Find the inflection points of the curve y =
Find the inflection points of the curve $y = \bigl(x-a\bigr)^{2n+1}+ax$, where $n\in\mathbb{N}$ and $a\neq0$.
Explanation of Method
An odd power always bends the curve through its centre.
Step 1: The derivatives
\[ y' = \bigl(2n+1\bigr)\bigl(x-a\bigr)^{2n}+a, \]
\[ y'' = 2n\bigl(2n+1\bigr)\bigl(x-a\bigr)^{2n-1}. \]
Step 2: The candidate
Since $n\ge1$, the coefficient $2n\bigl(2n+1\bigr)$ is positive, so
\[ y'' = 0 \;\Longleftrightarrow\; \bigl(x-a\bigr)^{2n-1} = 0 \;\Longleftrightarrow\; x = a. \]
Step 3: The sign changes
The exponent $2n-1$ is odd, so $\bigl(x-a\bigr)^{2n-1}$ carries the sign of $x-a$: $x\lt a:\ y''\lt 0; \qquad x\gt a:\ y''\gt 0$.
The concavity reverses, so $x = a$ is a genuine inflection for every $n\ge1$.
Step 4: The point
\[ y(a) = 0+a\cdot a = a^{2}. \]
Step 5: Verification
\[n = 1,\ a = 1:\ y = \bigl(x-1\bigr)^{3}+x,\ y'' = 6\bigl(x-1\bigr),\ \text{inflection } \bigl(1,1\bigr) \;\;\checkmark\]
\[n = 2,\ a = 1:\ y'' = 20\bigl(x-1\bigr)^{3},\ \text{the same point } \bigl(1,1\bigr) \;\;\checkmark\]
\[y'(a) = a:\ \text{the tangent there has slope } a,\ \text{so the inflection is not stationary} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl(a,\;a^{2}\bigr) \qquad\text{for every } n\in\mathbb{N},\ a\neq0\,}\]
Remark
Remark. The added term $ax$ is linear and contributes nothing to $y''$, so it cannot create or destroy an inflection — it only tilts the tangent there, from horizontal to slope $a$. As $n$ grows the curve becomes flatter and flatter near $x = a$ while the inflection stays put: the point $\bigl(a,a^{2}\bigr)$ lies on the parabola $y = x^{2}$ as $a$ varies.