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

Answer:

D. $f(x) = 2\cdot x^{2}-4\cdot x -16$

Step-by-step 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.