91Ó°ÊÓ

The concept of x-intercepts for a function is quite straightforward. X-intercepts are the points where the graph of a function crosses or touches the x-axis. Mathematically, it is where the output or y-value of the function is zero. To find the x-intercepts, we set the function equal to zero and solve for x.

For our function, \( f(x) = \frac{3}{2} \sqrt{4 - x^2} \), we begin by equating the entire function to zero:

• \( \frac{3}{2} \sqrt{4 - x^2} = 0 \) simplifies to \( \sqrt{4 - x^2} = 0 \)
• Solving \( \sqrt{4 - x^2} = 0 \) leads to \( 4 - x^2 = 0 \)
• Solving \( x^2 = 4 \) results in \( x = \pm 2 \)

This indicates that the x-intercepts of the function \( f(x) \) occur at \( x = 2 \) and \( x = -2 \). These points are (2, 0) and (-2, 0). Remember, x-intercepts illustrate crucial boundary points of a function, especially when dealing with semicircular functions like this one.

Y-intercepts are similarly important. They are found where the graph crosses the y-axis. At these points, \( x \) is zero. Thus, the y-value at \( x = 0 \) is our y-intercept. To find it in our given function, we substitute \( x = 0 \) and calculate:

• Plug \( x = 0 \) into \( f(x) = \frac{3}{2} \sqrt{4 - x^2} \)
• This gives us \( f(0) = \frac{3}{2} \sqrt{4 - 0^2} = \frac{3}{2} \times 2 = 3 \)

Thus, the y-intercept for the function \( f(x) \) is the point (0, 3). This information can be extremely valuable when visualizing the function, because the y-intercept is the function's output when the input starts from zero.

The function \( f(x) = \frac{3}{2} \sqrt{4 - x^2} \) is a perfect example of a semicircle function when analyzed within its domain. A semicircle function describes a part of a circle that is cut off by a line, usually its diameter. Here’s why this function represents a semicircle:

• The term \( \sqrt{4 - x^2} \) is the defining equation of a semicircle centered at the origin.
• This setup constrains \( x \) to values that keep the expression inside the square root (the radicand) non-negative: \( -2 \leq x \leq 2 \).
• The diameter, in this case, stretches from -2 to 2 on the x-axis, describing exactly half of a circle.

Visualizing this function graphically will show you a smooth arc, reminiscent of half a circle, with its highest point corresponding to the y-intercept. Understanding these properties can significantly help in both algebraic manipulations and geometric interpretations of semicircle functions.