Chapter 9.4, Problem 15CYU

To determine

Find the quadratic function.

Answer to Problem 15CYU

  $y=-{\frac{x}{4}}^2+\frac{x}{4}+3$

Explanation of Solution

Given Information: $(-3,0),(4,0),(8,-11)$

Formula Used: Elimination method

Calculation:

  $(-3,0),(4,0),(8,-11)$

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

Now,

  $(-3,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(-3)}^2+b(-3)+c$

So,

  $9a-3b+c=0$  (2)

Now,

  $(4,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(4)}^2+b(4)+c$

Hence,

  $16a+4b+c=0$

Now,

  $16a=-4b-c$

Hence,

  $a=\frac{-4b-c}{16}$  (3)

Now,

  $(8,-11)$ passes through equation $y=a{x}^2+bx+c$

So,

  $-11=a{(8)}^2+b(8)+c$

Hence,

  $64a+8b+c=-11$  (4)

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

  $9a-3b+c=0$  (2)

As, $a=\frac{-4b-c}{16}$

  $9\times \left(\frac{-4b-c}{16}\right)-3b+c=0$

So,

  $\left(\frac{-36b-9c}{16}\right)-3b+c=0$

So, simplify

  $\frac{-36b-9c-48b+16c}{16}=0$

Hence,

  $-84b+7c=0$  (5)

Put value of eq. (3) in eq. (4),

  $64a+8b+c=-11$  (4)

As, $a=\frac{-4b-c}{16}$

  $64\times \left(\frac{-4b-c}{16}\right)+8b+c=-11$

So,

  $-16b-4c+8b+c=-11$

So, simplify

  $-8b-3c=-11$

So, simplify

  $-\left(8b+3c\right)=-11$

Hence,

  $8b+3c=11$  (6)

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

  $-84b+7c=0$  (5)

and

  $8b+3c=11$  (6)

Now, for elimination method, multiply equation (5) by 2 and multiply equation (6) by 21,

  $-168b+14c=0$  (7)

and

  $168b+63c=231$  (8)

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

  $77c=231$

So, simplify

  $c=\frac{231}{77}$

Hence,

  $c=3$  (9)

Put value of c in equation (6),

  $8b+3c=11$  (6)

As, $c=3$

  $8b+3\times 3=11$

Now,

  $8b=11-9$

So, simplify

  $b=\frac{2}{8}$

Hence,

  $b=\frac{1}{4}$  (10)

Put value of b & c in equation (3),

  $a=\frac{-4b-c}{16}$  (3)

As, $b=\frac{1}{4}$ and $c=3$

  $a=\frac{-4\left(\frac{1}{4}\right)-3}{16}$

Now, simplify

  $a=\frac{-1-3}{16}=\frac{-4}{16}$

Hence,

  $a=-\frac{1}{4}$  (11)

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

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