X- and Y-InterceptsDate: 09/15/98 at 20:24:01 From: Mandy Haun Subject: Algebra - Quick Graphs Using Intercepts Dr. Math, I'm having a hard time finding out how to solve the equation and then do the graph. The intercepts confuse me also. One problem I'm having trouble with is: y = 4x + 8. I know you have to exchange the x with 0 so it would be y = 4(0) + 8. Now, I have no clue what to do. Can you help me? Date: 09/16/98 at 13:10:35 From: Doctor Peterson Subject: Re: Algebra - Quick Graphs Using Intercepts Hi, Mandy. What you are trying to do is to find two points on the line that you can graph easily, so you can draw the line between them and you're done. These two points are the x-intercept, where the line crosses the x axis, and the y-intercept, where it crosses the y axis. The y-intercept is easiest. When the graph crosses the y axis, its x coordinate is zero, so all we are asking is, "what is the value of y when x is zero?" All we have to do is put 0 for x and calculate y, as you started to do: y = 4x + 8; x = 0 y = 4*0 + 8 = 8 So the y-intercept is 8; that is, one point we can graph is (0, 8). The x-intercept is the opposite. It's the point on the line where y is zero. This time we have to set y to zero and solve for x: y = 4x + 8 and y = 0 0 = 4x + 8 -8 = 4x -2 = x Solving this kind of equation is very simple. All I had to do was to subtract the 8 from both sides, then divide both sides by the 4 to leave the x by itself. Now I know that the x-intercept is -2, so the point (-2, 0) is on the line. Now I can plot those points and draw the line: |/ *8 <--- y-intercept /| / | / | / | / | / | -*------+---- /-2 | ^ | x-intercept I hope that helps. Let me know if you need more help. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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