Chapter 7, Problem 45RE

To determine

To calculate: the equation of the parabola $y=a{x}^2+bx+c$ that passes through the given points and plot the points and graph the parabola.

Answer to Problem 45RE

The equation of the parabola is $y=3x^2+11x-14$

Graph is

  

Explanation of Solution

Given information:

The equation of the parabola is given as $y=a{x}^2+bx+c$ and it passes through the points $\left(1,0\right),\left(2,20\right),\left(-5,6\right)$ from the figure.

  

Calculation:

First substitute $\left(1,0\right)$ in the given equation

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

Now substitute $\left(2,20\right)$ to get other equation.

  $\begin{array}{l}20=a{\left(2\right)}^2+b\left(2\right)+c\\ 4a+2b+c=20\cdots \left(ii\right)\end{array}$

Now substitute $\left(-5,6\right)$ for another equation

  $\begin{array}{l}6=a{\left(-5\right)}^2+b\left(-5\right)+c\\ 25a-5b+c=6\cdots \left(iii\right)\end{array}$

Subtract equations $\left(i\right)$ and $\left(ii\right)$ we get

  $\begin{array}{l}a+b+c-4a-2b-c=0-20\\ -3a-b=-20\cdots \left(iv\right)\end{array}$

Subtract equations $\left(ii\right)$ and $\left(iii\right)$ we get

  $\begin{array}{l}4a+2b+c-25a+5b-c=20-6\\ -21a+7b=14\\ -3a+b=2\cdots \left(v\right)\end{array}$

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

  $\begin{array}{l}-3a-b+3a-b=-20-2\\ -2b=-22\\ b=11\end{array}$

Substitute $b=11$ in $\left(v\right)$ we get

  $\begin{array}{l}-3a+b=2\\ -3a+11=2\\ -3a=-9\\ a=3\end{array}$

Put $b=11$ and $a=3$ in $\left(iii\right)$ the value of c is

  $\begin{array}{l}a+b+c=0\\ 3+11+c=0\\ c=-14\end{array}$

The equation of the parabola is $y=3x^2+11x-14$

The graph is as follows

  

Conclusion:

The equation of the parabola is $y=3x^2+11x-14$