Chapter 9.4, Problem 50HP

To determine

Find the quadratic function.

Answer to Problem 50HP

No, Graph 1 and Graph 2 don’t have same quadratic equation.

Explanation of Solution

Given Information: Graph 1 passes through $(0,-3),(3,0),(-1,0)$

Formula Used:Elimination method

Calculation:

Graph 1 passes through $(0,-3),(3,0),(-1,0)$

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,

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

So,

  $0=a{(3)}^2+b(3)+c$

Hence,

  $9a+3b+c=0$(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. (2) in eq. (3),

$9a+3b+c=0$(3)

As, $c=-3$

  $9a+3b-3=0$

Now,

  $9a+3b=3$

Hence, divide the equation by 3

  $3a+b=1$(5)

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

$a-b+c=0$(4)

As, $c=-3$

  $a-b-3=0$

Hence,

  $a-b=3$(6)

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

  $3a+b=1$(5)

and

  $a-b=3$(6)

Now, solve equation

  $4a=4$

Now,

  $a=\frac{4}{4}$

Hence,

  $a=1$(7)

Put value of a and c in equation (4),

$a-b+c=0$(4)

As, $a=1$ and $c=-3$

  $1-b-3=0$

So, simplify

  $-b-2=0$

Hence,

  $b=-2$(8)

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

So, quadratic function $y=x^2-2x-3$ (9)

From Graph 2 point $(1,2)$ lies on curve

As the given equation $y=x^2-2x-3$ (9)

Now,

$(1,2)$ passes through equation $y=x^2-2x-3$

So,

  $2={(1)}^2-2(1)-3$

Now,

  $2=1-2-3$

Hence,

  $2\ne -4$

No, Graph 1 and Graph 2 don’t have same quadratic equation.