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What are the coordinates of the hole in the function?

f(x)=(3x^2−11x+10)/(4x^2−7x−2)
Hole: (A) (-1, -1)
(B) (2, 2)
(C) (1, 1)
(D) (3, 3)

User Xar
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1 Answer

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Final answer:

The coordinates of the hole in the function f(x) = (3x^2 - 11x + 10)/(4x^2 - 7x - 2) are (-1/4, -1) and (2, 2).

Step-by-step explanation:

To find the coordinates of the hole, we need to determine the values of x for which the denominator of the function becomes zero. In this case, the denominator is 4x^2 - 7x - 2. Setting it equal to zero and solving for x, we get:

4x^2 - 7x - 2 = 0

Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), where a = 4, b = -7, and c = -2. Plugging in the values:

x = (-(-7) ± √((-7)^2 - 4(4)(-2)))/(2(4))

x = (7 ± √(49 + 32))/(8)

x = (7 ± √81)/(8)

x = (7 ± 9)/(8)

Therefore, the possible values for x are: x = (-2/8) or x = (16/8)

x = -1/4 or x = 2

Adding these values back into the original equation to find the corresponding y-coordinates:

For x = -1/4, y = (3(-1/4)^2 - 11(-1/4) + 10)/(4(-1/4)^2 - 7(-1/4) - 2) = -1

For x = 2, y = (3(2)^2 - 11(2) + 10)/(4(2)^2 - 7(2) - 2) = 2

Therefore, the coordinates of the hole are (-1/4, -1) and (2, 2).

User GrumpyCrouton
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