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Chapter 3: Problem 18

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.) $$ f(x)=\sqrt{9-x^{2}}+c ; \quad c=-3,0,2 $$

Short Answer

Expert verified
Plot three semicircles with centers at (0,-3), (0,0), and (0,2).

Step by step solution

01

Understand the Basic Function

The function given is \( f(x) = \sqrt{9 - x^2} + c \). The term \( \sqrt{9 - x^2} \) represents a semicircle centered at the origin (0,0) with radius 3. Without any constant \( c \), the graph is a semicircle on the upper half-plane.
02

Analyze the Effect of c on the Graph

The constant \( c \) acts as a vertical shift for the semicircle. If \( c > 0 \), the semicircle shifts upwards by \( c \) units. If \( c < 0 \), the semicircle shifts downwards by \( |c| \) units. For \( c = 0 \), there is no vertical shift.
03

Draw the Graph for c = -3

For \( c = -3 \), the graph is a downwards shift of the semicircle centered at (0,0) by 3 units. Therefore, the center of the semicircle will be \((0, -3)\). The semicircle will still have a radius of 3.
04

Draw the Graph for c = 0

For \( c = 0 \), the graph remains the original semicircle with its center at the origin \((0,0)\). It is a semicircle on the upper half-plane, with a radius of 3.
05

Draw the Graph for c = 2

For \( c = 2 \), the graph shifts the semicircle upwards by 2 units. The center of the semicircle is now at \((0, 2)\), with the semicircle still having a radius of 3.
06

Combine the Graphs on One Plane

Place all three semicircles on the same coordinate plane. Each has a radius of 3 but are vertically shifted based on the value of \( c \). For \( c = -3 \), the semicircle is centered at \((0, -3)\). For \( c = 0 \), it is at the origin \((0,0)\). For \( c = 2 \), it is at \((0, 2)\). Ensure accurate plotting so that the symmetrical nature of the semicircles is clear.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Shifts
In graph transformations, vertical shifts play a significant role in moving the entire graph up or down on the coordinate plane. This movement is controlled by adding or subtracting a constant value from the function. In the expression given, namely \(f(x) = \sqrt{9 - x^2} + c\), the constant \(c\) determines how much and in which direction the graph is shifted vertically.
  • If \(c > 0\), every point on the semicircle will move up by \(c\) units. This is an upward shift.
  • For \(c < 0\), the entire graph shifts down, with each point moving \(|c|\) units lower.
  • When \(c = 0\), no vertical shift occurs, and the graph remains in its original position.
Understanding vertical shifts helps in predicting and visually displaying how graphs change when adjusted by different constant values.
Semicircle Graph
Graphs of semicircles are frequently encountered in mathematical problems involving geometric shapes. A semicircle graph represents half of a circle and is defined by equations involving square roots, as seen in \(f(x) = \sqrt{9-x^{2}}\).
  • This particular equation denotes a semicircle because it features the square root of an expression similar to the equation of a circle. The square root limits the graph to the upper half-plane.
  • The center of the semicircle is at the origin, (0, 0), before any transformation is applied.
  • All points on this graph are equidistant from the origin, forming a perfect half-dome shape that stops at the x-axis.
Such graphs are symmetrically balanced along the y-axis unless influenced by additional transformations like vertical shifts.
Radius of a Semicircle
The radius in a circle or semicircle remains a crucial measurement that determines the size of the graph. For the function \(f(x)=\sqrt{9-x^2}\), the number under the square root, 9, directly implies a radius of 3.
  • The reason is that the term \(9-x^2\) resembles the format \(r^2\) of a circle's equation \(x^2 + y^2 = r^2\). Here \(r = 3\) since \(\sqrt{9} = 3\).
  • The radius determines the maximum distance any point on the semicircle can have from its center. In this case, this radius ensures that the graph arch reaches x-coordinates -3 to 3.
  • Consistent radius across all transformation maintains the general size and shape of semicircles, regardless of shifting.
A clear understanding of radius aids in accurate graphing and recognizing the impact of different graph transformations.
Coordinate Plane Plotting
Plotting on a coordinate plane requires an understanding of both horizontal and vertical axes. To plot a function like \(f(x) = \sqrt{9-x^2}+c\), follow these steps:
  • Identify the center of the semicircle, which initially is at \(0, 0\).
  • Considering vertical shifts: move this center up or down based on the value of \(c\).
  • Plot the endpoints of the semicircle along the x-axis, ensuring they respect the radius. For a radius of 3, these endpoints would be at \(-3, 3\).
  • Draw the semicircle forming an arc from the left endpoint to the right, ensuring symmetry.
Accurate plotting on coordinate planes enables visualization of the function's transformation and the impact evident on the graph's shape.

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Most popular questions from this chapter

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.) $$ f(x)=c x^{3} ; \quad c=-\frac{1}{3}, 1,2 $$

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\sqrt{3-x}, \quad g(x)=\sqrt{x^{2}-16} $$

Exer. 53-54: The symbol \(\llbracket x \rrbracket\) denotes values of the greatest integer function. Sketch the graph of \(f\). (a) \(f(x)=\llbracket x-3 \rrbracket\) (b) \(f(x)=\llbracket x \rrbracket-3\) (c) \(f(x)=2 \llbracket x \rrbracket\) (d) \(f(x)=\llbracket 2 x \rrbracket\) (e) \(f(x)=\llbracket-x \rrbracket\)

Exer. 27-32: If the point \(P\) is on the graph of a function \(f\), find the corresponding point on the graph of the given function. $$ P(3,-1) ; \quad y=2 f(x)+4 $$

Let \(y=f(x)\) be a function with domain \(D=[-6,-2]\) and range \(R=[-10,-4]\). Find the domain \(D\) and range \(R\) for each function. (a) \(y=\frac{1}{2} f(x)\) (b) \(y=f(2 x)\) (c) \(y=f(x-2)+5\) (d) \(y=f(x+4)-1\) (e) \(y=f(-x)\) (f) \(y=-f(x)\) (g) \(y=f(|x|)\) (h) \(y=|f(x)|\)
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