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 May 5th, 2010, 04:30 AM #1 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Plotting Hi, I have two functions: y = 1/5 (4x -3) and its inverse function: y = 5x+3/4 -- How do I plot and represent these on an xy graph ? wulfgarpro. May 5th, 2010, 05:28 AM #2 Global Moderator   Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Plotting I prefer to start with the equations in "slope-intercept form" (i.e. "y = mx + b"). Consider the equation y = (3/4)x + (-5) The slope is 3/4. This means that a vertical increase of 3 units corresponds to a horizontal increase of 4 units. ("rise over run") The y-intercept is (-5). You can find this directly by evaluating when x = 0... Step 1: Plot the y-intercept. In this case, we go down 5 units from the origin and mark the point (0, -5). Step 2: Use the SLOPE to get from this point to another point, and then from your new point to another, and so on. So from (0,-5), we go UP 3 and RIGHT 4 units. Up 3 to (0,-2) and right 4 to (4,-2) (4, -2) is another point on the line. At this point you can draw the line containing the two points and you're done. Another method is to use the intercepts. Let's look at your "inverse" equation y = 5x + 3/4 When x = 0, y = 3/4 When y = 0, 5x = -3/4, so x = -3/20. This information can be consolidated to the two points: (0,3/4) and (-3/20, 0). Plot these points. Draw the line containing them. Boom. Tags plotting Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post benmuskler Algebra 1 September 28th, 2013 03:23 PM uperkurk Elementary Math 1 January 25th, 2012 02:55 PM mathsissmart Computer Science 2 August 26th, 2011 09:55 AM Matt Sullivan Math Software 2 January 4th, 2011 06:13 AM TranzRail New Users 6 October 8th, 2009 06:43 PM

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