Which Quadrants Does Y = 1/(x-3)+4 Occupy?

Understanding the behavior of functions and where their graphs lie on the Cartesian plane is a fundamental skill in mathematics. Today, we're going to dive deep into the function $y = \frac{1}{x-3}+4$ and determine exactly which quadrants its graph graces. This isn't just about memorizing rules; it's about understanding the underlying principles of graph transformations and asymptotes. By the end of this exploration, you'll be able to confidently pinpoint the location of such graphs and appreciate the elegance of how algebraic expressions translate into visual representations. We'll break down the function, analyze its components, and build our understanding step-by-step.

Deconstructing the Function: The Foundation

Let's start by looking closely at the function $y = \frac{1}{x-3}+4$. This equation is a transformation of the basic reciprocal function, $y = \frac{1}{x}$. To understand where our specific function lies, we first need to grasp the behavior of the parent function. The graph of $y = \frac{1}{x}$ has a vertical asymptote at $x=0$ and a horizontal asymptote at $y=0$. This means the graph never touches or crosses these lines. The graph of $y = \frac{1}{x}$ exists in the first and third quadrants. When $x$ is positive, $y$ is positive (Quadrant I). When $x$ is negative, $y$ is also negative (Quadrant III). The key is that $x$ and $y$ always have the same sign.

Now, let's consider the transformations applied to $y = \frac{1}{x}$ to get our target function $y = \frac{1}{x-3}+4$. There are two main transformations happening here:

1. Horizontal Shift: The term $(x-3)$ inside the denominator indicates a horizontal shift. Specifically, it shifts the graph of $y = \frac{1}{x}$ three units to the right. This means the vertical asymptote, which was originally at $x=0$, is now shifted to $x=3$.
2. Vertical Shift: The $+4$ outside the fraction indicates a vertical shift. It moves the graph of $y = \frac{1}{x}$ four units up. This means the horizontal asymptote, which was originally at $y=0$, is now shifted to $y=4$.

These shifts are crucial because they redefine the boundaries and the overall position of the graph. Understanding these transformations allows us to predict the graph's location without even needing to plot points. We can visualize how the original shape of $y = \frac{1}{x}$ is being moved around the coordinate plane.

Identifying Asymptotes: The Guiding Lines

The concept of asymptotes is absolutely vital when determining the location of rational functions like $y = \frac{1}{x-3}+4$. Asymptotes are lines that the graph of a function approaches but never touches or crosses. For our function, the vertical and horizontal asymptotes act as boundaries, essentially dividing the coordinate plane into regions where the graph can and cannot exist. Let's pinpoint these asymptotes:

• Vertical Asymptote: The vertical asymptote occurs where the denominator of the rational part of the function becomes zero. In $y = \frac{1}{x-3}+4$, the denominator is $x-3$. Setting this to zero, we get $x-3=0$, which means $x=3$. So, the vertical asymptote is the line $x=3$. This vertical line separates the left side of the graph from the right side. No point on the graph will ever have an x-coordinate of 3.

• Horizontal Asymptote: The horizontal asymptote is determined by the behavior of the function as $x$ approaches positive or negative infinity. For functions of the form $y = \frac{a}{x-h}+k$, the horizontal asymptote is always at $y=k$. In our case, $k=4$. Therefore, the horizontal asymptote is the line $y=4$. This horizontal line indicates the value that $y$ approaches as $x$ becomes very large (positive or negative). The graph will get closer and closer to $y=4$ but never reach it.

These two lines, $x=3$ and $y=4$, intersect at the point $(3, 4)$. This intersection point is where the 'center' of the hyperbola has effectively moved. The original hyperbola $y=1/x$ was centered at $(0,0)$. Now, our transformed hyperbola is centered at $(3,4)$. These asymptotes are the key to understanding the quadrant analysis because they dictate the regions the graph can occupy.

Analyzing the Behavior: Quadrant by Quadrant

Now that we've identified the vertical asymptote at $x=3$ and the horizontal asymptote at $y=4$, we can analyze the behavior of the function $y = \frac{1}{x-3}+4$ in relation to these lines. The graph of a function of the form $y = \frac{a}{x-h}+k$ with $a > 0$ (like our function where $a=1$) will have two branches that mimic the shape of $y = \frac{1}{x}$ but shifted. One branch will be in the region above and to the right of the intersection of the asymptotes, and the other branch will be in the region below and to the left of the intersection.

Let's consider the behavior of $y = \frac{1}{x-3}+4$ for values of $x$ greater than 3 (to the right of the vertical asymptote $x=3$).

• When $x > 3$, the term $(x-3)$ is positive. Therefore, $\frac{1}{x-3}$ is positive.
• Since $\frac{1}{x-3}$ is positive, when we add 4 to it, the value of $y$ will be greater than 4. That is, $y = \frac{1}{x-3}+4 > 4$.

So, for all $x > 3$, we have $y > 4$. This means the graph is located in the region where $x$ is positive and $y$ is greater than 4. This region is within the first quadrant. Specifically, it's the part of the first quadrant that is to the right of $x=3$ and above $y=4$.

Now, let's consider the behavior of $y = \frac{1}{x-3}+4$ for values of $x$ less than 3 (to the left of the vertical asymptote $x=3$).

• When $x < 3$, the term $(x-3)$ is negative. Therefore, $\frac{1}{x-3}$ is negative.
• Since $\frac{1}{x-3}$ is negative, when we add 4 to it, the value of $y$ will be less than 4. That is, $y = \frac{1}{x-3}+4 < 4$.

This part requires a bit more careful thought. We know $y < 4$. But does the graph cross the x-axis (where $y=0$)? To determine this, let's find the x-intercept by setting $y=0$:

$0 = \frac{1}{x-3}+4$

$-\frac{1}{x-3} = 4$

$-1 = 4(x-3)$

$-1 = 4x - 12$

$11 = 4x$

$x = \frac{11}{4} = 2.75$

Since $x=2.75$ is a valid x-value (it's not the vertical asymptote), the graph does cross the x-axis at $x=2.75$. This means that for values of $x$ just to the left of 3, $y$ is positive (e.g., if $x=2.9$, $1/(2.9-3) = 1/(-0.1) = -10$, so $y = -10+4 = -6$. Wait, I made a mistake in the logic above. Let's re-evaluate the second branch more carefully.

Let's re-examine the region where $x < 3$. We established that for $x < 3$, $\frac{1}{x-3}$ is negative. Consequently, $y = \frac{1}{x-3}+4$ will be less than 4. Now, we need to determine if $y$ can be positive or negative in this region.

Consider values of $x$ just to the left of $x=3$. For example, let $x=2.9$. Then $x-3 = -0.1$. So, $y = \frac{1}{-0.1} + 4 = -10 + 4 = -6$. In this case, $x$ is positive (2.9) and $y$ is negative (-6). This point lies in the fourth quadrant.

Consider values of $x$ that are much smaller, further to the left. For example, let $x=0$. Then $y = \frac{1}{0-3} + 4 = \frac{1}{-3} + 4 = -\frac{1}{3} + \frac{12}{3} = \frac{11}{3}$. In this case, $x$ is negative (0) and $y$ is positive ($\frac{11}{3}$). This point lies in the second quadrant.

Let's check the x-intercept again. We found $x=2.75$ when $y=0$. This means the graph crosses the x-axis at $(2.75, 0)$. For $x$ values between $2.75$ and $3$, $y$ will be negative. For example, if $x=2.8$, $y = 1/(2.8-3) + 4 = 1/(-0.2) + 4 = -5 + 4 = -1$. Here $x$ is positive, $y$ is negative, so it's in the fourth quadrant.

For $x$ values less than $2.75$, $y$ will be positive. For example, if $x=0$, $y=11/3$ (Quadrant II). If $x=-1$, $y = 1/(-1-3) + 4 = 1/(-4) + 4 = -0.25 + 4 = 3.75$. Here $x$ is negative, $y$ is positive, so it's in the second quadrant.

This analysis reveals that for $x < 3$, the graph behaves differently depending on whether $x$ is to the left or right of the x-intercept $x=2.75$.:

• Region 1: $x < 2.75$: Here, $x$ is negative, and $y$ is positive (since the x-intercept is at $x=2.75$). This part of the graph lies in the second quadrant.
• Region 2: $2.75 < x < 3$: Here, $x$ is positive, and $y$ is negative (since it's between the x-intercept and the vertical asymptote). This part of the graph lies in the fourth quadrant.

Wait, I need to be more precise about how the branches relate to the quadrants defined by the new asymptotes $x=3$ and $y=4$, not the original quadrants of the coordinate plane.

The graph of y = rac{1}{x-3}+4 is a hyperbola. Its branches lie in the regions defined by the intersection of its asymptotes. The vertical asymptote is $x=3$, and the horizontal asymptote is $y=4$. These lines divide the plane into four regions:

1. Region above $y=4$ and to the right of $x=3$.
2. Region above $y=4$ and to the left of $x=3$.
3. Region below $y=4$ and to the right of $x=3$.
4. Region below $y=4$ and to the left of $x=3$.

For functions of the form y = rac{a}{x-h}+k where $a > 0$, the branches lie in regions 1 and 4 relative to the asymptotes. This means:

• Branch 1: Above the horizontal asymptote ($y > 4$) and to the right of the vertical asymptote ($x > 3$). This region is entirely within the first quadrant of the coordinate plane.

• Branch 2: Below the horizontal asymptote ($y < 4$) and to the left of the vertical asymptote ($x < 3$). This region is more complex. Let's analyze it carefully.

When $x < 3$, we know $y < 4$. We found the x-intercept is at $x=2.75$. This means:

• If $2.75 < x < 3$: $x$ is positive (between 2.75 and 3), and $y$ is negative (between $y=0$ and $y=-\infty$). This part lies in the fourth quadrant.
• If $x < 2.75$: $x$ can be positive or negative, and $y$ is positive (since it's below the x-intercept $y=0$). Wait, this is also incorrect. Let's try a test point for $x<2.75$. Take $x=0$. $y = 1/(0-3) + 4 = -1/3 + 4 = 11/3$. Here, $x$ is negative, and $y$ is positive. This point $(0, 11/3)$ lies in the second quadrant.

So, the second branch (where $x<3$ and $y<4$) spans across the second and fourth quadrants.

Therefore, the graph of $y = \frac{1}{x-3}+4$ exists in the first, second, and fourth quadrants. This is a crucial detail often missed when only considering the general shape relative to the asymptotes without checking for intercepts.

Let's reconfirm. The function is y=rac{1}{x-3}+4. The asymptotes are $x=3$ and $y=4$. The graph is a hyperbola. Since the numerator (1) is positive, the branches are in the