Question

Ok, well -x^2 + 60x + 900
i can't really transform this into a vertex form
some help please
this is as far as i could go
-(x^2-60x-900)

Answer 1

\[-(x^2-60x)+900\]
\[-(x^2-60x+(\frac{60}{2})^2)+900+(\frac{60}{2})^2\]
\[-(x-30)^2+900+30^2\]
\[-(x-30)^2+1800\]

Answer 2

yeah half of -60 o=is -30 so write as
\[y=-(x-30)^2+k\]

Answer 3

then to find k you can do all the work my master myininaya did, or you can make the brilliant observation that if x = 30 you will get
\[y=-(30-30)^2+k=k\] so just put in 30 for x in the original equation and see what you get

Answer 4

o well actually at first it was x(60-x) = m
than 60x-x^2=m
than i got -x^2-60x+900-900 = m
than - (x^2-60x-900) -900 = m
sorry i didn't include that! :(

Answer 5

in other words compute
\[-30^2+60\times -30 +900\]

Answer 6

what is m?

Answer 7

m is maximum

Answer 8

are you solving a word problem of some sort?

Answer 9

so are you trying to find the maximum of
\[x(60-x)\]?

Answer 10

the problem is a word problem.
The sum of two numbers is 60. Find the numbers if their product is a maximum
X+Y=60
XY=M

Answer 11

because if you are, by symmetry you should see that the max is when x = 30

Answer 12

right

Answer 13

of course. if you want a rectangle with largest area make a square

Answer 14

\[M=x(60-x)=60x-x^2\] a parabola facing down. the vertex is at
\[-\frac{b}{2a}=-\frac{60}{2}=30\]

Answer 15

X+Y=60=>Y=60-X
so M=XY=X(60-X)=60X-X^2
so we have
M=-X^2+60X
M=-(X^2-60X)
\[M=-(X^2-60X+(\frac{60}{2})^2)+(\frac{60}{2})^2\]
\[M=-(X-30)^2+900\]

Answer 16

and if you replace x by 30 you get
\[30\times 30=900\]

Answer 17

i've said it before and i'll say it again: if i want the vertex of some quadratic i will compute
\[-\frac{b}{2a}\] and then i have to complete no square or try to reconcile what i have with what i want. it completes the square for me

Answer 18

lol

Answer 19

really!

Answer 20

automatic complete square machine. patent pending

Answer 21

whats going on down here? lool

Answer 22

btw buck it occurs to me that you have already solved the problem. when you asked "write it in vertex form" what you really meant was you had already figured out that the maximum area was 900 when x = 30

Answer 23

\[P(x)=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c\]
\[=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a*(\frac{b}{2a})^2\]
\[=a(x+\frac{b}{2a})^2+c-\frac{b^2}{2a}\]

Answer 24

the problem is a word problem. The sum of two numbers is 60. Find the numbers if their product is a maximum X+Y=60 XY=M
this is the problem you wrote yes?

Answer 25

oops i missed something

Answer 26

however you got form there to
\[-x^2+60+900\]
meant you had already completed the problem. you got the 900 from somewhere and that is your answer

Answer 27

:S WHATS HAPPENING?! :L

Answer 28

im good, i solved it, thank you guys!

Answer 29

vertex is
\[(-\frac{b}{2a},c-\frac{b^2}{4a})\]

Answer 30

your answer is:
the the product is 900 and both numbers are 30

Answer 31

\[P=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\]