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

Respuesta :

Answer:

D. [tex]f(x) = 2\cdot x^{2}-4\cdot x -16[/tex]

Step-by-step explanation:

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

[tex]y = a\cdot x^{2}+b\cdot x + c[/tex]

Where:

[tex]x[/tex] - Independent variable, dimensionless.

[tex]y[/tex] - Dependent variable, dimensionless.

[tex]a[/tex], [tex]b[/tex], [tex]c[/tex] - Coefficients, dimensionless.

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

(-2, 0)

[tex]4\cdot a -2\cdot b + c = 0[/tex] (Eq. 1)

(4, 0)

[tex]16\cdot a + 4\cdot b +c = 0[/tex] (Eq. 2)

(0,-16)

[tex]c = -16[/tex] (Eq. 3)

Then,

[tex]4\cdot a - 2\cdot b = 16[/tex] (Eq. 4)

[tex]16\cdot a + 4\cdot b = 16[/tex] (Eq. 5)

(Eq. 3 in Eqs. 1 - 2)

[tex]a - 0.5\cdot b = 4[/tex] By Eq. 4 (Eq. 4b)

[tex]a = 4 + 0.5\cdot b[/tex]

Then,

[tex]16\cdot (4+0.5\cdot b) + 4\cdot b = 16[/tex] (Eq. 4b in Eq. 5)

[tex]64 + 12\cdot b = 16[/tex]

[tex]12\cdot b = -48[/tex]

[tex]b = -4[/tex]

The remaining coeffcient is:

[tex]a = 4 + 0.5\cdot b[/tex]

[tex]a = 4 + 0.5\cdot (-4)[/tex]

[tex]a = 2[/tex]

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

Answer:

D ƒ(x) = 2x2 – 4x – 16

Step-by-step explanation: