Chapter 7.3, Problem 47E

To determine

To calculate: the equation of parabola that passes through the given points and plots the points and graph the parabola.

Answer to Problem 47E

The equation of parabola passing through $\left(1,3\right),\left(2,2\right),\left(3,3\right)$ is $y=x^2-4x+6$

The graph is

  

Explanation of Solution

Given information:

The given equation of the parabola is $y=a{x}^2+bx+c$ and it passes through $\left(1,3\right),\left(2,2\right),\left(3,3\right)$

Calculation:

Let us substitute $\left(1,3\right)$ in the equation $y=a{x}^2+bx+c$ ,

  $\begin{array}{l}3=a{\left(1\right)}^2+b\left(1\right)+c\\ a+b+c=3\cdots \left(i\right)\end{array}$

Substitute $\left(2,2\right)$ in the equation $y=a{x}^2+bx+c$

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

Substitute $\left(3,3\right)$ in the equation $y=a{x}^2+bx+c$

  $\begin{array}{l}3=a{\left(3\right)}^2+b\left(3\right)+c\\ 9a+3b+c=3\cdots \left(iii\right)\end{array}$

Now subtract equation $\left(i\right)$ and $\left(ii\right)$ ,

  $\begin{array}{l}a+b+c-4a-2b-c=3-2\\ -3a-b=1\cdots \left(iv\right)\end{array}$

Now subtract equation $\left(ii\right)$ and $\left(iii\right)$ ,

  $\begin{array}{l}4a+2b+c-9a-3b-c=2-3\\ -5a-b=-1\cdots \left(v\right)\end{array}$

Subtract equations $\left(iv\right)$ and $\left(v\right)$ to eliminate term.

  $\begin{array}{l}-3a-b+5a+b=1+1\\ 2a=2\\ a=1\end{array}$

Substitute this value in $\left(iv\right)$ to get b

  $\begin{array}{l}-3a-b=1\\ -3\left(1\right)-b=1\\ -3-b=1\\ b=-4\end{array}$

Substitute these values in $\left(i\right)$ to get c

  $\begin{array}{l}a+b+c=3\\ 1-4+c=3\\ c=6\end{array}$

By substituting these values we get the equation of parabola as

  $\begin{array}{l}y=1x^2-4x+6\\ y=x^2-4x+6\end{array}$

The graph of the parabola is as follows

  

Conclusion:

The equation of parabola passing through $\left(1,3\right),\left(2,2\right),\left(3,3\right)$ is $y=x^2-4x+6$