Chapter 9.4, Problem 5BGP

To determine

Find the quadratic function.

Answer to Problem 5BGP

  $y=\frac{x^2}{2}+3x+\frac{5}{2}$

Explanation of Solution

Given Information: $(-5,0),(-1,0),(1,6)$

Formula Used: Elimination method

Calculation:

  $(-5,0),(-1,0),(1,6)$

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

Now,

  $(-5,0)$ passes through equation $y=a{x}^2+bx+c$

So,

  $0=a{(-5)}^2+b(-5)+c$

Hence,

  $25a-5b+c=0$  (2)

Now,

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

So,

  $0=a{(-1)}^2+b(-1)+c$

Now,

  $a-b+c=0$

Hence,

  $a=b-c$  (3)

Now,

  $(1,6)$ passes through equation $y=a{x}^2+bx+c$

So,

  $6=a{(1)}^2+b(1)+c$

Hence,

  $a+b+c=6$  (4)

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

  $a+b+c=6$

As, $a=b-c$

  $b-c+b+c=6$

So,

  $2b=6$

So, simplify

  $b=\frac{6}{2}$

Hence,

  $b=3$  (5)

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

  $a+b+c=6$  (4)

As, $b=3$

  $a+3+c=6$

So, simplify

  $a+c=6-3$

Hence,

  $a+c=3$  (6)

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

  $25a-5b+c=0$  (2)

As, $b=3$

  $25a-5\times (3)+c=0$

So, simplify

  $25a-15+c=0$

Hence,

  $25a+c=15$  (7)

Now, solving the equation (6) & (7) by elimination method,

  $a+c=3$  (6)

and

  $25a+c=15$  (7)

Now, for elimination method, multiply equation (6) by -1,

  $-a-c=-3$  (8)

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

  $24a=12$

So, simplify

  $a=\frac{12}{24}$

Hence,

  $a=\frac{1}{2}$  (9)

Put value of equation (9) in equation (6),

  $a+c=3$  (6)

As, $a=\frac{1}{2}$

  $\frac{1}{2}+c=3$

Now, simplify

  $c=3-\frac{1}{2}$

Hence,

  $c=\frac{5}{2}$  (10)

Now, put value of equation (5), (9) and (10) in equation (1) to solve quadratic function.

Hence, quadratic function $y=\frac{x^2}{2}+3x+\frac{5}{2}$