There is a Creative Way To Solve This!

Point $A$ has coordinates $(2,11)$ and Point $A$ has coordinates $(2,11)$ .

Since the equation of the curve is $y=ax^2+b$ , we can say that

$\begin{aligned} & & 11 & ={ 2 }^{ 2 }a+b & \\ & \longrightarrow & 11 & =4a+b & \dashrightarrow (1) \\ & & 3.5 & =0.5^{ 2 }a+b & \\ & \longrightarrow & 3.5 & =0.25a+b & \dashrightarrow (2) \\ & (1)-(2): & 11-3.5 & =4a+b-(0.25a+b) & \\ & & 7.5 & =3.75a & \\ & & a & =\boxed{2} & \end{aligned}$

Substituting $a=2$ into $(1)$ ,

$\begin{aligned} 11 & =4(2)+b \\ \longrightarrow 11 & =8+b \\ \longrightarrow \quad b & =\boxed{3} \end{aligned}$

Substituting $a=2$ and $b=3$ into $y=ax^2+b$ , we have

$y=2x^2+3$

Since Point $C$ is on the y-axis , we know that its X -coordinate is $0$ .

Substituting $x=0$ into $y=2x^2+3$ ,

$\huge\begin{aligned} y & =2x^{ 2 }+3 \\ & =2(0)^{ 2 }+3 \\ & =\boxed{3} \end{aligned}$ .