Calculation:

First, obtain the three system of three linear equation by substituting the coordinate points in \(\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}\)

\(\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}{\left({1}\right)}\)

Consider the first coordinate point (1, 4).

Here, \(\displaystyle{x}={1}{\quad\text{and}\quad}{y}={4}\)

Substitute the values of x and y in the equation (1)

\(\displaystyle{4}={a}{\left({1}\right)}^{2}+{b}{\left({1}\right)}+{c}\)

\(\displaystyle{4}={a}+{b}+{c}\)

Consider the second coordinate point (2, 1)

Here,\(\displaystyle{x}={2}{\quad\text{and}\quad}{y}={1}\)

Substitute the values of x and y in the equation (1)

\(\displaystyle{1}={a}{\left({2}\right)}^{2}+{b}{\left({2}\right)}+{c}\)

\(\displaystyle{1}={4}{a}+{2}{b}+{c}\)

Consider the third coordinate point (3, 4).

Here, \(\displaystyle{x}={3}{\quad\text{and}\quad}{y}={4}\).

Substitute the values of x an y in the equation (1)

\(\displaystyle{4}={a}{\left({3}\right)}^{2}+{b}{\left({3}\right)}+{c}\)

\(\displaystyle{4}={9}{a}+{3}{b}+{c}\)

The obtained equations are,

\(\displaystyle{a}+{b}+{c}={4}{\left({2}\right)}\)

\(\displaystyle{4}{a}+{2}{b}+{c}={1}{\left({3}\right)}\)

\(\displaystyle{9}{a}+{3}{b}+{c}={4}{\left({4}\right)}\)

Step 1: Reduce the system to two equation in two variables.

Eliminate the variable c from (2) and (3) by multiplying (2) by -1 and adding with (3).

\(\displaystyle{\left({2}\right)}\times-{1}:-{a}-{b}-{c}=-{4}\)

\(\displaystyle{\left({3}\right)}:\frac{{{4}{a}+{2}{b}+{c}={1}}}{{{3}{a}+{b}=-{3}}}\)

That is, \(\displaystyle{3}{a}+{b}=-{3}\).

Now, eliminate c from (3) and (4) by multiplying (3) by -1 and adding with (4).

\(\displaystyle{\left({3}\right)}\times-{1}:-{4}{a}-{2}{b}-{c}=-{1}\)

\(\displaystyle{\left({4}\right)}:\frac{{{9}{a}+{3}{b}+{c}={4}}}{{{5}{a}+{b}={3}}}\)

That is, \(\displaystyle{5}{a}+{b}={3}\)

Hence, the system of two equation in two variable is obtained as,

\(\displaystyle{3}{a}+{b}=-{3}{\left({5}\right)}\)

\(\displaystyle{5}{a}+{b}={3}.{\left({6}\right)}\)

Step 2: Solve the resulting system of two equation in two variables.

Multiply equation (5) by -1 and then add with (6)

\(\displaystyle{\left({5}\right)}\times-{1}:-{3}{a}-{b}={3}\)

\(\displaystyle{\left({6}\right)}:\frac{{{5}{a}+{b}={3}}}{{{2}{a}={6}}}\)

Apply the multiplication property of equality and simplify it.

\(\displaystyle\frac{{{2}{a}}}{{2}}=\frac{6}{{2}}\)

\(\displaystyle{a}={3}\)

Step 3: Use back-substribution in one of the equations in two variables to find the value of the second vatiable.

Substitute \(\displaystyle{a}={3}\) in equation (5).

\(\displaystyle{3}{\left({3}\right)}+{b}=-{3}\)

\(\displaystyle{9}+{b}=-{3}\) [Multiply]

\(\displaystyle-{9}+{9}+{b}=-{3}-{9}\) [Subtract 9 from both the sides of the equation]

\(\displaystyle{b}=-{12}\)

Step 4: Back-substitute the values found for two variables into one of the original equations to find the value of the third variable.

Substitute \(\displaystyle{a}={3}{\quad\text{and}\quad}{b}=-{12}\) is (2),

\(\displaystyle{3}-{12}+{c}={4}\)

\(\displaystyle-{9}+{c}={4}\) [Combine the like terms]

\(\displaystyle-{9}+{9}+{c}={4}+{9}\) [ Add 9 to both sides of the equation]

\(\displaystyle{c}={13}\)

With \(\displaystyle{a}={3},{b}=-{12},{c}={13}\), the proposed solution is the ordered triple (3, -12, 13).

Substitute \(\displaystyle{a}={3},{b}=-{12},{c}={13}\) in (1),

\(\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}\)

\(\displaystyle={\left({9}\right)}{x}^{2}+{\left(-{12}\right)}{x}+{13}\)

\(\displaystyle={9}{x}^{2}-{12}{x}+{13}\)

So, the quadratic function whose graph passes through the point (1, 4), (2, 1) and (3, 4) is \(\displaystyle{y}={9}{x}^{2}-{12}{x}+{13}\)