Question

complete the square and then identify the ordered pair (h.k)
2y=5-3x^2-3x

Answer 1

is (h,k) the vertex?

Answer 2

you are going to have to write this as
\[y=\text{expression}\] as a first step, so divide by 2

Answer 3

the instructions for the problem are : for the following equation, complete the square and express the answer in appropriate form, either y=a(x-h)^2+k or x=a(y-k)^2+h. Then identify the ordered pair (h.k).
i know how to complete squares, but then?

Answer 4

\[y=-\frac{3}{2}x^2-\frac{3}{2}x+\frac{5}{2}\]

Answer 5

annoying but there is no way around it

Answer 6

ok im following

Answer 7

then you job is to rewrite as
\[y=a(x-h)^2+k\] and pretty clearly

\[a=-\frac{3}{2}\]

Answer 8

question is what is h, and you have a choice on how to compute it
you could write
\[y=-\frac{3}{2}(x^2+x)+\frac{5}{2}\] as a first step and then write
\[y=-\frac{3}{2}(x+\frac{1}{2})^2+\text{stuff}\] i will explain how to find the "stuff" easily

Answer 9

oh no please no myininaya

Answer 10

or you could do it the easy way and say the vertex is at
\[x=-\frac{b}{2a}=-\frac{1}{2}\] and be done

Answer 11

\[y=-\frac{3}{2}(x^2+x)+\frac{5}{2}=\frac{-3}{2}(x^2+x+(\frac{1}{2})^2)+\frac{5}{2}+\frac{3}{2}\cdot (\frac{1}{2})^2\]
\[=\frac{-3}{2}(x+\frac{1}{2})^2+\frac{5}{2}+\frac{3}{8}\]

Answer 12

\[=\frac{-3}{2}(x+\frac{1}{2})^2+\frac{23}{8}\]

Answer 13

now that i know the vertex has x coordinate
\[-\frac{1}{2}\] i know it looks like
\[y=-\frac{3}{2}(x+\frac{1}{2})^2+k\] and k is what i get when i replace x by
\[\frac{1}{2}\] in the original equation. you will get
\[\frac{23}{8}\]

Answer 14

if you can follow all that adding and subtracting an keeping track by my esteemed colleague myininaya that way will work.
i much prefer computing
\[-\frac{b}{2a}\] and then the vertex is
\[(-\frac{b}{2a},k)\] where k is what you get when you replace x by
\[-\frac{b}{2a}\] in the original expression

Answer 15

your choice

Answer 16

i know which one you choose!

Answer 17

\[k=f(-\frac{b}{2a})\]

Answer 18

how was dinner?

Answer 19

yummy :)

Answer 20

oooh yes! that is what i use.

Answer 21

but i guess you could just where k is what you get when you replace x by -b/(2a)
but thats so long when you can just say f(-b/(2a))

Answer 22

:)