The parity of the analytic rank of any elliptic curve $E_k: y^2 = x^3 + k$
over ${\bf Q}$ was determined in the paper

Liverance, Eric:
A formula for the root number of a family of elliptic curves.
*J. Number Theory* **51** #2, 288--305 (1995).

(This reference was posted a few weeks ago by Larry Washington
to the NMBRTHRY mailing list in response to another question on
the parity of some curves $E_k$.)

Suppose $k$ is a sixth-power-free integer, and write $k = 2^b 3^c k_1$
with $\gcd(k_1,6) = 1$. Then Liverance writes the sign of the
functional equation of $E_k$ as $-w^{\phantom.}_2 w^{\phantom.}_3 (-1)^r$, where:

$w^{\phantom.}_2 = 1$ or $-1$ depending only on $b$ and on $3^c k_1 \bmod 4$;

$w^{\phantom.}_3 = 1$ or $-1$ depending only on $c$ and on $2^b k_1 \bmod 9$;

and $r$ is the number (*without* multiplicity) of prime factors of $k_1$
congruent to $-1 \bmod 6$.

[NB Liverance's $w^{\phantom.}_2$ and $w^{\phantom.}_3$ are not quite the same as
the local root numbers of $E_k$ at $2$ and $3$, though they are
closely related with these root numbers.]

Therefore: if $k$ is not divisible by the square of any prime
congruent to $-1 \bmod 6$, then the parity is determined entirely by
$w^{\phantom.}_2$, $w^{\phantom.}_3$, and whether $k>0$ or $k<0$.
This happens in particular if $k$ has no $-1 \bmod 6$
factors at all, which is the case for the quadratic polynomials
$k = -108 t^2 + 36 t - 7$ and $k = -108 t^2 + 36 t - 67$
in K.Acres' self-answer, because their discriminants are of the form $-3d^2$.
Moreover, for each of these polynomials $k \bmod 36$ is constant with $\gcd(k,6) = 1$,
so the sign of the functional equation is the same for all $t$.
The smaller examples $-6t^2-8$ and $6t^2+2$ from K.Acres' comment
also have discriminants $-3d^2$, but showing that their sign is constant
requires some case analysis for the variation of $b$, $c$, and $k_1$ with $t$.

Even when the sign is not fully predictable there can be large biases.
For example, in the arithmetic progression $k=36n+1$ the sign is
*usually* $+1$ if $k>0$ and $-1$ if $k<0$. There are exceptions
(starting at $k=36\cdot8+1 = 17^2$ and $k=36\cdot9+1 = 5^2 13$),
but they require that $k$ be divisible by $p^2$ for some
$p \equiv -1 \bmod 6$, and that happens less than 6% of the time.

allthe curves $y^2 = x^3 + k$ of rank at least $8$ with $|k| \leq N$ and $N$ large enough to find $10^4$ such curves. – Noam D. Elkies Aug 2 '12 at 6:47