Question

complete the square to find vertex y=-x^2+5x−3

Answer 1

@surjithayer

Answer 2

@Mertsj

Answer 3

\[y= -x ^{2}+5x-3=-\left( x ^{2}-5x \right)-3\]
\[y=-\left( x ^{2} -5x+ \left( \frac{ 5 }{ 2 } \right)^{2}-\left( \frac{ 5 }{2} \right)^{2}\right)-3\]
\[y=-\left( x-\frac{ 5 }{ 2 } \right)^{2}+\frac{ 25 }{4 }-3\]
\[y=-\left( x-\frac{ 5 }{ 2} \right)^{2}+\frac{ 13 }{4 }\]

Answer 4

this is really hard to understand

Answer 5

but Im still kind of confused :(

Answer 6

keep -3 outside bracket

Answer 7

then??

Answer 8

the vertex??

Answer 9

(-5/2, 37/4)

Answer 10

thats the vertex right??

Answer 11

@Loser66

Answer 12

ok please re-state it

Answer 13

just tell me the vertex !!!

Answer 14

the vertex is ??? what

Answer 15

-5/2, 37/4

Answer 16

and??

Answer 17

thanks!!

Answer 18

it is not \[x ^{2}+5x-3\]
but it is \[-x ^{2}+5x-3\]

Answer 19

it is a downward parabola.
in this case vertex is \[\left( \frac{ 5 }{ 2} ,\frac{ 13 }{ 4 }\right)\]