Chapter 9.4, Problem 5AGP

To determine

Find the quadratic function.

Answer to Problem 5AGP

  $y=2x^2-6x-20$

Explanation of Solution

Given Information: $(-2,0),(4,-12),(5,0)$

Formula Used: Elimination method

Calculation:

  $(-2,0),(4,-12),(5,0)$

As the general equation $y=a{x}^2+bx+c$  (1)

Now,

  $(-2,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(-2)}^2+b(-2)+c$

So,

  $4a-2b+c=0$

Now,

  $4a=2b-c$

Hence,

  $a=\frac{2b-c}{4}$  (2)

Now,

  $(4,-12)$ passes through equation $y=a{x}^2+bx+c$

So,

  $-12=a{(4)}^2+b(4)+c$

Hence,

  $16a+4b+c=-12$  (3)

Now,

  $(5,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(5)}^2+b(5)+c$

Hence,

  $25a+5b+c=0$  (4)

Put value of eq. (2) in eq. (3),

  $16a+4b+c=-12$

As, $a=\frac{2b-c}{4}$

  $16\times \left(\frac{2b-c}{4}\right)+4b+c=-12$

So,

  $8b-4c+4b+c=-12$

So, simplify

  $12b-3c=-12$

Now, divide the equation by 3.

  $4b-c=-4$  (5)

Put value of eq. (2) in eq. (4),

  $25a+5b+c=0$

As, $a=\frac{2b-c}{4}$

  $25\times \left(\frac{2b-c}{4}\right)+5b+c=0$

So,

  $\left(\frac{50b-25c}{4}\right)+5b+c=0$

So,

  $\frac{50b-25c+20b+4c}{4}=0$

Now, simplify

  $50b-25c+20b+4c=0$

Hence,

  $70b-21c=0$  (6)

Now, solving the equation (5) & (6) by elimination method,

  $4b-c=-4$  (5)

and

  $70b-21c=0$  (6)

Now, for elimination method, multiply equation (5) by -21,

  $-84b+21c=84$  (7)

Now, solve equation (6) & (7),

  $-14b=84$

Hence,

  $b=-6$  (8)

Put value of b in equation (5),

  $4b-c=-4$  (5)

As, $b=-6$

  $4\times (-6)-c=-4$

Now,

  $c=-24+4$

Hence,

  $c=-20$  (9)

Put value of b & c in equation (2),

  $a=\frac{2b-c}{4}$  (2)

As, $b=-6$ and $c=-20$

  $a=\frac{2\times (-6)-(-20)}{4}$

Now, simplify

  $a=\frac{-12+20}{4}=\frac{8}{4}$

Hence,

  $a=2$  (10)

Now, put value of equation (8), (9) and (10) in equation (1) to solve quadratic function.

So, quadratic function $y=2x^2-6x-20$