I assume you mean "greatest $k$", since the problem has no solution if we're looking for a "least $k$" (because for any $k$ that works, so does $k - 1$).
The strategy involves noting that the minimal value of $f$ on the interval $[n, n+1]$ for $n$ integral is $-\frac{1}{4} 2^n$. This can be shown first by establishing the result on $[0, 1]$ (using some calculus or plain old completing the square) and then extending it to $[n, n+1]$ for positive $n$ by induction, then for negative $n$ by induction.
This means that the value in question must (if it exists) be in the interval $[2, 3]$. Over this interval, we find $f(x) = 4 (x - 2)(x - 3)$. Note that on the interval $[2, 5/2]$, $f$ is strictly decreasing. Further, note that $f(2) = 0$ and $f(5/2) = -1$. Find the $k$ in $(2, 5/2)$ such that $f(k) = -3/4$ (this $k$ exists by the intermediate value theorem). It turns out that this $k$ is $9/4$.
Now suppose that $x \leq k$. Case 1: $x \in [n, n+1]$ for some $n \leq 1$. Then $f(x) \geq -\frac{1}{4}2^n \geq -\frac{1}{2}$. Case 2: $x \in [2, k]$. Since $f$ is decreasing over $[2, k]$, we have $f(x) \geq f(k) = -3/4$. In either case, $x \leq k$ implies $f(x) \geq -3/4$.
Suppose we have some $k'$ where for all $x \leq k'$, we have $f(x) \geq -3/4$. Then suppose for sake of contradiction that $k' > k$. Then we can take some $x \in (k, 3] \cap [k, k')$. Since $f$ is strictly decreasing on $[k, 3]$, we would have $f(x) < f(k) = -3/4$. But $x < k'$ and thus $f(x) \geq -3/4$; contradiction. Then $k' \leq k$.
Thus, we have shown that $k = 9/4$ is the greatest $k$ such that for all $x \leq k$, $f(x) \geq -\frac{3}{4}$.
