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When I try and graph $y = |2x-3|$ I created a table of values ranging from $-3$ to $3$.

However, when I used an online graphing tool, and looked at the worked solution page, I noticed that the line reaches the $x$ axis at $1.5$.

I cannot know this using the table of values, so there must obviously be another way of fixing my graph. Is their a technique for graphing such inequities? Do I have to solve for $x$ first?

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  • $\begingroup$ where 2x-3 is positive? where negative? now use definition. $\endgroup$ – user147308 Jun 8 '14 at 7:21
  • $\begingroup$ When you create your table of values and then plot your graph, you should have a straight line decreasing at forty five degrees followed by a straight line increasing at forty five degrees. You perhaps haven't got quite enough values to see this clearly in your table. But if you extend the lines through the points you have and draw accurately it should be obvious that they meet at $(1.5,0)$ and then you can check that directly by substituting the values into the equation. $\endgroup$ – Mark Bennet Jun 8 '14 at 7:33
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    $\begingroup$ This is one of the reasons why sketching graphs using a table of values and nothing more is a very bad idea! You should always try to understand the general shape of the graph and then, if necessary, plot a small number of points in order to refine it. One thing you could do which is frequently useful is to find where the graph intersects the axes. In this case it is easy to solve $|2x-3|=0$. $\endgroup$ – David Jun 8 '14 at 8:17
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Solve $y=|2x-3|$ by substituting different values of $x$, following is the table you will get:

enter image description here

Now observe that $y$ is same for $x=1$ and $x=2$, so let us find what is the value of $y$ between $x=1$ and $x=2$, put $x=\frac{3}{2}$, that gives $y=0$

and the graph is as follows:

enter image description here

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enter image description here

Taking modulus of a graph essentially means reflecting that portion of the graph where y is -ve, using X-axis as the mirror, and then plotting the resultant image. The rest portion of graph (containing +ve values for y) remains as it is.

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You consider the function $$y = |2x-3|$$ When ($2x-3$) is positive, then $y=2x-3$ but when ($2x-3$) is negative, then $y=3-2x$. So, you have to plot two functions and overlap them.

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