 2

Answer 36

or square it***

Answer 37

lol my bad.

Answer 38

\[y=(-x^2-3x)+\frac{1}{2} \\ y=-(x^2+3x)+\frac{1}{2} \\ y=-(x^2+3x+?)+\frac{1}{2}+? \\ \text{ the ? is } (3/2)^2 \\ y=-(x^2+3x+(\frac{3}{2})^2)+\frac{1}{2}+(\frac{3}{2})^2 \\ y=-(x+\frac{3}{2})^2+\frac{1}{2}+\frac{3}{2} \cdot \frac{3}{2}\]

Answer 39

I wrote (3/2)^2 as 3/2*3/2
so you wouldn't think (3/2)^2 means 3/2*2

Answer 40

I just went blonde for the past 15 minutes. x'D

Answer 41

I got y= -(x+1.5)^2 -(5/2)

(5/2) = 5 over 2(fraction)

Answer 42

wait nevermind, i did it wrong.

Answer 43

\[\text{ These are the steps } \\ y=ax^2+bx+c \\ y=ax^2+\frac{a}{a}bx+c \text{ remember a/a=1} \\ y=a(x^2+\frac{1}{a}bx)+c \text{ I factored whatever was in front of the } \\ x^2 \text{ from both of the first two terms } \\ y=a(x^2+\frac{b}{a}x)+c \\ y=a(x^2+\frac{b}{a}x+?)+c-a \cdot ? \\ \text{ the } ? \text{ is } (\frac{b}{2a})^2 \\ y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a \cdot (\frac{b}{2a})^2 \\ y=a(x+\frac{b}{2a})^2+c-a \cdot \frac{b^2}{4a^2} \\ y=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ y=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]
Where the vertex is :
\[(\frac{-b}{2a},\frac{4ac-b^2}{4a})\]

Answer 44

I know how to find the vertex, but we got to set the quadratic equation to vertex form and with that, we get the vertex.

Answer 45

Well I showed you just now the process for getting both the vertex equation and the vertex form.

Answer 46

and i mean the vertex also

Answer 47

Yes thanks, but my school teaches it in a terrible way.