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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

User Snemarch
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2 Answers

2 votes

Answer:

D ƒ(x) = 2x2 – 4x – 16

Explanation:

User FLICKER
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6.5k points
2 votes

Answer:

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.

User Xavier Delaruelle
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