Recover a rational-curve parameter from an inflection

Build on derivative and quotient-graph foundations

Second derivatives locate candidates; quotient-graph analysis retains poles and the original domain. Here the inflection coordinate is given and a coefficient is unknown. Solve the resulting condition before verifying the geometry

An original parameter problem

The curve has an inflection at the stated coordinate. Recover the parameter, inflection ordinate and tangent

For u=x−1 and f(x)=u+3/u+B/u², which x-value is excluded from the curve?

Differentiate before substituting

Let u=x−1 and f(x)=u+3/u+B/u², with an inflection at x=−1. Because du/dx=1, f′=1−3/u²−2B/u³ and f″=6/u³+6B/u⁴. At u=−2, f″=0 gives 6(−2)+6B=0 after multiplication by u⁴. Thus B=2.

For u=x−1 and f(x)=u+3/u+B/u², the curve has an inflection at x=−1. Recover B.

Check a change of concavity

With B=2, f″=6(u+2)/u⁴. The denominator is positive for u≠0. On the continuous branch u<0, f″ changes from negative to positive at u=−2. Thus x=−1 is a genuine inflection

The curve is defined and continuous for u<0; u=0 is excluded. Why does the candidate qualify as an inflection?

Find the local point and tangent

Let u=x−1 and f(x)=u+3/u+2/u². At x=−1, u=−2. y=−2−3/2+2/4=−3. Also f′=1−3/4+4/8=3/4. Through (−1,−3), the tangent is y=(3/4)x−9/4.

Let u=x−1 and f(x)=u+3/u+2/u². Give the ordinate at the inflection x=−1.

Let u=x−1 and f(x)=u+3/u+2/u². Give the tangent gradient at x=−1.

Let u=x−1 and f(x)=u+3/u+2/u². Give the intercept q of the tangent at x=−1 in y=mx+q.

A second local check

Let u=x−1 and f(x)=u+3/u+2/u². At x=−1, u=−2. Differentiate again: f‴=−18/u⁴−48/u⁵. At u=−2 this is 3/8, which is nonzero. In this smooth family that verifies a simple zero of f″. This does not turn the excluded pole into a graph point

Let u=x−1 and f(x)=u+3/u+2/u². Give f‴ at the inflection x=−1.

Recover and justify the whole curve feature, ungraded

Let u=x−1 and f(x)=u+3/u+B/u², with an inflection at x=−1. For this curve, recover B without assuming f′=0, justify the actual concavity change and obtain the tangent. Explain why the excluded x=1 cannot be an inflection. Sketch the local crossing of the tangent using your derivatives

Model: parameter recovery and local geometry

Let u=x−1 and f(x)=u+3/u+B/u². At x=−1, u=−2≠0. Since f″=6(u+B)/u⁴, f″=0 gives B=2. Then f″<0 for u<−2 and f″>0 for −2<u<0: both intervals meet at a valid continuous point. The original rule gives y=−3; f′ gives 3/4, so the tangent is y=(3/4)x−9/4. The nonzero f‴ value 3/8 confirms the local change. The pole at x=1 is outside the curve, regardless of any sign change across it. A local sketch must show the point and tangent with concavity down to the left and up to the right