Question

Determine the equation of the quadratic with zeros x = 1 and x = 5 passing through the

point (3,-12).

Please help

Answer 1

`y=3x^(2) -18x+15` is the equation the quadratic as per the given conditions.

Explanation:

Let the equation of the quadratic be:

`y=ax^(2) +bx+c`

Given that it has zeroes at x = 1 and x = 5 i.e. y = 0 at both the values of x.

Also passes through point (3, -12). i.e. when x = 3, y = -12

Putting x = 1, y = 0:

`\Rightarrow a* 1^(2) +b* 1+c=0\\\Rightarrow a+b+c=0 ....... (1)`

Putting x = 5, y = 0:

`\Rightarrow a* 5^(2) +b* 5+c=0\\\Rightarrow 25a+5b+c=0 ....... (2)`

Putting x = 3, y = -12:

`\Rightarrow a* 3^(2) +b* 3+c=-12\\\Rightarrow 9a+3b+c=-12 ....... (3)`

Equation (2) - Equation (1):

`24a+4b=0\\\Rightarrow 6a +b=0 ...... (4)`

Equation (2) - Equation (3):

`16a+2b=12\\\Rightarrow 8a +b=6 ...... (5)`

Equation (5) - Equation (4):

`2a=6\\\Rightarrow a =3`

Putting value of a in equation (4):

`6* 3+b=0\\\Rightarrow b = -18`

Putting a and b in equation (1):

`3-18+c=0\\\Rightarrow c= 15`

So, the quadratic equation is:

`y=3x^(2) -18x+15`