Let's work through the function $f(x) = \frac{x^2 - 4}{x^2 - 3x + 2}$ to identify intercepts, asymptotes, and the general shape for sketching.

Step 1: Factor both the numerator and the denominator.

1. Numerator: $x^2 - 4 = (x - 2)(x + 2)$

2. Denominator: $x^2 - 3x + 2 = (x - 1)(x - 2)$

So the function becomes: $f(x) = \frac{(x - 2)(x + 2)}{(x - 1)(x - 2)}$

Step 2: Identify the Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero, except for any factors that cancel with the numerator.

• The denominator $(x - 1)(x - 2) = 0$ gives two possible points: $x = 1$ and $x = 2$.
• Notice that $(x - 2)$ cancels with the numerator, so there is no vertical asymptote at $x = 2$. However, $x = 2$ will be a hole in the graph.
• Therefore, the only vertical asymptote is at $x = 1$.

Step 3: Identify the Horizontal Asymptote

To find the horizontal asymptote, compare the degrees of the numerator and the denominator:

• Both the numerator and denominator are degree 2 polynomials.
• When the degrees are equal, the horizontal asymptote is found by dividing the leading coefficients. In this case, both leading coefficients are 1.

Thus, the horizontal asymptote is: $y = \frac{1}{1} = 1$

Step 4: Identify the Intercepts

• x-intercept(s): Set the numerator equal to zero. $(x - 2)(x + 2) = 0$ So, $x = 2$ and $x = -2$. However, $x = 2$ is a hole, not an intercept. Thus, the only x-intercept is at $x = -2$.

• y-intercept: Set $x = 0$ in the original function. $f(0) = \frac{0^2 - 4}{0^2 - 3(0) + 2} = \frac{-4}{2} = -2$ The y-intercept is at $y = -2$.

Step 5: Sketch the Graph

• Vertical Asymptote: $x = 1$
• Horizontal Asymptote: $y = 1$
• Hole: At $x = 2$
• Intercepts: x-intercept at $x = -2$, y-intercept at $y = -2$

Would you like me to generate a sketch of this graph or provide further details?

Here are some related questions for further understanding:

1. How do you determine where a function has a hole versus a vertical asymptote?
2. Why does the degree of the polynomial influence the horizontal asymptote?
3. What happens to the graph near a hole, like at $x = 2$ in this case?
4. How would the graph of this function behave at its intercepts?
5. Can you explain how to check the behavior of the graph near the asymptotes?

Tip: Canceling terms between the numerator and denominator simplifies the function, but remember that this creates a hole, not an asymptote, at the canceled point.