## HT3347 POINT OF INTERSECTION

751 Views 4 Replies Latest reply: Feb 1, 2013 11:07 AM by Badunit
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Jan 31, 2013 9:45 AM

In the NUMBERS programme, how do I find the point of intersection between a a plotted line and the x axis ( accurate point of int. not just looking at it obviously)

MacBook Pro with Retina display, OS X Mountain Lion (10.8.2)
• Level 7 (27,470 points)
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Jan 31, 2013 10:29 AM (in response to Alida-Sophie)

Alida-Sophie,

I don't believe that there is one correct answer to your question. Any line on your chart that connects two data points is just an estimate of what the data may have been at those other points. Since there probably is no one "right" ansewr, your "looking at it" method is probably the best way to go.

If you particularly like one of the "estimate" lines, you can "show equation" and then solve for Y=0. This will be easy only for a linear approximation, and the result is not likely to be better than just looking at the chart.

Jerry

• Level 4 (3,645 points)
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Feb 1, 2013 3:15 AM (in response to Alida-Sophie)

Hi AS,

I agree with everything that Jerry said. A fitted line is just a "best fit" to the data. And data (data are plural) always include unavoidable errors.

Are you sure that you mean "the point of intersection between a a plotted line and the x axis"

Usually, the x axis is the uncontrolled (and often uncontrollable) variable. The y axis is the measured variable. The intercept is usually an estimated value of y when x=0.

That said, you can create a chart (graph in my language). I suggest a scatter plot. Then add an *estimated* line of best fit. To add the equation of best fit, select the graph, then open Inspector > Chart Tab > Series > Advanced > Trendline. You can choose various types of trendlines, but I assume that you have a linear graph.

At the bottom of the Inspector box, click on Show Equation.

The equation shows slope and intercept. For example, if the equation is:

Y = 0.76x +0.13

The slope of the "best fit" line is 0.76 and the intercept on the **Y** axis (where x = 0) is 0.13

If your post was correctly phrased, and you do want to find the intercept on the **X** axis, solve the equation for x.

You can also show R-squared (at the bottom of the Inspector box, click on Show R^2). This is reliable only for linear graphs. Beware of R-squared for non-linear graphs!

Cheers,

Ian.

• Level 4 (3,645 points)
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Feb 1, 2013 3:36 AM (in response to Yellowbox)

P.S. I was wrong about X being the "uncontrolled" variable. Sorry.

• Level 6 (10,525 points)
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Feb 1, 2013 11:07 AM (in response to Alida-Sophie)

If it is a linear fit and you want to find the X intercept (the point where Y = 0):

y=m*x + b

0 = m*x + b

x = -b/m

So the X intercept = -INTERCEPT(y-values, x-values)/SLOPE(y-values, x-values)

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