First of all, since the dilation factor is equal to $1$, the equation automatically reduces to $y=(x-b)^3+c$. Now, since the equation only has a single factor, $(x-b)$, the vector $(b,c)$ is the result we're looking for. Now, the $b$ and $c$ can be found by plugging in the given intercepts:
$$0=(-4-b)^3+c$$$$28=(-b)^3+c$$
Since this is a system of nonlinear equations, we can solve it by substitution.
$$c=-(-4-b)^3=(4+b)^3$$$$28=(-b)^3+(4+b)^3$$
Since we can't easily combine the right-hand side of this in factored form, we'll expand it.
$$28=64+48b+12b^2+b^3-b^3=64+48b+12b^2$$
Well now, that's just a simple quadratic. Solving it, we get two solutions: $(-1, 0)$ and $(-3, 0)$. Let's try plugging both of those back into the first equation:
$$c=(4-1)^3=27$$$$c=(4-3)^3=1$$
So now we have two possible coordinate pairs, $(-1, 27)$ and $(-3, 1)$. If you graph these equations, you'll see that either coordinate works.
Let's just take $f_1(x) = x^4-x^3-4x^2 = x^2\ (x^2 - x - 4)$. Start by finding the roots (the values of $x$ where $f_1(x) = 0$). A quartic equation may have up to four real roots. From the $x^2$ factor, we see that two of them are $x = 0$ and $x = 0$ — a "double" root.
What about the other two roots? To solve $x^2 - x - 4 = 0$, we can use the quadratic equation:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{17}}{2} $$
So, $y = x^2 - x - 4$ would describe a parabola that crosses the $x$-axis at -1.562… and 2.562…. It is centered at $x = \frac{1}{2}$ (the midpoint between the roots), with a minimum at (0.5, -4.25).
Now let's consider what happens when we multiply that by $x^2$. That turns the parabola into a lopsided W-like curve (but smoother than a W), shifted to the right of the origin, and touching the $x$-axis at -1.562…, 0, 0, and 2.562….
To find interesting points such as the minima and inflection points, though, you would need calculus. To give you a taste of what you haven't learned… we take the first derivative ("$\frac{d}{dx}\ f_1(x)$"), which tells you the slope of the curve for any $x$.
$$\frac{d}{dx}\ f_1(x) = 4x^3 - 3x^2 - 4 \cdot 2 \cdot x = \frac{x}{16} (8x - (3 - \sqrt{137}))(8x - (3 + \sqrt{137}))$$
We find that the slope is 0 at $x = 0$, $x \approx -1.088$, and $x \approx 1.838$. We conclude that $f_1(x)$ has a local maximum at (0, 0), a local minimum at (-1.088, -2.0458), and a local minimum at (1.838, -8.3097).
To find the inflection points, we would take the derivative of the derivative. This second derivative is
$$\frac{d^2}{dx^2}\ f_1(x) = 4 \cdot 3 \cdot x^2 - 3 \cdot 2 \cdot x - 8 = \frac{1}{12} (12x - (3 - \sqrt{105}))(12x - (3 + \sqrt{105}))$$
The inflection points, where the second derivative is zero, are at $x \approx -0.604$ and $x \approx 1.104$. That is, $f_1(x)$ is $\cap$-shaped in the interval $\frac{3-\sqrt{105}}{12} \lt x \lt \frac{3+\sqrt{105}}{12}$, and $\cup$-shaped everywhere else.
All of the information above can be seen in the plot. The roots are the red points, the extrema are green, and the inflection points are purple.
Best Answer
Welcome! You simply forgot that having the turning points provides the derivative up to a nonzero constant factor, i.e. $$y'(x)=K(x+2)(x-4),\quad K\in \mathbf R^*, \quad \text{ whence }\;y(x)=K\biggl(\frac{x^3}3 -x^2-8x\biggr)+C.$$ You now have two constants to adjust the ordinates.