Posted by **~christina~** on Sunday, September 2, 2007 at 1:22am.

Q: How could the function y=3t^2 +4 be plotted on a cartesian graph to produce a straight line? What would be the numerical values of the slope and the intercept of the line?

For this it's explained in the lab book for physics as if the function is y=mx+b it would be a straight line but if it's y=mx^2+b then it is a parabola. It's explained that to get a straight line I should go and make the equation with the square y=mx'+b with x'= x^2. Then it says that if you plotted these numbers with the corresponding y values you'd get a straight line.

I'm not sure I understand how this would work.

For the 2nd part, wouldn't the slope be the same as the original equation (3)? and the y intercept be the same as the original equation too (4)?

Thanks =D

- graphing -
**drwls**, Sunday, September 2, 2007 at 4:05am
Let x = t^2 and use equal increments of t^2 along the x axis, when plotting y vs x.

For t = 0, x = 0.

For t = 1, x = 1.

For t = 1.414, x = 2

For t = 1.732, x = 3 .

For t= 2, x = 4, etc.

The slope of the line you get will be 3.

- graphing -
**~christina~**, Sunday, September 2, 2007 at 10:21am
Okay thanks but wouldn't the y intercept still equal 4 DrWls?

- graphing -
**drwls**, Sunday, September 2, 2007 at 5:48pm
Yes, when you plot y vs x (=t^2) . I forgot to mention the intercept. Sorry

- graphing -
**~christina~**, Sunday, September 2, 2007 at 9:29pm
Thanks DrWls =D

- graphing -
**G**, Wednesday, September 11, 2013 at 5:33pm
I have no idea how to do this

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