f(x) = x^2/(x+5) = ((x^2 + 5x) - (5x + 25) + 25)/(x+5)f(x)=x2x+5=(x2+5x)−(5x+25)+25x+5
=((x-5)(x+5)+25)/(x+5)=(x−5)(x+5)+25x+5
=x-5 + 25/(x+5)=x−5+25x+5
For large positive or negative xx this will be asymptotic to x - 5x−5
f(x)f(x) has a simple pole at x = -5x=−5 with f(-5-epsilon)f(−5−ε) being large and negative, and f(-5+epsilon)f(−5+ε) is large and positive for small epsilon > 0ε>0
f(0) = 0f(0)=0 so the curve passes through (0, 0)(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]}
