If the curve $y=a x^2-6 x+b$ passes through $(0,4)$ and has its tangent parallel to the x -axis at $x=\frac{3}{2}$, then the value of $a$ and b respectively are

The curve $y=ax^2-6x+b$ passes through the point $(0,4)$, so substituting gives $4=a(0)^2-6(0)+b$, which simplifies to $b=4$.

Since the tangent is parallel to the x-axis at $x=\frac{3}{2}$, the derivative $\frac{dy}{dx}=2ax-6$ must be zero at that point. Setting $2a\cdot\frac{3}{2}-6=0$ yields $3a-6=0$, so $a=2$.

The values are therefore $a=2$ and $b=4$, corresponding to option B.