Problem 26 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=-\sqrt{16-(c x)^{2}} ; \quad c=1, \frac{1}{2}, 4 $$

Step-by-Step Solution

Answer

Draw the semicircles centered at origin with endpoints: (-4,4), (-8,8), (-1,1).

Step 1: Understand the basic structure of the function

The given function is \( f(x) = -\sqrt{16-(c x)^2} \). This is a transformation of the basic function \(-\sqrt{16-x^2}\), which represents the lower semicircle with radius 4 centered at the origin, but only reflected over the x-axis because of the negative sign.

Step 2: Setup for graphing

For each value of \(c\), the expression \((c x)^2\) changes. Specifically, different \(c\) values result in compressing or stretching the graph horizontally. For instance, a larger \(c\) compresses the graph, and a smaller \(c\) stretches it.

Step 3: Graph for c = 1

Substituting \(c = 1\) into the function, we get \( f(x) = -\sqrt{16-x^2} \). This is the original semicircle reflecting downward with endpoints at \((-4,0)\) and \((4,0)\).

Step 4: Graph for c = \(\frac{1}{2}\)

When \(c = \frac{1}{2}\), the function becomes \( f(x) = -\sqrt{16-(\frac{1}{2}x)^2} \). This translates to \( f(x) = -\sqrt{16-\frac{1}{4}x^2} \). It is a horizontally stretched version of the semicircle with endpoints at \((-8,0)\) and \((8,0)\).

Step 5: Graph for c = 4

For \(c = 4\), the equation becomes \( f(x) = -\sqrt{16-(4x)^2} \), or \( f(x) = -\sqrt{16-16x^2} \). This is a compressed graph horizontally with endpoints at \((-1,0)\) and \((1,0)\).

Step 6: Combine the graphs

Draw all three functions \( f(x) = -\sqrt{16-x^2} \), \( f(x) = -\sqrt{16-\frac{1}{4}x^2} \), and \( f(x) = -\sqrt{16-16x^2} \). They will all start from the origin but differ in how wide they are across the x-axis, depending on the compression/stretch due to \(c\).

Key Concepts

Transformation of Functions

Understanding the transformation of functions is key when sketching graphs. Each function can be altered in various ways to transform its graph on the coordinate plane. In the exercise, you are given a base function,

• Transformation of functions involves changes such as shifting, stretching, reflecting, or compressing.
• Function transformations make use of parameters that scale, shift, or reflect the original graph.
• In our exercise, the value of the parameter \(c\) in the expression \((c x)^2\) directly influences the horizontal transformations.

By altering this parameter, we observe different transformations like horizontal stretching, horizontal compression, and reflection over the axis, each facilitating a unique understanding of how equations represent different graphical forms.

Semicircle Graph

A semicircle graph is derived from a complete circle but represents only half of it. The function \( f(x) = -\sqrt{16 - x^2} \) is a perfect example of a semicircle graph, specifically the lower half.

• The given semicircle has a radius of 4 and is centered at the origin.
• The range of values for \(x\) is restricted within \([-4, 4]\), confirming its semicircular nature.
• The negative sign before the square root reflects the semicircle below the x-axis.

In visualizing this, always remember that each point \((x, y)\) on the graph corresponds to the \(x\) value plugged into the equation, showcasing a quarter cycle from end to end. Thus, understanding these basic properties ensures you can correctly graph and interpret semicircle functions.

Function Reflection

Reflection in function graphing involves flipping the output of the function over a particular axis. In our exercise, the function \( f(x) = -\sqrt{16 - (c x)^2} \) represents the reflection of a semicircular function over the x-axis, due to the negative sign.

• Reflection over the x-axis changes each \(y\) value of the usual "upper" semicircle to its negative counterpart, effectively placing the graph below the x-axis.
• For example, where normally \(\sqrt{16-x^2}\) would draw a semicircle in the upper half for \(|x| \leq 4\), this reflection mirrors it down.
• The concept of reflection helps us understand symmetry in graphs and unlocks more options for analysis and graphing techniques.

Reflecting functions essentially maintains the same shape, but reorients it with a shift in direction making it a vital tool for graph transformations.

Horizontal Compression and Stretching

Horizontal compression and stretching occur when a function is transformed by varying parameters that affect its width on the graph. In our exercise, the parameter \(c\) is used to influence these changes.

• For \(c = 1\), the graph retains its standard width, ranging from \(-4\) to \(4\).
• If \(c < 1\), as with \(c = \frac{1}{2}\), the function experiences horizontal stretching, expanding its reach across the x-axis. The semicircle then spans from \(-8\) to \(8\).
• Conversely, \(c > 1\), as when \(c = 4\), causes horizontal compression, narrowing the semicircle to \([-1, 1]\).

These transformations help us see how equations can affect shapes in terms of domain and visual representation on a plane. By considering such manipulations, graphs can be adapted to express diverse real-world scenarios and other mathematical problems.