Question

Select the function that represents a parabola with zeros at x = –2 and x = 4, and y-intercept (0,–16). A ƒ(x) = x2 + 2x – 8 B ƒ(x) = 2x2 + 4x – 16 C ƒ(x) = x2 – 2x – 8 D ƒ(x) = 2x2 – 4x – 16

Answer 1

D ƒ(x) = 2x2 – 4x – 16

Explanation:

Answer 2

D.
`f(x) = 2\cdot x^(2)-4\cdot x -16`

Explanation:

Any parabola is modelled by a second-order polynomial, whose standard form is:

`y = a\cdot x^(2)+b\cdot x + c`

Where:

`x` - Independent variable, dimensionless.

`y` - Dependent variable, dimensionless.

`a`,
`b`,
`c` - Coefficients, dimensionless.

In addition, a system of three linear equations is constructed by using all known inputs:

(-2, 0)

`4\cdot a -2\cdot b + c = 0` (Eq. 1)

(4, 0)

`16\cdot a + 4\cdot b +c = 0` (Eq. 2)

(0,-16)

`c = -16` (Eq. 3)

Then,

`4\cdot a - 2\cdot b = 16` (Eq. 4)

`16\cdot a + 4\cdot b = 16` (Eq. 5)

(Eq. 3 in Eqs. 1 - 2)

`a - 0.5\cdot b = 4` By Eq. 4 (Eq. 4b)

`a = 4 + 0.5\cdot b`

Then,

`16\cdot (4+0.5\cdot b) + 4\cdot b = 16` (Eq. 4b in Eq. 5)

`64 + 12\cdot b = 16`

`12\cdot b = -48`

`b = -4`

The remaining coeffcient is:

`a = 4 + 0.5\cdot b`

`a = 4 + 0.5\cdot (-4)`

`a = 2`

The function that represents a parabola with zeroes at x = -2 and x = 4 and y-intercept (0,16) is
`f(x) = 2\cdot x^(2)-4\cdot x -16`. Thus, the right answer is D.