The vertical asymptote is x = 2, because when x approaches 2 from either side, the denominator becomes very close to zero and the function approaches positive or negative infinity.
To find the horizontal asymptote, we can use the fact that the highest power of x in the denominator is 1 (since it's just x - 2), and the highest power of x in the numerator is also 1 (since it's just 1). Therefore, as x approaches positive or negative infinity, the function approaches the ratio of the leading coefficients, which is 1/1 or just 1. Hence, the horizontal asymptote is y = 1.
To sketch the graph, we can start by plotting a few points. For example, when x = 0, y = 1/(-2) + 1 = 1/2 + 1 = 3/2. When x = 1, y = 1/(-1) + 1 = 0. And when x = 3, y = 1/1 + 1 = 2.
We can also note that the function can never be equal to or less than zero, because the denominator is always positive (since it's just x - 2) and the numerator is always positive (since it's just 1). Therefore, the range of the function is y > 0.
With these pieces of information, we can sketch the graph as follows:
- There is a vertical asymptote at x = 2.
- There is a horizontal asymptote at y = 1.
- The function is undefined at x = 2 (because of the vertical asymptote), so there is a "hole" in the graph there.
- The function is always positive, so the graph is above the x-axis.
- The graph passes through the point (0, 3/2).
- The graph approaches the horizontal asymptote as x approaches infinity in either direction.
Here's a rough sketch of the graph:
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Note: This is just a rough sketch, and the graph should actually have a "hole" at x = 2. Also, the scale of the graph may not be accurate.
Sketch the asymptotes and the graph of the function. Identify the domain and range.
y = 1/x-2 + 1
Identify the asymptotes.
There is a vertical asymptote at x = ?
There is a horizontal asymptote at y = ?
1 answer
5 answers