Chapter 11, Problem 99RE

To determine

Find the quadratic function $y=a{x}^2+bx+c$ .

Answer to Problem 99RE

  $\boxed{\frac{1}{3}{x}^2-\frac{2}{3}x-1=0}$

Explanation of Solution

Given information:

Find the quadratic function $y=a{x}^2+bx+c$ that passes through the three points $\left(0,1\right),\left(1,0\right)$ and $\left(-2,1\right)$ .

Calculation:

Consider quadratic function $y=a{x}^2+bx+c$ that passes through the three points must satisfy the equation.

Substitute each point $\left(0,1\right),\left(1,0\right)$ and $\left(-2,1\right)$ one by one in given function,

  $\begin{array}{l}y=a{x}^2+bx+c\\ 1=a{\left(0\right)}^2+b\left(0\right)+c\\ 1=0+0+c\\ 1=c\end{array}$

Now for point $\left(1,0\right)$ ,

  $\begin{array}{l}y=a{x}^2+bx+c\\ 0=a{\left(1\right)}^2+b\left(1\right)+c\\ 0=a+b+c\end{array}$

For point $\left(-2,1\right)$ ,

  $\begin{array}{l}y=a{x}^2+bx+c\\ 1=a{\left(-2\right)}^2+b\left(-2\right)+c\\ 1=4a-2b+c\end{array}$

A system of linear equation s is formed,

  $\{\begin{array}{ccccc}& & c& =& 1\\ a& +b& +c& =& 0\\ 4a& -2b& +c& =& 1\end{array}$

Now write the augmented matrix corresponding to the system of equations,

  $\left[\begin{array}{cccc}0& 0& 1& 1\\ 1& 1& 1& 0\\ 4& -2& 1& 1\end{array}\right]$

To solve the system reduce the matrix in to row echelon form so perform the row opration $R_1=r_2+r_1$

  $\left[\begin{array}{cccc}1+0& 1+0& 1+1& 1+0\\ 1& 1& 1& 0\\ 4& -2& 1& 1\end{array}\right]\to \left[\begin{array}{cccc}1& 1& 2& 1\\ 1& 1& 1& 0\\ 4& -2& 1& 1\end{array}\right]$

Now do operation $R_2=r_1-r_2$ .

  $\left[\begin{array}{cccc}1& 1& 2& 1\\ 1-1& 1-1& 2-1& 1-0\\ 4& -2& 1& 1\end{array}\right]\to \left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 0& 1& 1\\ 4& -2& 1& 1\end{array}\right]$

Now do operation $R_3=\frac{r_3}{-2}$ .

  $\left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 0& 1& 1\\ 4/-2& -2/-2& 1/-2& 1/-2\end{array}\right]\to \left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 0& 1& 1\\ -2& 1& -\frac{1}{2}& -\frac{1}{2}\end{array}\right]$

The system of equations corresponding to the resultant matrix is,

  $\{\begin{array}{ccccc}& & c& =& 1\\ a& +b& +2c& =& 1\\ -2a& +b& -\frac{1}{2}c& =& -\frac{1}{2}\end{array}$

Now eliminate $b$ to subtract third equation from the second equation,

  $\begin{array}{l}a+b+2c-\left(-2a+b-\frac{1}{2}c\right)=1-\left(-\frac{1}{2}\right)\\ a+\cancel{b}+2c+2a-\cancel{b}+\frac{1}{2}c=\frac{3}{2}\\ 3a+\frac{5}{2}c=\frac{3}{2}\\ put,c=1\\ 3a+\frac{5}{2}=\frac{3}{2}\\ 3a=\frac{3}{2}-\frac{5}{2}\\ 3a=\frac{-2}{2}\\ a=-\frac{1}{3}\end{array}$

Put $a=-\frac{1}{3}$ and $c=1$ in $a+b+2c=1$ we get,

  $\begin{array}{l}-\frac{1}{3}+b+2\left(1\right)=1\\ -\frac{1}{3}+b+2=1\\ b=1+\frac{1}{3}-2\\ b=-\frac{2}{3}\end{array}$

Now substitute the calculated values of $a,b,c$ in the given function,

  $\begin{array}{l}-\frac{1}{3}{x}^2+\left(-\frac{2}{3}\right)x+1=0\\ or\\ \frac{1}{3}{x}^2-\frac{2}{3}x-1=0\end{array}$

Hence the function is $\boxed{\frac{1}{3}{x}^2-\frac{2}{3}x-1=0}$