Question

I NEED HELP PLS, its due tonight :)

Answer 1

Explanation:

Step 1: Quadratic equation is y = a
`x^(2)` + bx + c

Step 2: As you can see one of the points from graph is (0, -2). Now we can plug in x = 0 and y = -2 in the equation

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

c = -2

One of the points given is (2, -6). Let’s plug in these values and make an equation. Call it equation A.

-6 = a
`(2)^2` + b(2) + c

-6 = 4a + 2b + c

Also, (5, 3) is another point given. Let’s plug in these values and make an equation. Call it equation B.

3 = a
`(5)^2` + b(5) + c

3 = 25a + 5b + c

Step 3: Substitute c = -2 in equation A and equation B

4a + 2b + c = -6

4a + 2b - 2 = -6

4a + 2b = -6 + 2

4a + 2b = -4

Divide both sides by 2

2a + b = -2

Now substitute c = -2 in equation B

25a + 5b + c = 3

25a + 5b - 2 = 3

25a + 5b = 5

Divide both sides by 5

5a + b = 1

Step 4: Solve the two equations from step 3 by subtracting

2a + b -5a - b = -2 - 1

-3a = -3

a = 1

Now plug in a=1 in any of the two equations to find b.

2(1) + b = -2

2 + b = -2

b = -4

Step 5: Replace a, b and c with their values in the quadratic equation

y = 1.
`x^(2)` - 4x - 2

y =
`x^(2)` - 4x -2

Answer: y =
`x^(2)` - 4x - 2