Question

Which of the following equations would have no real zeros?

Answer 1

|f(x)|+2 I wanna say this

it be a v at with vertex at the 2 of the y axix I think lol :P

Answer 2

well theres more to the problem but i dont know if it will help, do you want me to post a pic?

Answer 3

well i'll draw yo upic

Answer 4

never touches zero but i'll wait for second opinion if i were you.

Answer 5

heres the rest of the problem, if you could help me with any of it i would be greatful!!

Answer 6

and oh ok your graph make sense

Answer 7

oh that is easy to move a graph just do this f(x)

to move it up f(x)+c

to move it right f(x-c)

so now we need to find the vertex f(x-c1)+c2

i'll try it hold on

Answer 8

for part a

f(x-4)-3

Answer 9

how did you get that?

Answer 10

i explained above you just need to move the graph 1,32 to 5,-35

Answer 11

oh ok i get it, sorry i wasnt looking at the graph

Answer 12

it is just function manipulation :) f(x-4)-3
^ move right 4 units since w start at 1 and wanna end up at 5 :)

^ move down 3 units since we start at -32 and wanna wind up aat -35

Answer 13

dont let the f(x-c) confuse you it means to the right not left :) it is odd but it works :)

Answer 14

ya i remember doing that last year in my class lol. how do we do B?

Answer 15

be is just -32 just look for the lowest point :)

Answer 16

oh ok lol. do i just put -32 or do i include (1,-32)?

Answer 17

not sure but if it were up to me i just put -32

Answer 18

oh ok. and i think i have a misconception of C because i didnt think there was a minimum value. #confused

Answer 19

im thinking the answer is h(x) but idk

Answer 20

well by proccess of elimination i wanna say the first one

Answer 21

g(x)

-32/4 is 8 :)

Answer 22

-8

Answer 23

hmmm ok i see what you did. thanks. i think i got my graphs right on D but i dont know how to skeych the first one

Answer 24

now your q makes more sense since we know what f(x) looks like :)

Answer 25

haha ya!

Answer 26

the first graph is the one that confuses me though

Answer 27

here are my graphs for the other four

Answer 28

i wanna say the same still but it is odd

Answer 29

they look wrong

Answer 30

they aren't lines it is your graph in the top right manipulated and such :)

Answer 31

:(

Answer 32

what do you mean?

Answer 33

well i drew lines cuzz i assume your f(x) is x but it wasn't it is a weird curve.

i'll do one for you hold on

Answer 34

that is original