Question

Find the quadratic polynomial whose zeros are 3, - 2,the graph of which passes through (0,6)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer: The quadratic polynomial is $\mathbf{-x^{2}+x+6}$

Step-by-step explanation:

1. Let quadratic polynomial is

  $\mathbf{y= ax^{2}+bx+c}$      ...1)

2. Since quadratic polynomial have zeros 3 and -2. It means

   At x=3 , y=0       ...2)

   and also

  At  x= -2 , y=0       ...3)

3. It is also given that point (0,6) lie on curve. It means

   At x=0 , y =6      ...4)

4. Now from equation 1) and equation 4), we get

   $\mathbf{y= ax^{2}+bx+c}$

   $\mathbf{6= a(0)^{2}+b\times 0+c}$

 

   $\mathbf{6= 0+0+c}$

   So

   c=6

5. Now equation 1) can be written as

   $\mathbf{y= ax^{2}+bx+c}$

   $\mathbf{y= ax^{2}+bx+6}$        ...5)

6. Now from equation 2) and equation 5)

   $\mathbf{y= ax^{2}+bx+6}$

   $\mathbf{0= a(3)^{2}+b\times 3+6}$

   $\mathbf{0= 9a+3b+6}$

   $\mathbf{9a+3b=-6}$

   This can be written as

   $\mathbf{3a+b=-2}$        ...6)

7. Now from equation 3) and equation 5)

   $\mathbf{y= ax^{2}+bx+6}$

   $\mathbf{0= a(-2)^{2}+b\times (-2)+6}$

   $\mathbf{0= 4a-2b+6}$

   $\mathbf{4a-2b=-6}$

   This can be written as

   $\mathbf{2a-b=-3}$       ...7)

8. On adding equation 6) and equation 7)

   $\mathbf{5a=-5}$

   $\mathbf{a=-1}$        ...8)

9. Now from equation 8) and equation 7)

   $\mathbf{2a-b=-3}$

   $\mathbf{2(-1)-b=-3}$

   $\mathbf{-2-b=-3}$

  So

   $\mathbf{b=1}$         ...9)

10. Now from equation 9), equation 8) and equation 5)

    $\mathbf{y= ax^{2}+bx+6}$

    $\mathbf{y= -x^{2}+x+6}$     ...10)

   Equation 10) is required quadratic polynomial.