We are given a graph of a function and asked to find information about the reciprocal of this function. Specifically, we need to determine the asymptotes, intervals of increase and decrease, max/min points and four other points, and a sketch of the reciprocal function.

1. Problem Description

We are given a graph of a function and asked to find information about the reciprocal of this function. Specifically, we need to determine the asymptotes, intervals of increase and decrease, max/min points and four other points, and a sketch of the reciprocal function.

2. Solution Steps

First, let's analyze the given graph. It looks like a parabola with a vertex at $(0, 1)$. Thus, the equation of the given graph is $y = x^2 + 1$.

Now we want to consider the reciprocal function, $f(x) = \frac{1}{x^2 + 1}$.

a) Asymptotes:

The denominator of $f(x)$ is $x^2 + 1$.

Since $x^2 + 1$ is always positive and never equal to 0, there is no vertical asymptote.

As $x$ approaches infinity, $f(x)$ approaches

0. Therefore, there is a horizontal asymptote at $y = 0$.

b) Intervals of increase and decrease:

To find the intervals of increase and decrease, we can analyze the derivative of $f(x)$.

$f(x) = (x^2 + 1)^{-1}$

$f'(x) = -1(x^2 + 1)^{-2}(2x) = \frac{-2x}{(x^2 + 1)^2}$

If $f'(x) > 0$, then the function is increasing.

If $f'(x) < 0$, then the function is decreasing.

$f'(x) = \frac{-2x}{(x^2 + 1)^2} > 0$ when $-2x > 0$, which means $x < 0$. So, the function is increasing when $x < 0$.

$f'(x) = \frac{-2x}{(x^2 + 1)^2} < 0$ when $-2x < 0$, which means $x > 0$. So, the function is decreasing when $x > 0$.

Thus, the function is increasing on the interval $(-\infty, 0)$ and decreasing on the interval $(0, \infty)$.

c) State max/min points and four other points:

Since the function is increasing for $x < 0$ and decreasing for $x > 0$, there is a maximum at $x = 0$.

The value of the function at $x = 0$ is $f(0) = \frac{1}{0^2 + 1} = 1$. Therefore, there is a maximum point at $(0, 1)$.

Four other points:

$f(1) = \frac{1}{1^2 + 1} = \frac{1}{2}$ so $(1, \frac{1}{2})$.

$f(-1) = \frac{1}{(-1)^2 + 1} = \frac{1}{2}$ so $(-1, \frac{1}{2})$.

$f(2) = \frac{1}{2^2 + 1} = \frac{1}{5}$ so $(2, \frac{1}{5})$.

$f(-2) = \frac{1}{(-2)^2 + 1} = \frac{1}{5}$ so $(-2, \frac{1}{5})$.

d) Sketch the reciprocal function:

The reciprocal function has a horizontal asymptote at $y=0$. It is increasing for $x < 0$ and decreasing for $x > 0$. The maximum value is at $(0, 1)$. We also have the points $(-1, \frac{1}{2})$ and $(1, \frac{1}{2})$ and $(-2, \frac{1}{5})$ and $(2, \frac{1}{5})$.

3. Final Answer

a) Asymptotes: horizontal asymptote at $y=0$. No vertical asymptotes.

b) Intervals of increase and decrease: increasing on $(-\infty, 0)$, decreasing on $(0, \infty)$.

c) Max/min points and four other points: Maximum at $(0, 1)$, $(1, \frac{1}{2})$, $(-1, \frac{1}{2})$, $(2, \frac{1}{5})$, $(-2, \frac{1}{5})$.

d) Sketch: The graph starts close to $y=0$ for large negative $x$, increases to a maximum of 1 at $x=0$, then decreases back towards $y=0$ for large positive $x$.