Suppose a parabola passes through $\left(0,4\right),\left(1,9\right)$ and $\left(4,5\right)$ and has its axis parallel to the $y$-axis. Then the equation of the parabola is

Let the equation of parabola be $y=a{x}^2+bx+c$

Now given it passes through $\left(0,4\right), \left(1,9\right) \& \left(4,5\right)$,

Now on satisfying these points we get,

$4=0+0+c .......\left(1\right)$

$9=a+b+4 ......\left(2\right) \left\{\text{as} c=4\right\}$

$5=16a+4b+4 .....\left(3\right)$

Now solving equation equation $\left(2\right) \& \left(3\right)$ we get,

$a=\frac{-19}{12}, b=\frac{79}{12} \& c=4$

Now putting in the equation $y=a{x}^2+bx+c$ we get,

$19x^2+12y-79x-48=0$