Chapter 9.4, Problem 5CGP

To determine

Find the quadratic function.

Answer to Problem 5CGP

  $y=\frac{-x^2}{4}+4$

Explanation of Solution

Given Information: $(-4,0),(0,4),(4,0)$

Formula Used:Elimination method

Calculation:

  $(-4,0),(0,4),(4,0)$

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

Now,

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

So,

  $0=a{(-4)}^2+b(-4)+c$

Hence,

  $16a-4b+c=0$   (2)

Now,

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

So,

  $4=a{(0)}^2+b(0)+c$

Hence,

  $c=4$   (3)

Now,

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

So,

  $0=a{(4)}^2+b(4)+c$

Hence,

  $16a+4b+c=0$   (4)

Now, solving the equation (2) & (4) by elimination method,

  $16a-4b+c=0$   (2)

and

  $16a+4b+c=0$   (4)

Now,

  $32a+2c=0$

So, simplify

  $a=\frac{-2c}{32}$   (5)

Put value of eq. (3) in eq. (5),

$a=\frac{-2c}{32}$   (5)

As, $c=4$

Now,

  $a=\frac{-2\times 4}{32}=\frac{-8}{32}$

Hence,

  $a=\frac{-1}{4}$   (6)

Put value of equation (3) & equation (6) in eq. (4),

$16a+4b+c=0$   (4)

As, $a=\frac{-1}{4}$ and $c=4$

  $16\times \left(\frac{-1}{4}\right)+4b+4=0$

So,

  $-4+4b+4=0$

So, simplify

  $4b=0$

Hence,

  $b=0$   (7)

Now, put value of equation (3), (6) and (7) in equation (1) to solve quadratic function.

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