Question

HEY NEED HELP ASAP PLEASE:
Using the completing-the-square method, find the vertex of the function f(x) = –2x^2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.

Maximum at (–3, 5)
Minimum at (–3, 5)
Maximum at (3, 23)
Minimum at (3, 23)

Answer 1

\(f(x)=ax^2+bx+c\\f(x)=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\)

Has vertex \((-\frac{b}{2a}, c-\frac{b^2}{4a})\)

This is a maximum if \(a<0\) and a minimum if \(a>0\)

Answer 2

i am stuck between A and B

Answer 3

@zzr0ck3r

Answer 4

@Nnesha

Answer 5

@Hero

Answer 6

Do you not understand something I wrote?

Answer 7

NO i am just stuck between answer A and B

Answer 8

Show me your work. i.e. show me your vertex form.

Answer 9

i just want to know if i got it right and i am stuck between A and B!

Answer 10

For quadratic functions with the form f(x) = ax^2 +bx+c , if a is negative, then the function will have a maximum value at the vertex.

Answer 11

that is not what we are here for.

Answer 12

@hero must be on his phone, that took way to long and your just repeating :P

Answer 13

Yeah, I don't have the best tools right now.

Answer 14

s o it is A?

Answer 15

.Not tellin

Answer 16

i did the work but I'm just stuck between minimum and maximum which is A or B! p

Answer 17

We've given you enough info.

Answer 18

i don't know yet.

Answer 19

how do i tell between minimum and maximum

Answer 20

Read What I wrote about quadratic functions once more.

Answer 21

i did i still don't know A or B

Answer 22

@mathway

Answer 23

This is similar to what happens when you give someone money, but they don't Know how to spend it.

Answer 24

who wouldn't know how to spend money

Answer 25

@mathway please help

Answer 26

Have you found the vertex form?

Answer 27

well kind of, i am just stuck between a and B

Answer 28

Can you show me the steps you did?

Answer 29

i have it all written on paper i can't take a picture..

Answer 30

You can select the Draw button to write your work.

Answer 31

well for my last step i Got A!! finally figured how to do it:-)

Answer 32

So you will not show your work then? Okay.

Answer 33

no, i got the answer- thanks anyways though! @mathway

Answer 34

All right, but I won't say if you're right or wrong.

Answer 35

ok

Answer 36

i know I'm right so thank you though