Step 1: Choose the right form.
The axis is parallel to the $y$-axis, so the parabola is $y = ax^2 + bx + c$. We will find $a,b,c$ from the three given points.
Step 2: Use $(0,4)$.
Substituting $x=0,\,y=4$ gives $c = 4$.
Step 3: Use $(1,9)$.
Substituting $x=1,\,y=9$: $9 = a + b + 4$, so $a + b = 5$.
Step 4: Use $(4,5)$.
Substituting $x=4,\,y=5$: $5 = 16a + 4b + 4$, so $16a + 4b = 1$.
Step 5: Solve the system.
From $a+b=5$ we get $b = 5-a$. Substitute: $16a + 4(5-a) = 1$, i.e. $12a + 20 = 1$, so $12a = -19$ and $a = -\frac{19}{12}$. Then $b = 5 + \frac{19}{12} = \frac{79}{12}$.
Step 6: Assemble and clear fractions.
The parabola is $y = -\frac{19}{12}x^2 + \frac{79}{12}x + 4$. Multiply by $12$: $12y = -19x^2 + 79x + 48$, which rearranges to $19x^2 + 12y - 79x - 48 = 0$.
\[ \boxed{19x^2+12y-79x-48=0} \]