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Use the guidelines of this section to sketch the curve. y = x /x 2 − 25

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Answer:

Explanation:

To sketch the curve of the equation y = x / (x^2 - 25), we can start by analyzing the behavior of the function for different values of x.

First, let's identify any vertical asymptotes by finding the values of x that make the denominator equal to zero. In this case, the denominator x^2 - 25 becomes zero when x = ±5. Therefore, we have vertical asymptotes at x = -5 and x = 5.

Next, let's check the behavior of the function near these vertical asymptotes. We can do this by evaluating the function for values of x that approach the asymptotes from both sides.

As x approaches -5 from the left side (x < -5), the function becomes very negative since the numerator (-5) is negative, and the denominator becomes positive but small. As x approaches -5 from the right side (x > -5), the function becomes very positive since the numerator (+5) is positive, and the denominator becomes positive but small. This indicates that there is a vertical asymptote at x = -5.

Similarly, as x approaches 5 from the left side (x < 5), the function becomes very negative since the numerator (-5) is negative, and the denominator becomes positive but small. As x approaches 5 from the right side (x > 5), the function becomes very positive since the numerator (+5) is positive, and the denominator becomes positive but small. This indicates that there is also a vertical asymptote at x = 5.

Next, let's find the x-intercepts of the function. The x-intercepts occur when y = 0. Therefore, we can set the numerator x to zero and solve for x:

0 = x

x = 0

So, we have an x-intercept at x = 0.

Now, let's find the y-intercept of the function. The y-intercept occurs when x = 0. Plugging in x = 0 into the equation, we get:

y = 0 / (0^2 - 25) = 0 / (-25) = 0

So, we have a y-intercept at y = 0.

Based on these observations, we can sketch the curve as follows:

There are vertical asymptotes at x = -5 and x = 5.

There is an x-intercept at x = 0.

There is a y-intercept at y = 0.

The curve approaches the vertical asymptotes as x approaches -5 and 5 from both sides. The function value becomes very large in magnitude as x gets close to the vertical asymptotes.

Please note that without more specific information or additional points, it is challenging to precisely sketch the curve. The given guidelines provide a general idea of the behavior of the function.

User Marcus Frenkel
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