Problem 58 - Calculus Volume 1

Question

[T] Use a graphing calculator to graph the half-circle \(y=\sqrt{25-(x-4)^{2}} .\) Then, use the INTERCEPT feature to find the value of both the \(x\) - and \(y\) -intercepts.

Step-by-Step Solution

Answer

X-intercepts at -1 and 9; Y-intercept at 3.

Step 1: Understand the Equation

The given equation is \( y = \sqrt{25 - (x-4)^2} \). This represents the top half of a circle centered at \((4, 0)\) with radius 5. The circle's full equation would be \((x-4)^2 + y^2 = 25\).

Step 2: Set Up the Graphing Calculator

On your graphing calculator, enter the function \( y = \sqrt{25 - (x-4)^2} \). This will graph the top half of the circle. Ensure the window is set to show a range around the center (4,0) to capture the entire semicircle.

Step 3: Identify the X-Intercept

Using the graph on the calculator, enable the 'INTERCEPT' feature and navigate to find where the graph crosses the x-axis. One x-intercept will be to the left of the center (4,0) and the other to the right.

Step 4: Calculate the X-Intercepts

Set \( y = 0 \) in the equation to solve for \( x \): \( 0 = \sqrt{25 - (x-4)^2} \). Squaring both sides, we get: \( 25 = (x-4)^2 \). Solving this gives \( x-4 = \pm 5 \). Thus, the x-intercepts are at \( x = 9 \) and \( x = -1 \).

Step 5: Identify the Y-Intercept

Using the graph, observe where it crosses the y-axis. There is only one value where the graph will intersect the y-axis.

Step 6: Calculate the Y-Intercept

Setting \( x = 0 \) in the equation, solve for \( y \): \( y = \sqrt{25 - (0-4)^2} = \sqrt{25-16} = \sqrt{9} = 3 \). Therefore, the y-intercept is at \( y = 3 \).

Key Concepts

Half-Circle Equation

To understand the concept of a half-circle equation, it's helpful to start with the full circle equation. A circle centered at

• (4, 0)
• with a radius of 5,

can be described by the equation

• \((x-4)^2 + y^2 = 25\).

However, the exercise directs us to focus on the top half of this circle. This top half is represented by

• \(y = \sqrt{25 - (x-4)^2}\),

which describes only the positive values of y.
In practical terms, drawing this graph will yield a semicircle that sits above the x-axis. The equation includes a square root because we're only interested in the positive sqrt values (hence, the top half). This equation omits any negative y-values, effectively slicing the circle in half horizontally.

X-Intercepts

X-intercepts are the points where the curve crosses the x-axis. To find these, we set

• \(y = 0\)

in the equation and solve for x. In our half-circle equation

• \(y = \sqrt{25 - (x-4)^2}\),

setting

• \(y = 0\)

yields the equation

• \((x-4)^2 = 25\).

By solving this, we derive

• \(x-4 = \pm 5\).

This results in two solutions:

• \(x = 9\)
• and
• \(x = -1\),

representing the two x-intercepts of the half-circle. Hence, the graph intersects the x-axis at these points, defining the span of the semicircle along the x-axis.

Y-Intercept

The y-intercept is the point where the graph crosses the y-axis. To uncover this point, we set

• \(x = 0\)

in the half-circle equation. Solving

• \(y = \sqrt{25 - (0-4)^2}\)

yields

• \(y = \sqrt{25 - 16} = \sqrt{9}\).

This results in

• \(y = 3\),

determining that the y-intercept is at (0, 3). This intercept indicates at what point along the y-axis the top half of our circle reaches its peak as it extends upwards from the origin.

Graphing Techniques

Graphing a complex equation like a half-circle requires precise techniques. A graphing calculator is invaluable here, enabling you to visualize complicated functions swiftly. It's important to configure your graphing calculator correctly:

• Enter the Function: Input \(y = \sqrt{25 - (x-4)^2}\) into your calculator.
• Set the Window: Adjust your viewing window to ensure the center
• (4, 0) is visible, along with a wide enough range on both axes.

This helps capture the entire semicircle. Use the 'INTERCEPT' feature to pinpoint the exact x- and y-intercepts. Highlighting these on the graph can aid in solving the equation manually or verifying calculated values.Mastering these techniques not only clarifies the function's properties but also builds your general graphing skills. Graphing becomes an interpretive tool, allowing you to see beyond mere equations and understand the shapes mathematical concepts can form.