Chapter 7, Problem 20CT

To determine

To find: the equation of the given parabola,that passes through the points $\left(0,6\right),\left(-2,2\right)$ , and $\left(3,\frac{9}{3}\right)$ .

Answer to Problem 20CT

  $y=-\frac{1}{2}{x}^2+x+6$

Explanation of Solution

Given information:

The given parabola is $y=a{x}^2+bx+c$ ,

Given points are $\left(0.6\right),\left(-2,2\right),\left(3,\frac{9}{2}\right)$ .

Calculation:

Consider a parabola $y=a{x}^2+bx+c$which passes through the following points

  $\left(0.6\right),\left(-2,2\right),\left(3,\frac{9}{2}\right)$

Since the points lie on the parabola, therefore each the satisfy the equation of the parabola

Since the point $\left(0.6\right)$ satisfy $y=a{x}^2+bx+c$

  $\begin{array}{l}6=a{\left(0\right)}^2+b\left(0\right)+c\\ 6=c\mathrm{................}(1)\end{array}$

Since the point $\left(-2,2\right)$ satisfy $y=a{x}^2+bx+c$

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

Also, the point $\left(3,\frac{9}{2}\right)$ satisfy $y=a{x}^2+bx+c$

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

Substitute the value of $c$ from (1) in (2)

  $\begin{array}{l}2=4a-2b+6\\ -4=4a-2b\end{array}$

Divide both sides by 2

  $-2=2a-b\mathrm{.........}(4)$

Substitute the value of $c$ from (1) in (3)

  $\begin{array}{l}\frac{9}{2}=9a+3b+6\\ -\frac{3}{2}=9a+3b\end{array}$

Divide both sides by 3

  $-\frac{1}{2}=3a+b\mathrm{.............}(5)$

Add equations (4) and (5)

  $-\frac{5}{2}=5a$

Divide both sides by 5

  $-\frac{1}{2}=a$

Put $a=-\frac{1}{2}$in equation (4)

  $\begin{array}{c}-2=2\left(-\frac{1}{2}\right)-b\\ -2=-1-b\end{array}$

Add 1 to both sides

  $-1=-b$

Multiply both sides by -1

  $1=b$

Put values of $a,b,c$ into $y=a{x}^2+bx+c$to obtain the equation of parabola

  $y=-\frac{1}{2}{x}^2+x+6$ .