To find the intercepts of the function $f(x) = \frac{2x + 3}{x - 1}$, we'll calculate both the x-intercept and the y-intercept:

1. Finding the x-intercept:

The x-intercept occurs when $f(x) = 0$. This happens when the numerator of the function is equal to zero (since the denominator would make the function undefined if it were zero).

$\frac{2x + 3}{x - 1} = 0$

This implies that:

$2x + 3 = 0$

Solving for $x$:

$2x = -3$

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

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

2. Finding the y-intercept:

The y-intercept occurs when $x = 0$. We substitute $x = 0$ into the function to find $f(0)$:

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

Thus, the y-intercept is at $(0, -3)$.

Summary of intercepts:

• x-intercept: $\left( -\frac{3}{2}, 0 \right)$
• y-intercept: $(0, -3)$

Would you like more details or further clarification on this?

Related questions:

1. How do vertical asymptotes relate to the denominator in rational functions?
2. Can we find the horizontal asymptote of this function?
3. How would the graph of $f(x) = \frac{2x+3}{x-1}$ behave near $x = 1$?
4. What happens to the function as $x \to \infty$?
5. How do we find the slope of the tangent line at specific points on this function?

Tip:

When solving for intercepts in rational functions, always check if the denominator could be zero to avoid undefined values.