Formula with 2 points of inflection $x^3$ has a point of inflection at $x=0$. How will you modify the formula to add a 2nd point of inflection at $x=1$?
Plot of $x^3$

Plot of $x^3(x-1)^3$


Update
The plot I am aiming to achieve has a shape similar to the graph below. However I would like the 2 inflection points at $(0, 1)$ and $(1, 0.05)$, and intersects the x-axis at $(1.5,0)$ and y-axis at $(0,1)$.
Graph with similar target shape

Current attempt
The closest I can get is using $1 - [ 16x^3 - 23x^4 + 9x^5 ]$ using J.M.'s equation inside the $[]$ square brackets with $\alpha=2$ and $\beta=1$. How should I bring the point of inflection at $x=1$ up to around $y=0.05$? Its currently at $y=-1$


One rather general family is given by
$$2(5\alpha-2\beta)x^3+(7\beta-15\alpha)x^4+3(2\alpha-\beta)x^5$$
I obtained this through Hermite interpolation. That involves derivatives, so I'm not sure if this counts for a "precalculus" answer.


I would like the 2 inflection points at $(0,1)$ and $(1,0.05)$, and intersects the x-axis at $(1.5,0)$ and y-axis at $(0,1)$.

Using Hermite interpolation again, we obtain the polynomial
$$1-\frac{3977}{270}x^3+\frac{5389}{180}x^4-\frac{385}{18}x^5+\frac{706}{135}x^6$$
You can easily verify that this has the properties needed.
Another possibility, whose shape is a bit nearer to what the OP seems to want, is
$$1-\frac{6491}{270}x^3+\frac{34603}{540}x^4-\frac{1223}{18}x^5+\frac{4477}{135}x^6-\frac{838}{135}x^7$$
