Chapter 9.4, Problem 47PPS

To determine

Find the quadratic function.

Answer to Problem 47PPS

  $y=\frac{x^2}{4}-2x+3$

Explanation of Solution

Given Information: Graph passes through $(0,3),(2,0),(6,0)$

Formula Used:Elimination method

Calculation:

Graph passes through $(0,3),(2,0),(6,0)$

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

Now,

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

So,

  $3=a{(0)}^2+b(0)+c$

Hence,

  $c=3$(2)

Now,

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

So,

  $0=a{(2)}^2+b(2)+c$

Hence,

  $4a+2b+c=0$(3)

Now,

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

So,

  $0=a{(6)}^2+b(6)+c$

Hence,

  $36a+6b+c=0$(4)

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

$4a+2b+c=0$(3)

As, $c=3$

  $4a+2b+3=0$

Hence,

  $4a+2b=-3$(5)

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

  $36a+6b+c=0$(4)

As, $c=3$

  $36a+6b+3=0$

Now,

  $36a+6b=-3$

Hence,

  $12a+2b=-1$(6)

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

  $4a+2b=-3$(5)

and

  $12a+2b=-1$(6)

Now, for elimination method, multiply equation (5) by -1,

  $-4a-2b=3$(7)

Now, solve equation (6) & (7),

  $8a=2$

Now,

  $a=\frac{2}{8}$

Hence,

  $a=\frac{1}{4}$(8)

Put value of a and c in equation (3),

$4a+2b+c=0$(3)

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

  $4\times \frac{1}{4}+2b+3=0$

So, simplify

  $2b+4=0$

Now,

  $b=\frac{-4}{2}$

Hence,

  $b=-2$(9)

Now, put value of equation a, b and c in equation (1) to solve quadratic function.

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