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Points: (x1, y1) and (x2, y2)

Slope: y1-y2/x1-x2

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Q: How do you solve for slope from two points and a graph?
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What does a distance time graph show?

A distance-time graph shows the movement of an object with respect to time. The average slope between any two points on the graph is equal to the average velocity of the object between those two points. The instantaneous slope (or derivative) at a point on the graph is equal to the instantaneous velocity of the object at that point.


How do you find slope using x and y intercepts?

Your x and y intercepts give you two points on the line of the graph. Use these two points in the slope equation m = (y2-y1)/(x2-x1), and that gives you the slope.


How do you get the slope of a line on a graph?

Take a tangent at the point where you want the slope. Then the slope of the graph at that point is the slope of the tangent, which is found by taking another point on the tangent and then taking the change in y between the two points and divid it by the change in x.


How do you determine the slope of a line on an graph?

Rise divided by run. (Y2 - Y1) / (X2 - X1) - with (X1, Y1) and (X2, Y2) being two points on the graph.


How do you graph a line on a graph?

well you use a formula ..... the one that i use most is y=mx+b this formula is slope-intercept formula the slope of the line is m, and the intercept is the b(the y intercept). Say the slope(m) is 3/2 the starting from the y intercept (in this case let it be 2, which is the coordinate [0,2]) you would go to the right 2 units and up 3 units(3/2=rise/run) to get another point. With that point(which in this case would be [2,5]) you can make a straight line through the two points to get your line on the graph. you can make as many points as you want, just keep using the slope(in this case 3/2). Another way, with graph paper: if you are dealing with a straight line, then pick two values of x, and solve for the respective y values. Plot these two points on the graph paper, then using a straight edge, connect the dots, and extend beyond each point.