What is the equation, in standard form, of a parabola that models the values in the table?
x
–2
0
4
f(x)
0
–6
78
(1 point)
Responses
y = 6x2 + 5x – 4
y = 6 x 2 + 5 x – 4
y = –4x2 – 5x + 6
y = –4 x 2 – 5 x + 6
y = 5x2 + 4x – 6
y = 5 x 2 + 4 x – 6
y = 4x2 + 5x – 6
The correct equation, in standard form, of the parabola that models the values in the table is:
y = 4x^2 + 5x - 6
To find the equation of the parabola that models the values in the table, we need to use the standard form of a quadratic equation, which looks like y = ax^2 + bx + c.
Let's plug in the given values for x and f(x) from the table into the standard form equation to find the values of a, b, and c.
When x = -2, f(x) = 0:
0 = a(-2)^2 + b(-2) + c
0 = 4a - 2b + c ----(1)
When x = 0, f(x) = -6:
-6 = a(0)^2 + b(0) + c
-6 = c -----(2)
When x = 4, f(x) = 78:
78 = a(4)^2 + b(4) + c
78 = 16a + 4b + c -----(3)
We can use equations (1), (2), and (3) to solve for a, b, and c.
From equation (2), we know that c = -6. Let's substitute this value into equations (1) and (3).
0 = 4a - 2b - 6 -----(4)
78 = 16a + 4b - 6 -----(5)
Let's rearrange equation (4) to solve for a:
4a - 2b = 6
2a - b = 3
b = 2a - 3 -----(6)
Substitute the value of b from equation (6) into equation (5):
78 = 16a + 4(2a - 3)
78 = 16a + 8a - 12
78 = 24a - 12
90 = 24a
a = 90/24
a = 15/4
Now substitute the values of a and c into equation (1) to solve for b:
0 = 4(15/4) - 2b - 6
0 = 15 - 2b - 6
2b = 15 - 6
2b = 9
b = 9/2
b = 4.5
Therefore, the equation, in standard form, of the parabola that models the values in the table is:
y = (15/4)x^2 + (4.5)x - 6
To find the equation of a parabola in standard form, you need to use the general equation:
y = ax^2 + bx + c
where a, b, and c are constants.
To find the values of a, b, and c, we can use the given table of values.
Let's start by plugging in the values from the table into the equation:
For x = -2, f(x) = 0:
0 = a(-2)^2 + b(-2) + c
For x = 0, f(x) = -6:
-6 = a(0)^2 + b(0) + c
For x = 4, f(x) = 78:
78 = a(4)^2 + b(4) + c
Now we have a system of three equations with three unknowns (a, b, and c).
From the equation for x = -2:
4a - 2b + c = 0
From the equation for x = 0:
c = -6
From the equation for x = 4:
16a + 4b + c = 78
Now we need to solve this system of equations to find the values of a, b, and c.
Substituting c = -6 into the first equation:
4a - 2b - 6 = 0
4a - 2b = 6
Substituting c = -6 into the second equation:
16a + 4b - 6 = 78
16a + 4b = 84
Now we can solve this system of linear equations. Subtracting twice the first equation from the second equation:
16a + 4b - 2(4a - 2b) = 84 - 2(6)
16a + 4b - 8a + 4b = 84 - 12
8a + 8b = 72
a + b = 9
Subtracting a from both sides:
b = 9 - a
Now we can substitute this into the first equation:
4a - 2(9 - a) = 6
4a - 18 + 2a = 6
6a - 18 = 6
6a = 24
a = 4
Substituting the value of a into b = 9 - a:
b = 9 - 4
b = 5
So, we have found the values of a and b, which are a = 4 and b = 5. We also know that c = -6 from the equation for x = 0.
Therefore, the equation of the parabola in standard form is:
y = 4x^2 + 5x - 6
Therefore, the correct answer is:
y = 4x^2 + 5x - 6.