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

D: y = 3x² - 5x - 2

Step-by-step explanation:

The general form of a quadratic equation is: y = ax² + bx + c    

c is the y-intercept

We are given the point (0, -2) which is the y-intercept, so we can rewrite our general form into

y = ax² + bx - 2

We can create a system of equations to solve for a and b.  We are given two points.  

Equation 1:  Take the first point (-2, 20) and plug it into our general equation...

20 = a(-2)² + b(-2) - 2

   20 = 4a - 2b - 2

        22 = 4a - 2b         (add 2 to both sides)

           11 = 2a - b         (divide both sides by 2 since every coefficient is even)

Equation 2:   Take the point (1, -4) and plug it into the general equation

-4 = a(1)² + b(1) - 2

  -4 = a + b - 2

     -2 = a + b

Now we have our 2 equations:

11 = 2a - b

-2 = a + b

Since the coefficients of b are already have opposite signs, add the two equations together (elimination method)

Now we have

9 = 3a       now solve for a...

 3 = a   (divide by 3 on both sides)

If a = 3, then

-2 = 3 + b

    -5 = b

Our equation is

y = 3x² - 5x - 2

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

answer:

sometimes

*i think