To determine which rational function is graphed below, we analyze the key features of the graph and match them to the characteristics of rational functions. Rational functions are ratios of two polynomials, and their graphs often exhibit vertical asymptotes, horizontal asymptotes, holes, and x-intercepts.
Understanding the Graph
The graph in question has the following notable features:
- Vertical Asymptote at $ x = 2 $: This indicates that the denominator of the rational function is zero at $ x = 2 $, and the numerator is not zero at this point.
- Horizontal Asymptote at $ y = 1 $: This suggests that the degrees of the numerator and denominator are equal, and the leading coefficients are the same.
- X-intercept at $ x = -1 $: This means the numerator has a factor of $ (x + 1) $, and the denominator does not.
- No holes: This implies that the numerator and denominator do not share any common factors.
Constructing the Rational Function
Given the above observations, we can construct a rational function that fits the graph:
- Numerator: Must have a factor of $ (x + 1) $ to produce the x-intercept at $ x = -1 $. To match the horizontal asymptote at $ y = 1 $, the numerator and denominator must have the same degree and leading coefficients.
- Denominator: Must have a factor of $ (x - 2) $ to produce the vertical asymptote at $ x = 2 $. It should not share any factors with the numerator to avoid a hole.
A suitable rational function is:
$ f(x) = \frac{x + 1}{x - 2} $
- Numerator: $ x + 1 $ → degree 1, leading coefficient 1
- Denominator: $ x - 2 $ → degree 1, leading coefficient 1
This function has:
- A vertical asymptote at $ x = 2 $
- A horizontal asymptote at $ y = 1 $
- An x-intercept at $ x = -1 $
- No holes
Conclusion
The rational function that matches the graph is:
$ \boxed{\dfrac{x + 1}{x - 2}} $
To further verify that this function is the correct match, we can test a specific point on the graph. Take this: let's find the y-intercept by evaluating the function at $x = 0$:
$ f(0) = \frac{0 + 1}{0 - 2} = -\frac{1}{2} $
If the graph crosses the y-axis at $(0, -0.Conversely, as $x$ approaches $2$ from the left ($x \to 2^-$), the denominator is a very small negative number, causing $f(x)$ to approach $-\infty$. 5)$, this provides strong evidence that our constructed function is accurate. On top of that, as $x$ approaches $2$ from the right ($x \to 2^+$), the numerator is approximately $3$ and the denominator is a very small positive number, causing $f(x)$ to approach $+\infty$. Additionally, we can examine the behavior of the function as it approaches the vertical asymptote. This behavior matches the typical "split" characteristic of a vertical asymptote seen in the provided graph.
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By synthesizing the vertical and horizontal asymptotes, the x-intercept, and the specific point behavior, we have systematically narrowed down the possibilities to a single unique expression.
Final Summary
By identifying the zeros of the numerator to locate the x-intercept and the zeros of the denominator to locate the vertical asymptote, we established the basic structure of the function. The horizontal asymptote then allowed us to determine the ratio of the leading coefficients. Since all these conditions align perfectly with the visual data, we can confidently conclude that the function is:
$ \boxed{f(x) = \frac{x + 1}{x - 2}} $
Extending the Analysis: Domain, Range, and End‑Behavior
Having identified the algebraic form of the function, it is useful to explore a few additional properties that reinforce the match between the formula and the graph.
| Property | Computation | Result |
|---|---|---|
| Domain | All real numbers except where the denominator vanishes | ( \displaystyle \mathcal{D}=(-\infty,2)\cup(2,\infty) ) |
| Range | Solve (y=\frac{x+1}{x-2}) for (x) and exclude values that make the denominator zero | ( \displaystyle \mathcal{R}=(-\infty,1)\cup(1,\infty) ) (the line (y=1) is never attained) |
| Horizontal asymptote | (\displaystyle \lim_{x\to\pm\infty}\frac{x+1}{x-2}=1) | (y=1) |
| Vertical asymptote | Set denominator to zero: (x-2=0) | (x=2) |
| x‑intercept | Set numerator to zero: (x+1=0) | ((-1,0)) |
| y‑intercept | Evaluate at (x=0): (f(0)=-\frac12) | ((0,-\tfrac12)) |
Notice that the range excludes the value (y=1); the graph never actually touches the horizontal asymptote, which is precisely what the picture shows.
Verifying the Sketch with Calculus (Optional)
If you wish to confirm the curvature of the graph, a quick derivative check is illuminating:
[ f'(x)=\frac{(1)(x-2)-(x+1)(1)}{(x-2)^2} =\frac{x-2-x-1}{(x-2)^2} =\frac{-3}{(x-2)^2}. ]
Since ((x-2)^2>0) for every (x\neq2), we have (f'(x)=-\frac{3}{(x-2)^2}<0).
Thus the function is strictly decreasing on each interval of its domain, which matches the monotonic decline seen on both sides of the vertical asymptote in the plot.
A Quick Check with a Test Point
Pick a point that lies comfortably within one of the intervals, say (x=3):
[ f(3)=\frac{3+1}{3-2}=4. ]
On the graph, the curve to the right of the asymptote indeed rises above the horizontal asymptote and passes through ((3,4)). Similarly, for a point left of the asymptote, (x=-2):
[ f(-2)=\frac{-2+1}{-2-2}=\frac{-1}{-4}= \frac14. ]
The left‑hand branch of the sketch sits just above the (x)-axis and below the line (y=1), exactly as the calculated value predicts.
Closing Thoughts
All the salient features of the original diagram—vertical asymptote at (x=2), horizontal asymptote at (y=1), an (x)-intercept at ((-1,0)), a (y)-intercept at ((0,-\tfrac12)), and the monotonic decrease on each side of the asymptote—are captured by the simple rational expression
[ \boxed{,f(x)=\dfrac{x+1}{,x-2,},}. ]
Because the numerator and denominator share no common factors, the function has no removable discontinuities (holes), which aligns with the clean break observed in the graph. The analysis of domain, range, limits, and derivative further corroborates that this formula is not only a plausible candidate but the exact match for the plotted curve Most people skip this — try not to. And it works..
In summary, by systematically translating the visual cues into algebraic constraints and then verifying those constraints with elementary calculus, we have arrived at a single, well‑justified rational function that reproduces the given graph in its entirety.
Exploring Transformations for Insight (Optional)
Another powerful way to understand the graph is by rewriting the function in a transformed form. Observe that:
[ f(x) = \frac{x+1}{x-2} = \frac{(x-2)+3}{x-2} = 1 + \frac{3}{x-2}. ]
This expression reveals the graph as a transformation of the parent function ( g(x) = \frac{1}{x} ). Think about it: specifically:
- Horizontal shift: Replacing ( x ) with ( x-2 ) shifts the graph right by 2 units, moving the vertical asymptote from ( x=0 ) to ( x=2 ). - Vertical scaling: Multiplying by 3 stretches the graph vertically by a factor of 3, making the branches steeper.
- Vertical shift: Adding 1 shifts the entire graph up by 1 unit, raising the horizontal asymptote from ( y=0 ) to ( y=1 ).
This transformational perspective not only explains the asymptotes but also highlights why the function approaches ( y=1 ) as ( x \to \pm\infty ), since ( \frac{3}{x-2} \to 0 ) in those limits Practical, not theoretical..
Confirming the Range
From the transformed form ( f(x) = 1 + \frac{3}{x-2} ), we can deduce the range of the function. Since ( \frac{3}{x-2} ) can take any real value except 0 (as ( x \neq 2 )), adding 1 shifts all possible outputs to exclude ( y=1 ). Thus, the range is:
[ \text{Range: } (-\infty, 1) \cup (1, \infty). ]
This aligns perfectly with the earlier observation that the graph never touches the
