Consider the function below.

f(x) = (x^2)/(x−9)^2

(a) Find the vertical and horizontal asymptotes. x=? y=?

(b) Find the interval where the function is increasing. (Enter your answer using interval notation.)

Find the interval where the function is decreasing. (Enter your answer using interval notation.)

(c) Find the local minimum value.

(d) Find the inflection point.(x,y)=

Find the interval where the function is concave up. (Enter your answer using interval notation.)

Find the intervals where the function is concave down. (Enter your answer using interval notation.)

a

when x = 9, function is vertical y--->oo
when x is large negative or positive then y = 1
TRY NUMBERS!

b
dy/dx = [2x(x-9)^2 - x^2(2)]/(x-9)^4

= [2x^3 -38x^2 + 162 x]/(x-9)^2
the bottom is always +
so the top is the whole thing
if + then slope up, if - then slope down
turns out top is divisible by (x-9) so easier than it looks
Find zeros
oh well, use algorithm
http://www.wolframalpha.com/input/?i=x^2%2F%28x-9%29^2

I don't really think of a function as my gf, sorry. I can't consider them-

To find the vertical asymptotes of the function, we need to determine the values of x for which the denominator of the function becomes zero.

(a) Vertical asymptotes:
Set the denominator equal to zero and solve for x:
x - 9 = 0
x = 9

Therefore, the vertical asymptote is x = 9.

To find the horizontal asymptote, we need to examine the behavior of the function as x approaches positive and negative infinity. Let's consider the degree of the numerator and denominator.

The degree of the numerator is 2 and the degree of the denominator is also 2. Since the degrees are the same, we can use the ratio of the leading coefficients to determine the horizontal asymptote.

The leading coefficient of the numerator is 1, and the leading coefficient of the denominator is also 1. The ratio of the leading coefficients is 1/1 = 1.

Therefore, the horizontal asymptote is y = 1.

(b) To find the intervals where the function is increasing or decreasing, we need to analyze the sign of the derivative of the function.

First, let's find the derivative:
f'(x) = (2x(x-9)^2 - 2(x^2)(2(x-9)))/((x-9)^4)
Simplifying further:
f'(x) = (-18x^2 + 324x - 1458)/((x-9)^4)

To find the intervals where the function is increasing or decreasing, we need to determine the x-values that make the derivative positive or negative. We do this by finding the critical points.

Setting the derivative equal to zero and solving for x:
-18x^2 + 324x - 1458 = 0
Divide by -18 to simplify:
x^2 - 18x + 81 = 0
(x-9)(x-9) = 0
x = 9

The only critical point is x = 9.

To determine the intervals of increasing and decreasing, we can use test values in the derivative.

If we choose a value less than 9, like x = 8, and substitute it into f'(x), we get a negative result. This indicates that the function is decreasing for x < 9.

If we choose a value greater than 9, like x = 10, and substitute it into f'(x), we get a positive result. This indicates that the function is increasing for x > 9.

Therefore, the function is increasing on the interval (9, ∞) and decreasing on the interval (-∞, 9).

(c) To find the local minimum value, we need to locate the points where the derivative changes sign from negative to positive.

From the previous analysis, we know that the derivative changes sign at x = 9. So, we can conclude that the function has a local minimum at x = 9. To find the corresponding y-value, we substitute x = 9 into the original function:
f(9) = (9^2) / (9 - 9)^2
f(9) = 81 / 0

Since the denominator is zero, the function is undefined at x = 9, and there is no local minimum value.

(d) To find the inflection point(s), we need to locate the values of x for which the concavity of the function changes.

To analyze the concavity, we need to find the second derivative of the function:
f''(x) = (d^2/dx^2)[(2x(x-9)^2 - 2(x^2)(2(x-9)))/((x-9)^4)]
After simplification:
f''(x) = (36x - 306)/((x-9)^4)

To find inflection points, we set the second derivative equal to zero and solve for x:
36x - 306 = 0
36x = 306
x = 8.5

The inflection point is at (x,y) = (8.5, f(8.5)).

To determine the intervals of concave up and concave down, we can use test values in the second derivative.

If we choose a value less than 8.5, like x = 8, and substitute it into f''(x), we get a positive result. This indicates that the function is concave up for x < 8.5.

If we choose a value greater than 8.5, like x = 9, and substitute it into f''(x), we get a negative result. This indicates that the function is concave down for x > 8.5.

Therefore, the function is concave up on the interval (-∞, 8.5) and concave down on the interval (8.5, ∞).

To answer these questions about the function f(x) = (x^2)/(x−9)^2, we need to analyze its properties step by step.

(a) To find the vertical asymptote of a rational function, we look for values of x that make the denominator zero. In this case, the denominator is (x-9)^2, so x = 9 is the vertical asymptote.

To find the horizontal asymptote, we compare the degrees of the numerator and denominator. In this case, the degree of the numerator is 2 (since it's x^2) and the degree of the denominator is also 2 (since it's (x-9)^2). When the degrees are the same, we divide the leading coefficients. The leading coefficient of the numerator is 1 and the leading coefficient of the denominator is 1 as well. Therefore, the horizontal asymptote is y = 1 (or the line y = 1x^0).

(b) To find where the function is increasing and decreasing, we need to find the intervals where the first derivative is positive or negative. Let's calculate the first derivative:

f'(x) = [(2x)(x - 9)^2 - (x^2)(2(x - 9))]/(x - 9)^4
= (2x(x - 9)^2 - 2x^2(x - 9))/(x - 9)^4
= 2x(x - 9)[(x - 9) - x]/(x - 9)^4
= 2x(x - 9)(-9)/(x - 9)^4
= -18x/(x - 9)^3

The first derivative is negative for x < 0 and positive for x > 0. However, since we have a vertical asymptote at x = 9, we need to exclude that point from our intervals. Therefore, the function is increasing on the interval (-∞, 9) and decreasing on the interval (9, ∞).

(c) To find the local minimum value, we need to find the critical points and determine whether they correspond to a minimum. A critical point occurs when the first derivative is zero or undefined. In our case, the first derivative is never undefined (as long as x ≠ 9). To find when it's zero, we set -18x/(x - 9)^3 = 0 and solve for x:

-18x = 0
x = 0

So the critical point is x = 0. To determine if it's a minimum or maximum, we can use the second derivative test or observe the behavior of the function around the critical point.

(d) To find the inflection point, we need to find where the second derivative is zero or undefined. The second derivative can be found by differentiating the first derivative:

f''(x) = [-18(x - 9)^3 + 18x(3(x - 9)^2)]/(x - 9)^6
= -18(x - 9)^3 + 54x(x - 9)^2/(x - 9)^6
= -18(x - 9)^3 + 54x/(x - 9)^4
= -18(x - 9)/(x - 9)^2 + 54x/(x - 9)^4
= -18/(x - 9) + 54x/(x - 9)^4

To find when the second derivative is zero or undefined, we set the numerator equal to zero:

-18 = 54x/(x - 9)^4
54x = -18(x - 9)^4
3x = -(x - 9)^4
3x = -(x^4 - 36x^3 + 486x^2 - 2916x + 6561)
3x = -x^4 + 36x^3 - 486x^2 + 2916x - 6561
x^4 - 36x^3 + 486x^2 - 2916x + 6561 = 0

Solving this equation is non-trivial, so we'll use numerical methods to find the approximate solution: x ≈ 8.59. Plugging this value back into the function will give us the corresponding y-coordinate.

To find when the function is concave up or down, we need to determine the intervals where the second derivative is positive or negative. Since we already found that the second derivative equals zero at approximately x = 8.59, we need to test the behavior of the second derivative on either side of this point. For example, if we plug in x = 10, the second derivative is positive, indicating the function is concave up. Likewise, if we plug in x = 8, the second derivative is negative, indicating the function is concave down.

Thus, the interval where the function is concave up is (8.59, ∞) and the intervals where it is concave down are (-∞, 8) and (8, 8.59).

In summary:
(a) The vertical asymptote is x = 9. The horizontal asymptote is y = 1.
(b) The function is increasing on the interval (-∞, 9) and decreasing on the interval (9, ∞).
(c) The local minimum value occurs at x = 0.
(d) The inflection point occurs approximately at (x, y) ≈ (8.59, f(8.59)). The function is concave up on the interval (8.59, ∞) and concave down on the intervals (-∞, 8) and (8, 8.59).