How can I determine this polynomial of degree 3?

Question

I want to find a polynomial of degree 3 which passes through $(2,0) ,(-2,4),(-4,8)$ and has a minimum on the y-axis.

I wrote $f(x)=ax^3+bx^2+cx+d$, then we need to satisfy

1. $f(2)=0$
2. $f(-2)=4$
3. $f(-4)=8$
4. $f'(0)=0$

However, solving this system of equations gives me $$f(x)=-\frac{1}{4}x^3-\frac{5}{6}x^2+\frac{16}{3}$$ I know this is wrong because $$f ''(0)=2(-5/6)<0$$ means we don‘t have a minimum on the y axis.

Where is my mistake?

Answer 1

Your algebra is correct - it tells you there is no such polynomial.

There's a critical point on the $y$ axis but it's a local maximum. Here's the picture from desmos

Answer 2

Clearly there is a mistake — but it's not yours. The given cubic polynomial satisfies the stated constraints, which apparently include a typographical error. If the minimum was meant to be on the x-axis, then the cubic obtained with that constraint (which is different from the one here, due to the different minimum criterion) will work.