Hi I’m can I please receive help as soon as possible please

Question

Identify a quadratic function that fits the points (0, 3), (1, 2) , and (3, - 12).

A.) f(x) = 2x² + x + 3

B.) f(x) = -2x² + x - 3

C.) f(x) = 2x² - x + 3

D.) f(x) = -2x² + x + 3

Answer 1

Your quadratic function passes through (0, 3), which means when x = 0, f(x) = 3. When x is 0, the x² and x terms are all going to be zero. So the constant must be +3, not -3. This rules out (B).

Now I'm going to plug "when x = 1, f(x) = 2" into (A), (C), and (D).

(A): 2 = 2 + 1 + 3, doesn't make sense.

(C): 2 = 2 - 1 + 3, also doesn't make sense.

(D): 2 = -2 + 1 + 3

So the answer is (D).

I just want to check if (D) also satisfies "when x = 3, f(x) = -12" in case.

It gives: -12 = -18 + 3 + 3. Still sounds fine.

Answer 2

Let us suppose that choice A is the correct answer.

Let us test it out by substituting the given (x,y) to see if it makes a true equation.

A. f(x) = 2x² + x + 3 ; Consider (0,3)

Is f(0) =3 ?

f(0) = 2(0)² + 0 +3

= 0 + 0 + 3

3 = 3 is a true equation. (0,3) is a solution to the function.

Try the same approach with(1,2) and (3,-12).

If you get a true equation in both cases, then A is the correct answer.

Even if of the 3 given orders pairs is not a solution to that answer choice, you drop that choice and see if B or C or D could be satisfied .

Share your work in the comment and I will take a look at it ,if you do so within the next 30 minutes!