One interval where the graphs concave upwards, and one where it is concave downwards

**Where is the function Concave Up and Down?**

Hi im trying to find one interval where the function is concave up and one interval where it is concave down. The function is: f(x)=(x+5)/(x^2-16). Heres my information. What points do i use from my notes?

would it be between the inflection point and -4 asymptote (theres a inflextion)?

Asymptotoes are x=4 x=-4

Local minimum/maximum:

f'(x) = 0

−(xｲ+10x+16)/(xｲ−16)ｲ = 0

xｲ+10x+16 = 0

(x+8) (x+2) = 0

x = −8, x = −2

f(−2) = (−2+5)/(4−16) = −1/4

f''(x) = 2(xｳ+15xｲ+48x+80)/(xｲ−16)ｳ

f''(−8) = 2(−512+960−384+80)/(64−16)ｳ = 1/384 > 0 -----> local minimum

f''(−2) = 2(−8+60−96+80)/(4−16)ｳ = −1/24 < 0 ------------> local maximum

Local minimum at point (−8, −1/16)

Local maximum at point (−2, −1/4)

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Points of inflection:

f''(x) = 0

2(xｳ+15xｲ+48x+80)/(xｲ−16)ｳ = 0

xｳ + 15xｲ + 48x + 80 = 0

There is no easy way to solve this. Using graphing calculator, I found:

x = −5 − 3^(4/3) − 3^(2/3) ≈ −11.406832534

f(−5 − 3^(4/3) − 3^(2/3)) ≈ −0.056143242

Point of inflection: (−11.406832534, −0.056143242)