Chapter 4.3, Problem 1LC

Solution

To find: The equation of the parabola passing through three points.

The points $\left(1,0\right),\left(2,-3\right)$ and $\left(3,-10\right)$ do not lie on a parabola, i.e., no parabola passes through points $\left(1,0\right),\left(2,-3\right)$ and $\left(3,-10\right)$ .

Given data: The given three points are $\left(1,0\right),\left(2,-3\right)$ and $\left(3,-10\right)$ .

Formula used:

The equation of a parabola is a quadratic function $y=a{x}^2+bx+c$ .

Calculation:

Let the equation of parabola passing through points $\left(1,0\right),\left(2,-3\right)$ and $\left(3,-10\right)$ is

  $y=a{x}^2+bx+c$ … (1)

The parabola (1) passes through point $\left(1,0\right)$ , therefore point should satisfy equation (1).

Substitute $x=1,y=0$

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

OR

  $a+b+c=0$ … (2)

The parabola (1) passes through point $\left(2,-3\right)$ , therefore point should satisfy equation (1).

Substitute $x=2,y=-3$

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

OR

  $4a+2b+c=-3$ … (3)

The parabola (1) passes through point $\left(3,-10\right)$ , therefore point should satisfy equation (1).

Substitute $x=3,y=-10$ ,

  $\begin{array}{l}-10=a{\left(3\right)}^2+b\left(3\right)+c\\ -10=9a+3b+c\end{array}$

OR

  $9a+3b+c=-10$ … (4)

Subtract equation (2) from equation (3),

  $\left.\begin{array}{c}\left(4a+2b+c\right)-\left(a+b+c\right)=\left(-3\right)-\left(0\right)\\ 4a+2b+c-a-b-c=-3\\ 3a+b=-3\end{array}\right\}$ … (5)

From equation (4)

  $3\left(3a+b\right)+c=-10$

Substitute -3 for $3a+b$

  $\begin{array}{c}3\left(-3\right)+c=-10\\ -9+c=-10\\ c=-10+9\\ c=-1\end{array}$

Substitute −1 for c in equation (2)

  $\begin{array}{c}a+b-1=0\\ a+b=-1\end{array}$

From equation (3)

  $\left.\begin{array}{c}4a+2b+c=-3\\ 2a+\left(2a+2b\right)+c=-3\\ 2a+2\left(a+b\right)+c=-3\end{array}\right\}$ … (6)

Substitute -1 for c and -1 for $a+b$ in result (6),

  $\begin{array}{c}2a+2\left(-1\right)-1=-3\\ 2a-3=-3\\ 2a=0\\ a=0\end{array}$

Since, $a+b=-1$ , therefore, $0+b=-1$ or $b=-1$ .

Substitute, $a=0,b=-1,c=-1$ in equation (1),

  $\begin{array}{c}y=\left(0\right)x^2+\left(-1\right)x+\left(-1\right)\\ y=-\left(x+1\right)\end{array}$

But $y=-\left(x+1\right)$ is equation of a straight line.

Therefore, no parabola passes through points $\left(1,0\right),\left(2,-3\right)$ and $\left(3,-10\right)$ .