jmc

The line $x=k$ intersects the graph of the parabola $x=-2y^2-3y+5$ at exactly one point if and only if the equation $-2y^2-3y+5=k$ has exactly one real solution. This equation is equivalent to $$2y^2+3y+(k-5)=0,$$ and this equation has exactly one real solution if and only if the discriminant is 0. The discriminant of this quadratic is $3^2-4\cdot 2\cdot (k-5)$. Setting this equal to 0 and solving for $k$, we find $k=\boxed{\frac{49}{8}}$. (Note that this is the $x$-coordinate of the vertex of the parabola.)