Chapter 5, Problem 5GR

Solution

To find: the standard form of parabola.

Equation of parabola passing through points $\left(3,4\right),\left(-2,9\right),\left(2,1\right)$ in standard form is as:

  $y=x^2-2x+1$

Given: Points are $\left(3,4\right),\left(-2,9\right),\left(2,1\right)$ .

Concept use:

Standard form of parabola that use to find equation of parabola passing through three points is

  $y=a{x}^2+bx+c$

Calculation:

Standard form is

  $y=a{x}^2+bx+c\mathrm{.......}\left(1\right)$

Substitute all points in equation $\left(1\right)$ .

If $x=3\&y=4$ then from equation $\left(1\right)$

  $\begin{array}{c}a\times {\left(3\right)}^2+b\times \left(3\right)+c=4\\ 9a+3b+c=4\mathrm{.......}\left(2\right)\end{array}$

If $x=-2\&y=9$ then from equation $\left(1\right)$

  $\begin{array}{c}a\times {\left(-2\right)}^2+b\times \left(-2\right)+c=9\\ 4a-2b+c=9\mathrm{.......}\left(3\right)\end{array}$

If $x=2\&y=1$ then from equation $\left(1\right)$

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

From equation $\left(3\right)\&\left(4\right)$

  $\begin{array}{c}8a+2c=10\\ 2c=10-8a\\ c=5-4a\mathrm{.......}\left(5\right)\end{array}$

Substitute ‘c’ value in equation $\left(3\right)$ then

  $\begin{array}{c}4a-2b+5-4a=9\\ -2b=9-5\\ -2b=4\\ b=-2\mathrm{......}\left(6\right)\end{array}$

Substitute ‘c’ and ‘b’ values in equation $\left(2\right)$ then

  $\begin{array}{c}9a+3\times \left(-2\right)+5-4a=4\\ 9a-6+5-4a=4\\ 5a-1=4\\ 5a=5\\ a=1\mathrm{......}\left(7\right)\end{array}$

Now from equation $\left(5\right)$

  $\begin{array}{l}c=5-\left(4\times 1\right)\\ c=1\end{array}$

Thus, from equation $\left(1\right)$

  $y=x^2-2x+1$