Chapter 7, Problem 59RE

To determine

To find: The equation of the parabola.

Answer to Problem 59RE

The required equation is $y=2x^2+x-5$ .

Explanation of Solution

Given information:

The given equation is $y=a{x}^2+bx+c$ .

Given points are $\left(-1,-4\right),\left(1,-2\right)\text{and}\left(2,5\right)$ .

The given equation is $y=a{x}^2+bx+c$ .

Substitute the value $\left(-1,-4\right)$ for $\left(x,y\right)$ .

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

Substitute the value $\left(1,-2\right)$ for $\left(x,y\right)$ .

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

Similarly, substitute the value $\left(2,5\right)$ for $\left(x,y\right)$ .

  $\begin{array}{c}5=a\times {\left(2\right)}^2+b\times \left(2\right)+c\\ 5=4a+2b+c\\ 4a+2b+c=5\end{array}$  ...... (3)

Write in the matrix format.

  $\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right]=\left[\begin{array}{c}-4\\ -2\\ 5\end{array}\right]$

Now, solve the equation (1), (2) and (3) for $a,b\text{and}c$ . Use Cramer’s rule.

  $a=\frac{{\Delta }_1}{\Delta }$

Find the determinant value of $\Delta$ .

  $\Delta =\left[\begin{array}{ccc}1& -1& 1\\ 1& 1& 1\\ 4& 2& 1\end{array}\right]=1$

Noe, Find the determinant value of ${\Delta }_1$ .

  $\begin{array}{c}{\Delta }_1=\left[\begin{array}{ccc}-4& -1& 1\\ -2& 1& 1\\ 5& 2& 1\end{array}\right]\\ =2\end{array}$

Thus, substitute the value of $\Delta \text{and}{\Delta }_1$ .

  $\begin{array}{c}a=\frac{{\Delta }_1}{\Delta }\\ =\frac{2}{1}\\ =2\end{array}$

Similarly ${\Delta }_2\text{and}{\Delta }_3$ for $b\text{and}c$ .

  $\begin{array}{c}{\Delta }_2=\left[\begin{array}{ccc}1& -4& 1\\ 1& -2& 1\\ 4& 5& 1\end{array}\right]\\ =1\end{array}$

And,

  $\begin{array}{c}{\Delta }_3=\left[\begin{array}{ccc}1& -1& -4\\ 1& 1& -2\\ 4& 2& 5\end{array}\right]\\ =-5\end{array}$

Now, substitute the values.

  $\begin{array}{c}b=\frac{{\Delta }_2}{\Delta }\\ =\frac{1}{1}\\ =1\end{array}$

And,

  $\begin{array}{c}c=\frac{{\Delta }_3}{\Delta }\\ =\frac{-5}{1}\\ =-5\end{array}$

Substitute the values of $a,b\text{and}c$ in the given equation.

  $y=2x^2+x-5$

Draw the graph.

  

Hence, the required equation is $y=2x^2+x-5$ .