Question

help pleaseee

Answer 1

you know what \(y=x^2+1\) looks like? it is a parabola that opens up, with vertex at \((0,1)\)

Answer 2

No matter if you substitute a negative value or positive one for x, you get the same number so its not a function

Answer 3

so for this just plug in your values for \[x ^{2}+1\] so for 1
1^2+1=1+1=2 for 2
2^2+1=4+1=5 and so on

Answer 4

the inserting a negative is to tell if it is even or odd

Answer 5

not if it is a function

Answer 6

so after you plug in your values and graph it you do the vertical line test

Answer 7

thats for part 1

Answer 8

if you can draw a vertical line on every point of the graph without it crossing the graph more than once you have a function if for any point on the graph you cross the graph more than once it is not a function

Answer 9

so what do i do now

Answer 10

well what have you done so far? did you graph it and then determine if it is function?

Answer 11

After you do that, for part 3 you want to determine domain which asks from what x value to what other x value does this graph go to, and the range asks from what y value to what y value does this go

Answer 12

2 and 5?

Answer 13

I want to ask. Are there any x values that you can plug into y=x^2+1 and you will not get a y value?

Answer 14

-1?

Answer 15

well lets try it y=-1^2+1
-1^2=1 because a negative squared is a positive
1+1=2

Answer 16

more clearly it asks can you plug in anything into the x of
y=x^2+1
and it will not give you a y value

Answer 17

can you think of anything that will not give you a y?

Answer 18

um i really dont know im confused now

Answer 19

ok so you cannot find any because every x value will give you y so if you pluged in 24378 into the x of this equation or -456783 you will still get a y so this means every value for x works

Answer 20

oo ok

Answer 21

so your domain is all reals or if you teacher wants interval notation it is \[(-\infty,+\infty)\]

Answer 22

so the relation is a function?

Answer 23

so now do the same for y it might help to rearrange the equation so it becomes a x=y instead of y=x so you would do this
y=x2+1
subtract 1
\[y-1=x ^{2}\] take the square root to get rid of the exponent and you have
\[x=\sqrt{y-1}\]

Answer 24

Now. Is there any value I can put in for y that will not give me an x?

Answer 25

none right?

Answer 26

so think about a square root are there any numbers you can put under a square root that you cannot take the square root of? It also helps to look at the graph. Does it look like the graph will touch every possible y number there is possible?

Answer 27

1.5 or 1?

Answer 28

plug them in \[\sqrt{1-1}=\sqrt{0}=0\] and 0 is a number so it works
look at the graph it curves down touches 1 and then curves back up so it does not touch -1 plug that in
\[\sqrt{0-1}=\sqrt{-1}\] and remember we cannot take the square root of a negative number so dont get a number with that so everything from 0 down is not in the range of this graph but from 1 to infinity is in the range so the range is from 1 to infinity or in interval notation
\[[1,\infty)\]

Answer 29

o ok

Answer 30

do you understand now?

Answer 31

ok glad I could help