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 $y=-1$

 A: 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
$$1-\frac{6491}{270}x^3+\frac{34603}{540}x^4-\frac{1223}{18}x^5+\frac{4477}{135}x^6-\frac{838}{135}x^7$$
A: Use a fourth-degree polynomial $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, and get that:
$$ f''(x) = 12ax^2 + 6bx + c = 0 \text{ at } x = 0 \text{ and } 1$$ 
So we get that:
$$ c = 0 \text{ and } 12a + 6b = 0$$
Well that's still not enough! Based on what you said, we also know that $e = 1$ (since $f(0) = 1$). We also need that:
$$f(1) = a+b+c+d+e = \\ a + b + d + 1 = 0.05$$
So close, and your last criterion, that $f(1.5) = 0$, will let us finish it:
$$ 5.0625 a + 3.375 b + 1.5d + 1 = 0$$
We have the following system of equations:
$$ 2a + b = 0 \\ a + b + d + 1 = 0.05 \\ 5.0625 a + 3.375 b + 1.5d + 1 = 0 $$
Solving the system of equations, we get:
$$a = -\frac{34}{15} \\
  b =  \frac{68}{15} \\
  d = -\frac{193}{60} $$
Graphing the result:

A: A fourth degree polynomial can have two points of inflection. Let's pick an arbitrary one: $f(x) = ax^4 + bx^3 + cx^2 + dx + e$.
Inflection points occur when the second derivative is $0$: $f^{\prime\prime}(x) = 12ax^2 + 6bx^2 + c = 0$. We want these roots to occur at 0 and 1.
Substituting $x=0$, we get $c = 0$. Substituting $x=1$, we can pick $a = 1, b = -2$. Put these back into the original equation: $x^4 - 2x^3 + dx + e$. One solution is simply $x^4 - 2x^3$, but a prettier one is $x^4 - 2x^3 + x + \frac{1}{4}$
