# The quadratic function \displaystyle{y}={a}{x}^{2}+{b}{x}+{c} whose graph passes through the points (1, 4), (2, 1) and (3, 4).

Question
Alternate coordinate systems

The quadratic function $$\displaystyle{y}={a}{x}^{2}+{b}{x}+{c}$$ whose graph passes through the points (1, 4), (2, 1) and (3, 4).

2021-02-01

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 $$\displaystyle{y}={9}{x}^{2}-{12}{x}+{13}$$

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