Chapter 9.9, Problem 10P

To determine

To determine the equation of a parabola that passes through the points $(1,3)$ , $(2,10)$ and $(-2,-6)$ .

Answer to Problem 10P

  $y=x^2+4x-2$ is the solution.

Explanation of Solution

Given:

A parabola that passes through the points $(1,3)$ , $(2,10)$ and $(-2,-6)$ .

Formula Used:

The equation of a parabola is:

  $y=a{x}^2+bx+c$ where a, b and c are the co-efficient.

Calculation:

Given the parabola that passes through the points $(1,3)$ , $(2,10)$ and $(-2,-6)$ .

We need to calculate the co-efficient for each of the three points by substituting the values of x and y for each point.

Hence, for the point $(1,3)$ :

  $3=a\times {(1)}^2+b\times 1+c$

  $3=a+b+c$

For the point $(2,10)$ :

  $10=a\times {(2)}^2+b\times 2+c$

  $10=4a+2b+c$

For the point $(-2,-6)$ :

  $-6=a\times {(-2)}^2+b\times (-2)+c$

  $-6=4a-2b+c$

Solving the equation $3=a+b+c$ and $10=4a+2b+c$ we have:

  $-7=-3a-b$

  $b=7-3a$

Solving the equation $10=4a+2b+c$ and $-6=4a-2b+c$ we have:

  $16=4b$

  $b=4$

Putting the value of $b=4$ in the equation $b=7-3a$ for getting the value of a, we have:

  $4=7-3a$

  $3a=3$

  $a=1$

Putting the value of $b=4$ and $a=1$ in the equation $3=a+b+c$ for getting the value of c, we have:

  $3=1+4+c$

  $c=-2$

Now, putting the values of a, b and c in the general equation of the parabola, we get:

  $y=1\times x^2+4\times x-2$

  $y=x^2+4x-2$

Hence, the answer is $y=x^2+4x-2$

Conclusion:

  $y=x^2+4x-2$ is the solution.