Understanding Y² Graphing and Confusing Points on the X-Axis

  • Thread starter Thread starter qazxsw11111
  • Start date Start date
  • Tags Tags
    Graphing Graphs
qazxsw11111
Messages
95
Reaction score
0
Hi everyone. I am familiar with the graphing of y² graph which is essentially sqrt of the graph then reflected in the x-axis.

However, I am confused at the points where the y= graph cuts the x-axis. According to differentiation, the gradient of the sqrt y graph should have a sqrt y at the denominator using quotient rule. If y=0, the gradient should be infinity. However, I do see some graphs with some other shapes (some crossing, some flat) at the x-axis so I am wondering about this.

Hope someone can help to clarify my problems.
 
Physics news on Phys.org
Have you tried simply plugging in points and making a rough graph?
 
Yeah of course that would work but I was wondering if there is a general rule to see aside from just plotting it out since some question don't give the equation but just give a pictorial graph and ask you to transform.
 
qazxsw11111 said:
Hi everyone. I am familiar with the graphing of y² graph which is essentially sqrt of the graph then reflected in the x-axis.
Can you be more specific about what you're trying to do? This doesn't make any sense to me. An ordinary graph of a function y = f(x) is a plot of the points (x, y) that satisfy the equation y = f(x). Are you trying to see what you get when you plot the pairs (x, y^2)?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top