ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sun, 01 Sep 2013 11:16:45 +0200The plot of x=ln(y) looks just like y=ln(x)https://ask.sagemath.org/question/10480/the-plot-of-xlny-looks-just-like-ylnx/According to the author, x=ln(y) should go through y=1, not x=1, making the plot look similar to a half U on the positive side of the x-axis, instead of a half C, on the the positive side of the y-axis.
var('y')
f=ln(y)
plot(f, (x,-5,5), ymin=-5, ymax= 5, aspect_ratio=1)
bxdinSun, 01 Sep 2013 11:16:45 +0200https://ask.sagemath.org/question/10480/Graph based on y value as the input, and x as the output?https://ask.sagemath.org/question/10307/graph-based-on-y-value-as-the-input-and-x-as-the-output/I need to identify the vertex of the following parabola:
3x-7 = y^2-5y
According to the book's author, since this contains a y^2 term rather than an x^2 term; I must obtain an end result with the standard form x = a(y - k)^2 + h
The answer, i.e. vertex, is (1/4, 5/2)
I did the following in sage 5.9, and as you can see the vertex point isn't located at the parabola, since the x in the 1st statement's equation, should be y. But sage says y is undefined:
p1 = plot( (1/3)*((x - 5/2)^2) + 1/4, (x,-15, 15), ymin =-5 , ymax = 25)
p3 = point((1/4, 5/2), size = 25)
p0 = p1 + p3
show(p0)
Is it possible for Sage to graph the equation based f(y), as opposed to the standard f(x), so i can see the vertex match up with the parabola.bxdinMon, 01 Jul 2013 13:39:49 +0200https://ask.sagemath.org/question/10307/