Suppose $a, b$, and $c$ are real numbers with $a<b<0<c$. Let $f(x)$ be the quadratic function $f(x)=(x-a)(x-c)$ and $g(x)$ be the cubic function $g(x)=(x-a)(x-b)(x-c)$. Both $f(x)$ and $g(x)$ have the same $y$-intercept of -8 and $g(x)$ passes through the point $(-a, 8)$. Determine the value of $c$.

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Since $f(0)=-8$, we have $-8=(0-a)(0-c)=a c$.

Since $g(0)=-8$, we have $-8=(0-a)(0-b)(0-c)=-a b c$ or $a b c=8$.

Substituting $a c=-8$ into $a b c=8$ and solving for $b$, we get $b=-1$.

We also are given that $g(-a)=8$. Since $b=-1$, this means $8=(-a-a)(-a+1)(-a-c)$ which can be simplified to $8=-(-2 a)(1-a)(a+c)$ and then to $a(a+c)(1-a)=4$.

Expanding the above equation, we get $4=a\left(a-a^{2}+c-a c\right)=a^{2}-a^{3}+a c-a(a c)$.

From earlier, we have that $a c=-8$, so this equation simplifies to $4=a^{2}-a^{3}-8-a(-8)$ which can be rearranged to $a^{3}-a^{2}-8 a+12=0$.

We can factor $(a-2)$ out of the expression on the left to get $(a-2)\left(a^{2}+a-6\right)=0$ which further factors as $(a-2)(a-2)(a+3)=0$.

This means that either $a=2$ or $a=-3$. Since $a<0, a \neq 2$, so $a=-3$.

Since $a c=-8$, this gives $c=\frac{8}{3}$.

ANSWER: $\frac{8}{3}$