Question

Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)

Answer 1

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

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

It is passes through the point (0,11). So, substitute
`x=0,y=11` in (i).

`11=a(0)^2+b(0)+c`

`11=c`

Putting
`c=11` in (i), we get

`y=ax^2+bx+11` ...(ii)

The quadratic function passes through the point (5,31). So, substitute
`x=5,y=31` in (ii).

`31=a(5)^2+b(5)+11`

`31-11=a(25)+5b`

`20=25a+5b`

Divide both sides by 5.

`4=5a+b` ...(iii)

The quadratic function passes through the point (3,11). So, substitute
`x=3,y=11` in (ii).

`11=a(3)^2+b(3)+11`

`11-11=a(9)+3b`

`0=9a+3b`

Divide both sides by 3.

`0=3a+b` ...(iv)

Subtracting (iv) from (iii), we get

`4-0=5a+b-3a-b`

`4=2a`

`(4)/(2)=a`

`2=a`

Putting
`a=2` in (iv), we get

`0=3(2)+b`

`0=6+b`

`-6=b`

Putting
`a=2,b=-6` in (ii), we get

`y=(2)x^2+(-6)x+11`

`y=2x^2-6x+11`

Therefore, the required quadratic equation is
`y=2x^2-6x+11`.