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Let \(f(x) = \frac{x^2 - 6x + 6}{2x - 4}\) and \(g(x) = \frac{ax^2 + bx + c}{x - d}.\)


You are given the following properties:

 The graphs of \(f(x)\) and \(g(x)\) have the same vertical asymptote.

 The oblique asymptotes of \(f(x)\) and \(g(x)\) are perpendicular, and they intersect on the y-axis.

 The graphs of \(f(x)\) and \(g(x)\) have two intersection points, one of which is on the line \(x = -2.\)

Find the point of intersection of the graphs of \(f(x)\) and \(g(x)\) that does not lie on the line \(x = -2.\)

 Jun 15, 2023
 #1
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+1

The vertical asymptote of f(x) is x = 2/2 = 1. The vertical asymptote of g(x) is x = d. Since the graphs of f(x) and g(x) have the same vertical asymptote, then d = 1.

The oblique asymptotes of f(x) and g(x) are perpendicular, and they intersect on the y-axis. This means that the slopes of the oblique asymptotes are negative reciprocals of each other. The slope of the oblique asymptote of f(x) is -1/2. The slope of the oblique asymptote of g(x) is 2a. Since the slopes are negative reciprocals of each other, then 2a = -1/2. This gives us a = -1/4.

The point of intersection of the graphs of f(x) and g(x) that does not lie on the line x = -2 is the point where the oblique asymptotes intersect. This is the point where x = 1.

Plugging x = 1 into f(x) and g(x), we get:

f(1) = (1^2 - 6(1) + 6)/(2(1) - 4) = -1/2 g(1) = (a(1)^2 + b(1) + c)/(1 - d) = (-1/4(1) + b + c)/(1 - 1) = b + c

Since the oblique asymptotes of f(x) and g(x) intersect at y = -1/2, then b + c = -1/2.

We are given that the graphs of f(x) and g(x) have two intersection points, one of which is on the line x = -2. The point of intersection of the graphs of f(x) and g(x) on the line x = -2 is (-2, 3).

Plugging x = -2 into f(x) and g(x), we get:

f(-2) = (-2)^2 - 6(-2) + 6)/(2(-2) - 4) = 12/-4 = -3 g(-2) = (a(-2)^2 + b(-2) + c)/(-2 - 1) = (4a - 2b + c)/-3

Since the graphs of f(x) and g(x) have two intersection points, then the equations f(-2) = -3 and g(-2) = 3 must be equal. This gives us:

(4a - 2b + c)/-3 = 3

Solving for b and c in terms of a, we get:

b = 3a/2 c = -9a/2

Plugging b = 3a/2 and c = -9a/2 into b + c = -1/2, we get:

3a/2 - 9a/2 = -1/2

This gives us a = -1/12.

Plugging a = -1/12 into b = 3a/2, we get:

b = 3(-1/12)/2 = -1/8

Plugging a = -1/12 and b = -1/8 into g(1) = b + c, we get:

g(1) = (-1/12) + (-1/8) = -7/24

Therefore, the point of intersection of the graphs of f(x) and g(x) that does not lie on the line x = -2 is (1, -7/24).

 Jun 16, 2023
 #2
avatar+14 
+1

Sorry, but the answer is wrong. Thank you for trying though! This is the explanation for the correct answer:

 

The vertical asymptote of \(f(x)\) is \(x=2.\) Hence, \(d=2.\)

By long division,\(f(x) = \frac{1}{2} x - 2 - \frac{2}{2x - 4}.\)
Thus, the oblique asymptote of \(f(x)\) is \(y = \frac{1}{2} x - 2,\) which passes through \((0,-2).\) Therefore, the oblique asymptote of \(g(x)\) is \(y = -2x - 2.\)
Therefore, \(g(x) = -2x - 2 + \frac{k}{x - 2}\) for some constant \(k.\)
Finally, \(f(-2) = \frac{(-2)^2 - 6(-2) + 6}{2(-6) - 4} = -\frac{11}{4},\) so \(g(-2) = -2(-2) - 2 + \frac{k}{-2 - 2} = -\frac{11}{4}.\)
Solving, we find \(k = 19.\) Hence, \(g(x) = -2x - 2 + \frac{19}{x - 2} = \frac{-2x^2 + 2x + 23}{x - 2}.\)
We want to solve \(\frac{x^2 - 6x + 6}{2x - 4} = \frac{-2x^2 + 2x + 23}{x - 2}.\)
Then \(x^2 - 6x + 6 = -4x^2 + 4x + 46,\) or \(5x^2 - 10x - 40 = 0.\) This factors as \(5(x + 2)(x - 4) = 0,\) so the other point of intersection occurs at \(x = 4.\) Since \(f(4) = \frac{4^2 - 6 \cdot 4 + 6}{2(4) - 4} = -\frac{1}{2},\) the other point of intersection is \(\boxed{\left( 4, -\frac{1}{2} \right)}.\)

 

Still, thank you for trying, Guest!

humanbeing  Jun 18, 2023
 #3
avatar+397 
0

The oblique asymptote of f(x) is y = x/2 - 3, not y = x/2 - 2.

 Jun 18, 2023

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