How do you graph #f(x)=x^2/(x+5)#?

Question

Abigail Allen · Accepted Answer

undefined To graph the function f(x) = x^2/(x+5), follow these steps:

1. Determine the domain of the function by identifying any values of x that would make the denominator (x+5) equal to zero. In this case, x cannot be -5.

2. Find the y-intercept by substituting x=0 into the function: f(0) = 0^2/(0+5) = 0.

3. Determine the x-intercept(s) by setting the numerator (x^2) equal to zero and solving for x. In this case, x=0 is the only x-intercept.

4. Analyze the end behavior of the function. As x approaches positive or negative infinity, the function approaches zero.

5. Plot the points obtained from the y-intercept, x-intercept, and any additional points you choose.

6. Draw a smooth curve passing through the plotted points, considering the behavior of the function as x approaches -5 and infinity.

This will give you the graph of f(x) = x^2/(x+5).

Eva Adamson · Answer

#f(x) = x^2/(x+5) = ((x^2 + 5x) - (5x + 25) + 25)/(x+5)# #=((x-5)(x+5)+25)/(x+5)# #=x-5 + 25/(x+5)# For large positive or negative #x# this will be asymptotic to #x - 5# #f(x)# has a simple pole at #x = -5# with #f(-5-epsilon)# being large and negative, and #f(-5+epsilon)# is large and positive for small #epsilon > 0# #f(0) = 0# so the curve passes through #(0, 0)# #f'(x) = (2x)/(x+5)-x^2/(x+5)^2# #=((2x)(x+5)-x^2)/(x+5)^2# #=(x^2+10x)/(x+5)^2# #=(x(x+10))/(x+5)^2# So #f'(x) = 0# when #x = 0# and #x = -10# So there is a local minimum at #(0, 0)# and a local maximum at #(-10, f(-10)) = (-10, -20)# graph{x^2/(x+5) [-86.4, 73.6, -45.1, 34.9]}