Chapter 9.4, Problem 16CYU

To determine

Find the quadratic function.

Answer to Problem 16CYU

  $y=\frac{-2x^2}{3}+2x+36$

Explanation of Solution

Given Information: $(-6,0),(6,24),(9,0)$

Formula Used: Elimination method

Calculation:

  $(-6,0),(6,24),(9,0)$

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

Now,

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

So,

  $0=a{(-6)}^2+b(-6)+c$

So,

  $36a-6b+c=0$

Now,

  $36a=6b-c$

Hence,

  $a=\frac{6b-c}{36}$  (2)

Now,

  $(6,24)$ passes through equation $y=a{x}^2+bx+c$

So,

  $24=a{(6)}^2+b(6)+c$

Hence,

  $36a+6b+c=24$  (3)

Now,

  $(9,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(9)}^2+b(9)+c$

Hence,

  $81a+9b+c=0$  (4)

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

  $36a+6b+c=24$  (3)

As, $a=\frac{6b-c}{36}$

  $36\times \left(\frac{6b-c}{36}\right)+6b+c=24$

So,

  $6b-c+6b+c=24$

So, simplify

  $12b=24$

Now,

  $b=\frac{24}{12}$

Hence,

  $b=2$  (5)

Put value of eq. (2) and eq. (5) in eq. (4),

  $81a+9b+c=0$  (4)

As, $a=\frac{6b-c}{36}$ and $b=2$

  $81\times \left(\frac{6b-c}{36}\right)+9\times 2+c=0$

So,

  $\frac{81}{36}\times \left(6\times 2-c\right)+9\times 2+c=0$

So,

  $\frac{9}{4}\times \left(12-c\right)+18+c=0$

Now, simplify

  $27-\frac{9}{4}c+18+c=0$

So,

  $45-\frac{5}{4}c=0$

Now,

  $\frac{5}{4}c=45$

Now,

  $c=\frac{45\times 4}{5}$

So, simplify

  $c=\frac{180}{5}$

Hence,

  $c=36$  (6)

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

  $a=\frac{6b-c}{36}$  (2)

As, $b=2$ and $c=36$

  $a=\frac{6\times 2-36}{36}$

Now, simplify

  $a=\frac{12-36}{36}=\frac{-24}{36}$

Hence,

  $a=\frac{-2}{3}$  (7)

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

So, quadratic function $y=\frac{-2x^2}{3}+2x+36$