Chapter 7.4, Problem 85E

To determine

To calculate:

To find the quadratic function $f\left(x\right)=a{x}^2+bx+c$ that satisfies the equations.

Answer to Problem 85E

  $f\left(x\right)=-9x^2-5x+11$ is the quadratic function.

Explanation of Solution

Given information:

The given quadratic function is $f\left(x\right)=a{x}^2+bx+c$ and the given points are

  $\begin{array}{l}f\left(-2\right)=-15\\ f\left(-1\right)=7\\ f\left(1\right)=-3\end{array}$

Calculation:

Consider, the general form of the quadratic function $f\left(x\right)=a{x}^2+bx+c$ and the given points

  $\begin{array}{l}f\left(-2\right)=-15\\ f\left(-1\right)=7\\ f\left(1\right)=-3\end{array}$

Since the points lie on the quadratic function, they satisfy when we substitute them.

Substituting each of the three points in the equations gives us three equations in three variables.

  $\begin{array}{l}f\left(-2\right)=4a-2b+c=-15\to \left(1\right)\\ f\left(-1\right)=a-b+c=7\to \left(2\right)\\ f\left(1\right)=a+b+c=-3\to \left(3\right)\end{array}$

The given system of equations can be written in matrix notation as shown below,

  $\left(\begin{array}{ccc}4& -2& 1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right)\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}-15\\ 7\\ -3\end{array}\right)$

Multiply both sides by the inverse of the coefficient matrix.

Note that the product of a matrix with its inverse is an identity matrix of the same order.

Hence, we will get two column matrix on both sides, comparing them will give us the values of a,b,c.

  $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)={\left(\begin{array}{ccc}4& -2& 1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right)}^{-1}\left(\begin{array}{c}-15\\ 7\\ -3\end{array}\right)$

Consider, ${\left(\begin{array}{ccc}4& -2& 1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right)}^{-1}$

Augment with a $3\times 3$ identity matrix,

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

Reduce the matrix to row echelon form,

  $=\left[\begin{array}{ccccccc}4& -2& 1& \mid & 1& 0& 0\\ 0& \frac{3}{2}& \frac{3}{4}& \mid & -\frac{1}{4}& 0& 1\\ 0& 0& 1& \mid & -\frac{1}{3}& 1& \frac{1}{3}\end{array}\right]$

Reduce further,

  $=\left[\begin{array}{ccccccc}1& 0& 0& \mid & \frac{1}{3}& -\frac{1}{2}& \frac{1}{6}\\ 0& 1& 0& \mid & 0& -\frac{1}{2}& \frac{1}{2}\\ 0& 0& 1& \mid & -\frac{1}{3}& 1& \frac{1}{3}\end{array}\right]$

Hence, we get

  $=\left(\begin{array}{ccc}\frac{1}{3}& -\frac{1}{2}& \frac{1}{6}\\ 0& -\frac{1}{2}& \frac{1}{2}\\ -\frac{1}{3}& 1& \frac{1}{3}\end{array}\right)$

Now, $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)={\left(\begin{array}{ccc}4& -2& 1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right)}^{-1}\left(\begin{array}{c}-15\\ 7\\ -3\end{array}\right)$

  $=\left(\begin{array}{ccc}\frac{1}{3}& -\frac{1}{2}& \frac{1}{6}\\ 0& -\frac{1}{2}& \frac{1}{2}\\ -\frac{1}{3}& 1& \frac{1}{3}\end{array}\right)\left(\begin{array}{c}-15\\ 7\\ -3\end{array}\right)$

  $=\left(\begin{array}{c}-9\\ -5\\ 11\end{array}\right)$

Therefore,

  $a=-9,b=-5,c=11$

Hence, the equation of the quadratic function is $f\left(x\right)=-9x^2-5x+11$