Problem 24 - Calculus with Concepts in Calculus

Question

Sketch the graph. List the intercepts and describe the symmetry (if any) of the graph. $$ y=\sqrt{25-x^{2}} $$

Step-by-Step Solution

Answer

The graph is the upper semicircle of a circle with radius 5. It has intercepts at (0, 5), (-5, 0), and (5, 0), and is symmetric about the y-axis.

Step 1: Recognize the Graph Type

The equation \( y = \sqrt{25 - x^2} \) resembles the upper half of a circle. This is because \( y^2 = 25 - x^2 \) can be rearranged to \( x^2 + y^2 = 25 \), which is the equation of a circle with radius 5 centered at the origin \((0,0)\). Since \( y \) is expressed as the square root, we only consider the non-negative values of \( y \).

Step 2: Determine the Function Domain

The term inside the square root, \( 25 - x^2 \), must be non-negative. Therefore, solve \( 25 - x^2 \geq 0 \) leading to \( -5 \leq x \leq 5 \). This means the domain of \( x \) for the function \( y = \sqrt{25 - x^2} \) is \([-5, 5]\).

Step 3: Find the Intercepts

To find the y-intercept, set \( x = 0 \), giving \( y = \sqrt{25} = 5 \). Thus, the y-intercept is at \((0, 5)\). For the x-intercepts, set \( y = 0 \), yielding \( \sqrt{25 - x^2} = 0 \). This gives \( 25 = x^2 \), so \( x = \pm 5 \). The x-intercepts are \((-5, 0)\) and \((5, 0)\).

Step 4: Analyze Graph Symmetry

A graph is symmetric with respect to the y-axis if replacing \( x \) with \( -x \) gives the same equation. Substitute \( -x \) for \( x \) in \( y = \sqrt{25 - x^2} \), which results in the same equation. Therefore, the graph is symmetric with respect to the y-axis.

Step 5: Sketch the Graph

Sketch the portion of the circle with radius 5 centered at the origin. Only draw the top half due to the square root, covering from \( x = -5 \) to \( x = 5 \) along the x-axis. The highest point on the graph is \( (0, 5) \), and the graph intercepts the x-axis at \((-5, 0)\) and \((5, 0)\).

Key Concepts

Intercepts

Intercepts are points where the graph of a function crosses the x-axis or y-axis. In simpler terms, these are the points where the value of one of the variables becomes zero. For the function \[ y = \sqrt{25 - x^2} \]we find both x-intercepts and the y-intercept.

• **Y-Intercept**: Set \( x = 0 \). Plugging it into the equation, we get \( y = \sqrt{25} = 5 \). So, the y-intercept is at the point \( (0, 5) \).
• **X-Intercepts**: Set \( y = 0 \). Then \( \sqrt{25 - x^2} = 0 \), leading us to \( 25 = x^2 \). Solving this gives \( x = \pm 5 \), so the x-intercepts are at the points \( (-5, 0) \) and \( (5, 0) \).

In summary, intercepts tell us crucial points where the graph crosses axes, providing a useful insight into the graph's geometry.

Symmetry

Symmetry in a graph can help simplify graphing and analyzing mathematical functions. Specifically, a graph can exhibit symmetry with respect to the x-axis, y-axis, or the origin.In the case of the function\[ y = \sqrt{25 - x^2} \]we particularly focus on y-axis symmetry. This occurs if replacing \( x \) with \( -x \) results in the same equation. When we substitute \(-x\) for \(x\) in the equation, the expression remains unchanged.\[ y = \sqrt{25 - (-x)^2} = \sqrt{25 - x^2} \]Thus, this function is symmetric about the y-axis. Recognizing this symmetry allows us to only compute part of the graph, knowing that the other part mirrors the first about the y-axis. This fact significantly eases the sketching process and provides deeper geometric insights.

Equation of a Circle

The equation of a circle in the standard form is given by:\[ x^2 + y^2 = r^2 \]where \( r \) is the radius of the circle. The center is typically at the origin, \((0, 0)\) unless otherwise stated.For the function\[ y = \sqrt{25 - x^2} \]we can rearrange the terms to the standard circle equation:\[ x^2 + y^2 = 25 \]This is a circle with a radius of 5, centered at the origin. However, due to the square root in the function definition, we only consider the top half of the circle. This constraint arises because square roots yield non-negative results. Hence, the graph represents the semicircle that lies above the x-axis. This understanding of the circle’s equation aids in recognizing and visualizing its geometric structure.

Domain and Range

The domain and range of a function give us the possible values for its input and output, respectively. Specifically, for the function \[ y = \sqrt{25 - x^2} \]we derive the domain from the condition inside the square root being non-negative:\[ 25 - x^2 \geq 0 \]Solving this inequality, we find that \( -5 \leq x \leq 5 \). This means the domain, or all possible x-values, is \([-5, 5]\).Next, the range is determined by the possible y-values. Since we are dealing with a semicircle, the minimum y-value is 0 (when \( x = \pm 5 \)), and the maximum value is 5 (when \( x = 0 \)). Thus, the range is \([0, 5]\).Understanding these constraints helps define the specific portion of a graph you will plot. It's essential for fully capturing the behavior of the function on the graph.