Math Group
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Posted by chikazu on Monday November 9, 2009 at 1:50 PM
Y is equal to mx+b, with m being the slope and b being the spot on the Y axis that the line intercepts at. So, for Y=2x+4, it would have a slope of 2/1, the line would intercept at 4, and it would go to the right. Does that help?
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eNotes Editor
Posted by neela on Monday November 9, 2009 at 4:52 PMy=mx+b is the slope intercept form of a straight line.
When x=0, y= m*0+b=b. Therefore, (0,b) which is a point on y axis and it intercepts y axis at a distance b from the origin.
Now Let us plug y=0 in the equation, 0=m+b. So, x = -b/m. So,
In the triangle ABC, A = (-b/m,0), O, the origin =(0,0) and C= (0,b), the slope of the line y=mx+b is given by:
slope = tan angleOAC = OC/AO= (b-0)/[0-(-b/m)]= b/(b/m)=m.
Therefore, m is the slope and and b is the intercept of the line y = mx+b. Or y = slope times x + y intercept.
Similarly, for a given line, y=x+10 or y = 1*x+10, 1 is the slope and 10 is the intercept on y axis.
Similarly y=2x+4 is a line whose slope is 2 and the line cuts y axis at (0,4) or the y intercept of this line is 4.
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Posted by subbundd on Tuesday November 10, 2009 at 6:48 AM
Slope intercept form is x/a+y/b=1 where a ix intercept on x- axis and b is intercept on y -axis.slope m= TanA =b/a
The givem Eqis Y=2x+4 or 2x-y=-4 Divide both sides by -4 We get
x/(-2)+y/(-4)=1
Therefore a=-2 and b= -4.and slope = b/a = 2.
Since both intercepts are negative the graph lies in IIIrd the Quadrant.

