Chapter 9.4, Problem 41PPS

To determine

Find the quadratic function.

Answer to Problem 41PPS

  $y=\frac{3x^2}{4}-3x-9$

Explanation of Solution

Given Information: $(-2,0),(6,0),(8,15)$

Formula Used: Elimination method

Calculation:

  $(-2,0),(6,0),(8,15)$

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

Now,

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

So,

  $0=a{(-2)}^2+b(-2)+c$

Now,

  $4a-2b+c=0$

Hence,

  $a=\frac{2b-c}{4}$  (2)

Now,

  $(6,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(6)}^2+b(6)+c$

Now,

  $36a+6b+c=0$  (3)

Now,

  $(8,15)$ passes through equation $y=a{x}^2+bx+c$

So,

  $15=a{(8)}^2+b(8)+c$

Hence,

  $64a+8b+c=15$  (4)

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

  $36a+6b+c=0$  (3)

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

  $36\times \left(\frac{2b-c}{4}\right)+6b+c=0$

So,

  $18b-9c+6b+c=0$

So, simplify

  $24b-8c=0$  (5)

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

  $64a+8b+c=15$  (4)

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

  $64\times \left(\frac{2b-c}{4}\right)+8b+c=15$

So,

  $32b-16c+8b+c=15$

So, simplify

  $40b-15c=15$  (6)

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

  $24b-8c=0$  (5)

and

  $40b-15c=15$  (6)

Now, for elimination method, multiply equation (5) by -5 and multiply equation (6) by 3,

  $-120b+40c=0$  (7)

and

  $120b-45c=45$  (8)

Now, solve equation (7) & (8),

  $-5c=45$

So, simplify

  $c=-\frac{45}{5}$

Hence,

  $c=-9$  (9)

Put value of equation (9) in equation (5),

  $24b-8c=0$  (5)

As, $c=-9$

  $24b-8\times -9=0$

So, simplify

  $24b+72=0$

Now,

  $b=\frac{-72}{24}$

Hence,

  $b=-3$  (10)

Put value of equation (9) and (10) in equation (2),

  $a=\frac{2b-c}{4}$  (2)

As, $c=-9$ and $b=-3$

  $a=\frac{2\times (-3)-(-9)}{4}$

So,

  $a=\frac{-6+9}{4}$

Hence,

  $a=\frac{3}{4}$  (11)

Now, put value of equation (9), (10) and (11) in equation (1) to solve quadratic function.

Hence, quadratic function $y=\frac{3x^2}{4}-3x-9$