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OpenStudy (buttermequeasy):
Describe how you would graph this line using the slope-intercept method.
Y= 1 1 /2x +4
Please help me I don't understand how to do this.
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myininaya (myininaya):
so this in form y=mx+b
where m is the slope and b is the y-intercept
myininaya (myininaya):
so (0,4) is the y-intercept
myininaya (myininaya):
and slope is m=3/2 ( is that a mix fraction you have there; if so i just rewrote as an improper fraction)
myininaya (myininaya):
so using (0,4) I would count up three from that point and the count to the right twice
(using our slope)
myininaya (myininaya):
then I would connect those dots
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myininaya (myininaya):
to draw my line
OpenStudy (buttermequeasy):
It 1.5 in decimal form so it is one one half.
myininaya (myininaya):
ok well i would write it as an improper fraction like i did above
myininaya (myininaya):
|dw:1327186116598:dw|
myininaya (myininaya):
|dw:1327186131699:dw|
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myininaya (myininaya):
so i'm going to plot the point (0,4) this is our y-intercept
myininaya (myininaya):
|dw:1327186183438:dw|
myininaya (myininaya):
i should have put some more lines oh well
instead of looking at 3/2 we could look at -3/(-2) since
3/2=-3/(-2)
myininaya (myininaya):
so -3/(-2) means to count down three and over to the left 2 like this:
myininaya (myininaya):
now we connect the dots
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myininaya (myininaya):
just remember the slope is rise/run
thats how come i knew to fall 3 and go left 2
myininaya (myininaya):
|dw:1327186353139:dw|
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