Question

Which point lies on the graph of the following piecewise function?
y = x + 2 if x < 1
y = x2 if x greater than or equal to 1

(-1, -1)
(-2, 0)
(-2, 4)
(3, 6)

Answer 1

@jim_thompson5910

Answer 2

how far did you get?

Answer 3

well to be honest I only know to plug it in and see which is true, but I am not sure how to solve it the right way.

Answer 4

that's how you would do this problem

or you could graph the entire piecewise function and each point to see which point lies on the graph

Answer 5

oh ok so that's the only way to solve the problem? give me a sec to solve it and check with ya :)

Answer 6

this is for A

Answer 7

wait but answer choice A B and C are not greater than or equal to 1.

Answer 8

And answer choice D doesn't work cuz (3)^2=9 it needs to be 6. But not true

Answer 9

I am really confused because when I plug each answer choice in they don't work

Answer 10

what did you get for C?

Answer 11

let me draw it...

Answer 12

how about B

Answer 13

I meant the second part*

Answer 14

focusing on choice B, which piece does x = -2 satisfy?

Answer 15

I not sure what you're asking ;)

Answer 16

choice B is the point (-2,0)

x = -2
y = 0

Answer 17

yes

Answer 18

when x = -2, which piece of the piecewise function below

\[
f(x) = \begin{cases}
x+2 &\mbox{if } x < 1 \\
x^2 & \mbox{if } x \ge 1
\end{cases}
\]

does it satisfy?

Answer 19

but wait, -2 is not greater than or equal to 1

\[\Large -2 \ge 1 \ \ \text{ is a false statement}\]

Answer 20

that's what I said, lol ;)

Answer 21

so because it's false, you don't even consider the piece x^2

you only focus on x+2 because this is when x < 1

Answer 22

so that shows you B is correct

this graph confirms it

Answer 23

I see, but I don't really get how we plug in the point for option B to get the answer to be true. If the x^2 part is false shouldn't the whole become eliminated out?

Answer 24

you only plug x = -2 into the piece where x = -2 makes the inequality true

x = -2 makes x >= 1 false, so we ignore that second piece

Answer 25

x = -2 makes x < 1 true, which is why we focus on y = x+2

Answer 26

sorry to take up your time, but still confused? so we only need to look at which ever one makes the statement true even if it doesn't support the whole x^2 or x+2 ?

Answer 27

\[ f(x) = \begin{cases} x+2 &\mbox{if } x < 1 \\ x^2 & \mbox{if } x \ge 1 \end{cases} \]

really means

y = x+2 if x < 1

OR

y = x^2 if x >= 1

Answer 28

it's effectively "gluing" two different equations together to form a compound equation in a sense

when you plug in x values, you only pick on one piece at a time (otherwise, it's not a function)

Answer 29

so that means if you wanted to plug in x = 0, you'd look at the conditions

is x = 0 going to satisfy x < 1? or x >= 1?

since 0 < 1, this means it makes x < 1 true

therefore, we plug x = 0 into y = x+2 only and NOT y = x^2

Answer 30

thank you for explaining, it's more clear :) why would we not look at y=x^2 if it's not necessary why is it included in the problem?

Answer 31

it's only looked at when you plug in x values that are either 1 or larger than 1

so say x = 1, x = 2, x = 3, etc etc

Answer 32

ohhhhhhhhhhhhhhhhhhhhhhhhhhh you've it gold to my brain!!!! that was my issue. I get what is done......thanks again for your time. :)))))