Chapter 7.3, Problem 48E

To determine

To calculate:Find the equation $y=a{x}^2+bx+c$ of the parabola that passes through the points. To verify the result, use a graphing utility to plot the points and graph the parabola.

  $\left(0,3\right),\left(1,4\right),\left(2,3\right)$

Answer to Problem 48E

The equation of parabola passing through $\left(0,3\right),\left(1,4\right),\left(2,3\right)$ is $y=-x^2+2x+3$

Explanation of Solution

Given information:

The given equation of the parabola is $y=a{x}^2+bx+c$ and it passes through $\left(0,3\right),\left(1,4\right),\left(2,3\right)$

Calculation:

Let us substitute $\left(0,3\right)$ in the equation $y=a{x}^2+bx+c$ ,

  $\begin{array}{l}0=a{\left(3\right)}^2+b\left(3\right)+c\\ 9a+3b+c=0\to \left(1\right)\end{array}$

Substitute $\left(1,4\right)$ in the equation $y=a{x}^2+bx+c$ with c value also.

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

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

  $\begin{array}{l}3=a{\left(2\right)}^2+b\left(2\right)+c\\ 4a+2b+c=3\to \left(3\right)\end{array}$

Now subtract equation 1 and 2,

  $\begin{array}{l}9a+3b+c-a-b-c=0-4\\ 8a+2b=-4\\ 4a+b=-2\to \left(4\right)\end{array}$

Subtract equation 2 and 3

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

Add equations 4 and 5

  $\begin{array}{l}4a+b-3a-b=-2+1\\ a=-1\end{array}$

Now substitute this value in any of the equations to get b value.

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

The c value will be,

  $\begin{array}{l}a+b+c=4\\ -1+2+c=4\\ c=3\end{array}$

By substituting these values we get the equation of parabola as

  $\begin{array}{l}y=-1x^2+2x+3\\ y=-x^2+2x+3\end{array}$

The graph of the parabola is as follows:

  

Conclusion:

The equation of parabola passing through $\left(0,3\right),\left(1,4\right),\left(2,3\right)$ is $y=-x^2+2x+3$