Chapter 9.4, Problem 42PPS

To determine

Find the quadratic function.

Answer to Problem 42PPS

  $y=2x^2+14x-16$

Explanation of Solution

Given Information: $(-8,0),(0,-16),(1,0)$

Formula Used:Elimination method

Calculation:

  $(-8,0),(0,-16),(1,0)$

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

Now,

$(-8,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(-8)}^2+b(-8)+c$

Now,

  $64a-8b+c=0$   (2)

Now,

$(0,-16)$ passes through equation $y=a{x}^2+bx+c$

So,

  $-16=a{(0)}^2+b(0)+c$

Now,

  $c=-16$   (3)

Now,

$(1,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(1)}^2+b(1)+c$

Hence,

  $a+b+c=0$   (4)

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

$64a-8b+c=0$   (2)

As, $c=-16$

  $64a-8b-16=0$

So, divide the equation by 8

  $8a-b-2=0$

So, simplify

  $8a-b=2$   (5)

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

$a+b+c=0$   (4)

As, $c=-16$

  $a+b-16=0$

So, simplify

  $a+b=16$   (6)

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

  $8a-b=2$   (5)

and

  $a+b=16$   (6)

Now,

  $9a=18$

So, simplify

  $a=\frac{18}{9}$

Hence,

  $a=2$   (7)

Put value of equation (3) and (7) in equation (4),

$a+b+c=0$   (4)

As, $a=2$ and $c=-16$

  $2+b-16=0$

So,

  $b-14=0$

Hence,

  $b=14$   (8)

Now, put value of equation (3), (7) and (8) in equation (1) to solve quadratic function.

Hence, quadratic function $y=2x^2+14x-16$