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

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Answer 1 · 2015-06-07T21:48:38

$f(x) = x^2/(x+5) = ((x^2 + 5x) - (5x + 25) + 25)/(x+5)$

$=((x-5)(x+5)+25)/(x+5)$

$=x-5 + 25/(x+5)$

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]}

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