Chapter 7.3, Problem 85E

Solution

To state: The value of $f$ so that the graph contains the points $\left(-1,-4\right),\left(1,-2\right)$ and the family of curves is $a{x}^2+bx+c$ .

The resultant answer is $f\left(x\right)=\left(-c-3\right)x^2+x+c$ .

Given information:

The given points $\left(-1,-4\right),\left(1,-2\right)$ and the family of curves is $a{x}^2+bx+c$ .

Explanation:

Consider the given points:

  $\left(-1,-4\right),\left(1,-2\right)$

Now set up a system of equations using given points.

Point $\left(-1,-4\right)$ ,

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

Point $\left(1,-2\right)$ ,

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

It can be seen that there are 3 variables but only 2 equations, it means there are infinitely many solutions

Write the above system of equation in matrix and solve the system of equations using matrix:

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

Apply the operations $R_2\to -R_1-R_2$ ,

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

Apply the operations $R_2\to -\frac{1}{2}{R}_2$ ,

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

Apply the operations $R_1\to R_2+R_1$ ,

  $\left[\begin{array}{cccc}\begin{array}{l}1\\ 0\end{array}& \begin{array}{l}0\\ 1\end{array}& \begin{array}{l}1\\ 0\end{array}& \begin{array}{l}-3\\ 1\end{array}\end{array}\right]$

This correspond the system: $a+c=-3\text{and}b=1$

The solution of the system is $a=-c-3,b=1,\text{and}c=c$ and $f\left(x\right)=\left(-c-3\right)x^2+x+c$ .