Let's solve the given problems by finding the x-intercepts and y-intercepts of the functions.

1. $f(x) = \frac{x - 25}{x - 5}$

X-Intercept:

To find the x-intercept, set $f(x) = 0$.

$\frac{x - 25}{x - 5} = 0$

This happens when the numerator is 0:

$x - 25 = 0$

So, $x = 25$.

Thus, the x-intercept is at $(25, 0)$.

Y-Intercept:

To find the y-intercept, set $x = 0$ and solve for $f(x)$.

$f(0) = \frac{0 - 25}{0 - 5} = \frac{-25}{-5} = 5$

Thus, the y-intercept is at $(0, 5)$.

2. $f(x) = \frac{2x + 3}{x - 3}$

X-Intercept:

Set $f(x) = 0$:

$\frac{2x + 3}{x - 3} = 0$

This occurs when the numerator is zero:

$2x + 3 = 0$

Solving for $x$:

$2x = -3$ $x = -\frac{3}{2}$

Thus, the x-intercept is at $\left( -\frac{3}{2}, 0 \right)$.

Y-Intercept:

Set $x = 0$ to find the y-intercept:

$f(0) = \frac{2(0) + 3}{0 - 3} = \frac{3}{-3} = -1$

Thus, the y-intercept is at $(0, -1)$.

Would you like further explanation or additional details?

Here are 5 related questions for practice:

1. How do you find the vertical asymptote of these rational functions?
2. What happens if the denominator equals zero in a rational function?
3. Can you find the domain of each function?
4. How do you determine the horizontal asymptote of a rational function?
5. What is the significance of finding x- and y-intercepts in graphing functions?

Tip: When dealing with rational functions, always check for undefined points by setting the denominator equal to zero to find vertical asymptotes!