the interval in which

y = f(x)= x-1 / x^2 –3x+3

tansforms the real axis is

To determine the interval in which the function \( y = f(x) = \frac{x - 1}{x^2 - 3x + 3} \) transforms the real axis, we need to analyze the behavior of the function, particularly focusing on its critical points, asymptotes, and where it intersects the x-axis. Let's break this down step by step.

Identifying Key Features of the Function

First, we should identify the components of the function:

• The numerator is \( x - 1 \).
• The denominator is \( x^2 - 3x + 3 \).

Finding the Roots of the Function

To find where the function intersects the x-axis, we set the numerator equal to zero:

\( x - 1 = 0 \)

This gives us \( x = 1 \).

Analyzing the Denominator

Next, we need to examine the denominator \( x^2 - 3x + 3 \) to ensure it does not equal zero, as this would indicate vertical asymptotes. We can find the roots of the denominator using the quadratic formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a = 1 \), \( b = -3 \), and \( c = 3 \). Plugging in these values:

\( x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 12}}{2} = \frac{3 \pm \sqrt{-3}}{2} \)

Since the discriminant is negative (\(-3\)), the quadratic has no real roots. This means the denominator is always positive, and there are no vertical asymptotes.

Behavior at Infinity

Next, we should consider the behavior of the function as \( x \) approaches positive and negative infinity:

As \( x \to \infty \), \( f(x) \approx \frac{x}{x^2} = \frac{1}{x} \to 0 \).

As \( x \to -\infty \), \( f(x) \approx \frac{x}{x^2} = \frac{1}{x} \to 0 \).

Determining the Interval of Transformation

Now that we have established that the function intersects the x-axis at \( x = 1 \) and approaches zero as \( x \) goes to both positive and negative infinity, we can analyze the intervals:

• For \( x < 1 \), \( f(x) \) is negative since the numerator \( x - 1 < 0 \) and the denominator is positive.
• At \( x = 1 \), \( f(x) = 0 \).
• For \( x > 1 \), \( f(x) \) is positive since both the numerator and denominator are positive.

This leads us to conclude that the function transforms the real axis in the interval:

\( (-\infty, 1) \) where it is negative, and \( (1, \infty) \) where it is positive.

Final Thoughts

In summary, the function \( y = \frac{x - 1}{x^2 - 3x + 3} \) transforms the real axis at the point \( x = 1 \). It is negative for \( x < 1 \) and positive for \( x > 1 \), indicating a clear transition at this point. Understanding these intervals helps in graphing the function and analyzing its behavior across the real number line.