91Ó°ÊÓ

The domain of a function includes all the possible input values (x-values) that can go into the function without causing any mathematical problems such as division by zero. For the function \( f(x) = \frac{x-3}{x^2+2x-15} \), we need to carefully examine the denominator.
We set \( x^2 + 2x - 15 = 0 \) to find where the denominator becomes zero. This expression can be factorized into \( (x+5)(x-3) \). These factors tell us that the denominator is zero at \( x = -5 \) and \( x = 3 \).
Thus, these values are not included in the domain. Therefore, the domain of \( f(x) \) is all real numbers except \( x = -5 \) and \( x = 3 \). You can express this in interval notation as:

• \( (-\infty, -5) \cup (-5, 3) \cup (3, \infty) \)

Vertical asymptotes occur where a rational function's denominator is zero and the numerator is not zero. For the function \( f(x) = \frac{x-3}{x^2+2x-15} \), after factorizing, we previously found the critical x-values of \( x = -5 \) and \( x = 3 \). At \( x = -5 \), there is a vertical asymptote because the numerator is not zero. However, at \( x = 3 \), both the numerator and denominator are zero, meaning \( x = 3 \) is not a vertical asymptote.
Horizontal asymptotes depend on the degrees of the polynomials in the numerator and the denominator. Here, the degree of the numerator is 1, and the degree of the denominator is 2. Since the degree of the numerator is less than that of the denominator, there is a horizontal asymptote on the x-axis, or \( y = 0 \).

• Vertical Asymptote: \( x = -5 \)
• Horizontal Asymptote: \( y = 0 \)

To determine the intervals where the function is increasing or decreasing, we analyze the first derivative \( f'(x) \). Using the quotient rule, the first derivative is calculated. We find critical points by setting \( f'(x) = 0 \) and identifying where \( f'(x) \) is undefined.
We identify changes in the sign of \( f'(x) \) by testing values around these critical points. If \( f'(x) \) changes from positive to negative, the function decreases; if it changes from negative to positive, the function increases. This identification will give us intervals where the function is either increasing or decreasing.

• Check the sign changes at and around the critical points.
• Mark intervals to identify areas of increase/decrease clearly on your sketch.

The concavity of a function tells us how the curve bends. To assess this, we use the second derivative, \( f''(x) \). A positive \( f''(x) \) indicates that the graph is concave upwards like a cup, while a negative \( f''(x) \) suggests it is concave downwards like a cap.
Points of inflection occur where the concavity changes, meaning \( f''(x) \) changes sign. To find these points, set \( f''(x) = 0 \) or look for undefined points in \( f''(x) \) that cause a sign change. Calculate \( f(x) \) at these points to see the graph's behavior, ensuring these points are marked clearly on the sketch.

• Identify intervals of unknown concavity by testing signs of \( f''(x) \)
• Label the inflection points to reflect the shape of the graph accurately.