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Mathematics 26 Online
OpenStudy (pphalke):

i need help to do this the last one was marked as wrong i didnt know a dashed line was needed plz help

OpenStudy (pphalke):

myininaya (myininaya):

Do you know how to graph the line y=x+3 ?

myininaya (myininaya):

Do you know how to identify the slope and y-intercept of the graph that will given by the function f(x)=x+3 ?

OpenStudy (pphalke):

i dont know

myininaya (myininaya):

f(x)=mx+b tells us the slope is m and the y-intercept is (0,b)

myininaya (myininaya):

so can you identify m and b?

OpenStudy (pphalke):

for that dont you have to substitute 0 for x ?

myininaya (myininaya):

to identity m and b no but you could do that to find the y-intercept but you do not have do any math to compare f(x)=mx+b to f(x)=x+3 to identify m and b

myininaya (myininaya):

what values for m and b would make mx+b the same as x+3 ?

OpenStudy (pphalke):

slope 1 and y int 3?

myininaya (myininaya):

yes the slope m is 1 and the y-intercept is (0,b) or (0,3) in this case

myininaya (myininaya):

So plot the point (0,3) first

myininaya (myininaya):

That will a point on the y-axis between 2 and 4

myininaya (myininaya):

then use your slope to plot another point

OpenStudy (pphalke):

that would be (0,1)?

myininaya (myininaya):

and connect the dots with a dashed line since you have < or > you would use a solid line if you had a line underneath the inquality like \[\ge \text{ or } \le\]

myininaya (myininaya):

how did you get the point (0,1)?

OpenStudy (pphalke):

because of the slope 1 i thought (0,1)?

myininaya (myininaya):

slope=rise/run

OpenStudy (pphalke):

sorry sorry its (1,0)

myininaya (myininaya):

you have slope=1=1/1=rise/run rise once from your y-intercept and run once then you have your next point

OpenStudy (pphalke):

like this @myininaya

myininaya (myininaya):

no you have drawn a line with a negative slope we have a positive slope

myininaya (myininaya):

1/1 can you put your finger on (0,3)?

myininaya (myininaya):

now use your finger and go up once and to the right once

myininaya (myininaya):

since the slope is 1/1

OpenStudy (pphalke):

this?

OpenStudy (pphalke):

i re made it does it look right

myininaya (myininaya):

you are still now doing what I ask :( put your finger on (0,3) and we are going to find another point by using the slope which is 1/1 which tells us from the y-intercept we are going to rise up once and move right once can you do this?

myininaya (myininaya):

tell me what position your finger lays when you do this

myininaya (myininaya):

position/coordinates

OpenStudy (pphalke):

is that (2,5)?

myininaya (myininaya):

what is 3+1?

OpenStudy (pphalke):

4 so its now set at (1,4)

myininaya (myininaya):

you did it yes if you use the slope 1/1 to find another point from (0,3) you should get (1,4) conned those dots with a broken line

myininaya (myininaya):

connect*

myininaya (myininaya):

there is one part to this graph

myininaya (myininaya):

y<x+3 means we want to shade everything below the line y=x+3 since < means less than if we had y>x+3 instead this would mean we want to shade everything above the line y=x+3 since > means greater than

OpenStudy (pphalke):

its incorrect according to the homework

myininaya (myininaya):

can i see what you entered?

OpenStudy (pphalke):

what i had done before and the message

myininaya (myininaya):

where is the shading part?

myininaya (myininaya):

did you read what I said about the shading part above?

myininaya (myininaya):

y<x+3 means y-coordinates of every point is less than y=x+3 this means you need to shade everything below the broken line you have for y=x+3

OpenStudy (pphalke):

so how would i do that (0,3) and do i go down to the left

myininaya (myininaya):

if you had y>x+3 this means we would shade everything above the broken line y=x+3

myininaya (myininaya):

you already drew a broken line at y=x+3 ...

myininaya (myininaya):

you only need to shade below the line

myininaya (myininaya):

|dw:1461517969899:dw| you still need to shade below the line to show all the points that are a solution of y<x+3

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