Question

Given three points of a quadratic function, find the equation that defines the function: (-2, 20)(0, -2)(1, -4)

Answer 1

Answer:

$y=3x^2-5x-2$

Step-by-step explanation:

Let the quadratic function be

$y=ax^2+bx+c$

The given points must satisfy this function;

Substitute $(-2,20)$.

$20=4a-2b+c..(1)$

Substitute (0,-2)

$c=-2...(2)$

Substitute (1,-4);

$-4=a+b+c...(3)$

Put equation (2) into equation (1)

$20=4a-2b+-2$

$4a-2b=22$

$2a-b=11...(4)$

Put equation (2) into equation (3)

$-4=a+b-2...(3)$

$a+b=-2...(5)$

Add equation (4) and (5)

$3a=9$

Divide both sides by 3

$a=3$

Put a=3 in  equation 5.

$3+b=-2$

$b=-2-3$

$b=-5$

The quadratic function is;

$y=3x^2-5x-2$