How do you graph and find the discontinuities of #(x^2-25)/(x^2+5x)#?

Question

Eva Abramson · Accepted Answer

We first factorise.

#=((x+5)(x-5))/(x(x+5))# The disconituities happen when the denominator #=0# So at #x=0andx=-5# The discontinuity at #x=5# is removable as both limits go to the same value: #lim_(x->1^-) f(x)=lim_(x->1^+) f(x)=2# In other cases we can cancel out the #(x+5)#'s: #=(x-5)/x=x/x-5/x=1-5/x# This means that if #x# gets larger #+or-# the function goes to #1# (from up or down) graph{1-5/x [-16.02, 16, -8, 8.02]}

Eva Adamson · Answer

undefined To graph the function (x^2-25)/(x^2+5x) and find its discontinuities, follow these steps:

1. Factor the numerator and denominator:
   (x^2-25) can be factored as (x+5)(x-5)
   (x^2+5x) can be factored as x(x+5)

2. Simplify the function by canceling out common factors:
   (x+5)(x-5) / x(x+5)

3. Identify any values of x that would make the denominator equal to zero:
   x = 0 and x = -5 are the values that make the denominator zero.

4. Determine the vertical asymptotes:
   Vertical asymptotes occur at the values of x that make the denominator zero, so in this case, x = 0 and x = -5.

5. Plot the points on the graph:
   Plot the points (0, undefined) and (-5, undefined) to represent the vertical asymptotes.

6. Determine the horizontal asymptote:
   To find the horizontal asymptote, compare the degrees of the numerator and denominator. In this case, both have a degree of 2, so the horizontal asymptote is y = 1.

7. Plot the horizontal asymptote on the graph.

8. Determine any other points of interest:
   You can choose additional x-values to evaluate the function and plot corresponding points on the graph.

9. Connect the points smoothly to create the graph.

10. The discontinuities of the function are at x = 0 and x = -5, where the function is undefined.