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

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

$4=2a$

$\dfrac{4}{2}=a$

$2=a$

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 $y=2x^2-6x+11$.