Algebra
posted by delmore
Given a line containing the points(1,4), (2,7) and (3,10) determine that slopeintercept form of the equation, provide one additional point on this line, and graph the funtion.
Start by putting the first point into pointslope form: yy1=m(xx1), so y4=m(x1). Slope, or m, is y2y1 over x2x1. With the first two points, slope=74 over 21, or 3 over 1, which is 3. The equation is then y4=3(x1). After doing basic algebra, the slopeintercept form of the equation is y=3x+1.
Using the slope of 3, add three to the yvalue of the last point, (3,10), and 1 to the xvalue; you are basically doing the slope equation backwards here. Your new point could be (4,13).
To graph the function, just plot the points you were given; in fact, you only need 2 of them, and then draw your line.
i would say (4,13) or (0,1) depending which direction you want to go.
the way i see it is like this.
for the point (1,4) to get to the next point (2,7) they are going up 1 number on the x coordinate point and 3 on the y coordinate point.I know that theres a formula for this but i can't remember what it is but i just know that you line is a diagnol line.
I just set up two equations and solved them simultaneously.
y = mx + b
==============
substitute points 1,4 and 2,7
eqn 1: 4=m(1)+b
eqn 2: 7=m(2)+b
solve for m and b.
m=3 and b=1
from work done by the two previous posters, both 4,13 and 0,1 are on the line as well as the third point given in the problem of 3,10. And y = 3x+1 is the slope intercept form.
I just thought I would give an alternative method of solving the problem.
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