Mathematica Help: Plotting y=k-x/k for All -10<k<10

  • Thread starter Thread starter hang2010wang
  • Start date Start date
  • Tags Tags
    Mathematica
AI Thread Summary
The discussion focuses on plotting the function y=k-x/k in Mathematica for values of k ranging from -10 to 10, excluding k=0. Users share various methods to achieve this, including using the Plot and Table functions while ensuring k does not equal zero. One suggested approach involves using Cases to filter out k=0, enhancing the plot's accuracy. Another method combines two separate plots for positive and negative k values. The thread concludes with a user discovering an improved syntax for the plot, demonstrating the collaborative problem-solving nature of the forum.
hang2010wang
Messages
2
Reaction score
0
Hi, Guys
Can someone help me about this question?
I am trying to use Mathematica to plot all the functions: y=k-x/k for all -10<k<10 with interval of 0.1
what I wrote is like this:
Plot[Table[k-x/k,{k,-10,10,0.1}],{x,-10,10}]

However, because the function doesn't have meaning when k=0.

So how can I type the comment to eliminate k=0
 
Mathematics news on Phys.org
I cheated and tried this: Plot[Table[{k - x/k, -k + x/k}, {k, 1, 10, 1}], {x, -10, 10}]
 
You could also try this:
s1=Plot[Table[k-x/k,{k,0.1,10,0.1}],{x,-10,10}]
s2=Plot[Table[k-x/k,{k,-10,-0.1,0.1}],{x,-10,10}]
Show[s1,s2]
Enjoy!
 
HAHA,many thanks for all
I have just discovered that it's even better if type in:
Plot[Table[-x/k+k,Cases[{k,-10,10,0.1},Except[0]]],{x,-10,10}]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top