Exercises: systems with three variables

The function graph has the equation $y=a{x}^2+bx+c$.

If a point lies on a graph, it means that the equation gives a true statement.

From the given three points, you can set up three equations, all of which must be satisfied.

Plug in point $\mathrm{R}$

$y=a{x}^2+bx+c$

Plug the point $\mathrm{R}(1|2)$ into the equation.

$2=a\cdot 1^2+b\cdot 1+c$

You get your first equation.

$\mathrm{I}2=a+b+c$

Plug in point $\mathrm{Q}$

$y=a{x}^2+bx+c$

Plug the point $\mathrm{Q}(-1|3)$ into the equation.

$3=a\cdot (-1{)}^2+b\cdot (-1)+c$

You get your second equation.

$\mathrm{II}3=a-b+c$

Plug in point $\mathrm{S}$

$y=a{x}^2+bx+c$

Plug the point $\mathrm{Q}(0|1)$ into the equation.

$1=a\cdot (0{)}^2+b\cdot 0+c$

You get your third equation.

$\mathrm{III}1=c$

So the system of equations is:

$\begin{array}{rrlllllll}\mathrm{I}& 2& =& a& +& b& +& c& \\ \mathrm{II}& 3& =& a& -& b& +& c& \\ \mathrm{III}& 1& =& & & & & c& \end{array}$

The following system of equations is given:

$\begin{array}{rrlllllll}\mathrm{I}& 2& =& a& +& b& +& c& \\ \mathrm{II}& 3& =& a& -& b& +& c& \\ \mathrm{III}& 1& =& & & & & c& \end{array}$

It follows directly from the third equation that:

$c=1$

Substitute $c=1$ into the other two equations:

$\begin{array}{rrlllllll}{\mathrm{I}}^{\prime }& 2& =& a& +& b& +& 1& \\ \mathrm{I}{\mathrm{I}}^{\prime }& 3& =& a& -& b& +& 1& \end{array}$

Careful: The information of the third equation is now contained in ${\mathrm{I}}^{\prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime }$.

Now solve the new system of equations. Since the coefficients before the variable $b$ have different signs and the same absolute value, the addition method can be used.

Add the equation ${\mathrm{I}}^{\prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime }$. You get a new equation ${\mathrm{I}}^{\prime \prime }$.

$\begin{array}{rlrllll}{\mathrm{I}}^{\prime }+\mathrm{I}{\mathrm{I}}^{\prime }\to {\mathrm{I}}^{\prime \prime }& |& 5& =& 2a& +& 2\end{array}$

Solve for a.

$a=\frac{3}{2}$

Substitute $a=\frac{3}{2}$ into equation ${\mathrm{I}}^{\prime }$ to determine the parameter $b$.

$\begin{array}{rrlllllll}{\mathrm{I}}^{\prime }& 2& =& \frac{3}{2}& +& b& +& 1& \end{array}$

Solve for $b$.

$b=-\frac{1}{2}$

Write down the solution set.

$𝕃=\{(a|b|c)\in {ℝ}^{\mathrm{𝟛}}|a=\frac{3}{2};b=-\frac{1}{2};c=1\}$

The function equation is $f(x)=\frac{3}{2}\cdot x^2-\frac{1}{2}\cdot x+1$.