How do you find the points where the graph of the function #(x-2)(x^2+1) / (x+3)# has horizontal tangents and what is the equation?

Question

1142 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-30T15:00:22

See explanation

graph{y(x+3)-(x^2+1)(x-2)=0 [-5, 5, -2.5, 2.5]}
graph{y(x+3)-(1+x^2)(x-2)=0 [-320, 320, -160, 160]}

It is evident from the first graph that

there is a point of inflexion near x = 1.

The second contracted graph reveals the missing part on the left

that discloses a horizontal tangent that is close to y = 95.

By actual division. #y = x^2-5x+16-50/(x+3)#

#y'=2x-5+50/(x+3)^2#=0, when

#2x^3+7x^2-12x+5=0#

A zero of y' is bracketed in # (-6, -5)#.

If #(alpha, beta)# is this point, #y = beta in (91, 99)# is the equation of

this horizontal tangent.