Find the equation of parabola whose axis is parallel to y-axis and which passes through the points (2, 5) (1, 4)(–1, 8)

Let A(2, 5), B(1, 4), C(–1, 8)
The equation of parabola whose axis  parallel to y-axis is $\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\to \left(1\right)$
(1) passes through (2, 5)
$5=4\mathrm{a}+2\mathrm{b}+\mathrm{c}\to \left(2\right)$

(1) passes through (1, 4)
$4=\mathrm{a}+\mathrm{b}+\mathrm{c}\to \left(3\right)$
(1) passes through (–1, 8)
$8=\mathrm{a}-\mathrm{b}+\mathrm{c}\to \left(4\right)$
$\because$from (2) and (3)
$$\begin{gathered} \begin{array}{l}4a+2b+c=5\\ \frac{a+b+c = 4 \\ - - - - }{ }\\ \frac{3a+b=1\to \left(5\right)}{ }\end{array} \end{gathered}$$
from (3) and (4)
$\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}=4\\ \mathrm{a}-\mathrm{b}+\mathrm{c}=8\\ \frac{- + - - }{2\mathrm{b}=-4}\end{array}$
b =–2
$$\begin{gathered} \begin{array}{l}\mathrm{from} \left(5\right) \\ \Rightarrow 3\mathrm{a}+\mathrm{b}=1 \\ \Rightarrow 3\mathrm{a}-2=1 \\ \Rightarrow \mathrm{a}=1 \\ \\ from \left(3\right)\\ \begin{array}{l}\Rightarrow \mathrm{c}=4-\mathrm{a}-\mathrm{b} \\ \Rightarrow \mathrm{c}=4-1+2 \\ \Rightarrow c=5\\ \\ \end{array}\end{array} \end{gathered}$$
$\text{∴ Equation of parabola is }\mathrm{y}={\mathrm{x}}^2-2\mathrm{x}+5$
$\mathrm{y}={\mathrm{x}}^2-2\mathrm{x}+5$