Problem 36
Question
Sketch the graph of the equation. In each case determine whether the graph is that of a function. $$ x^{2}+y^{2}=9 \text { for } x \geq 0 $$
Step-by-Step Solution
Verified Answer
The graph depicts a semicircle (radius 3, centered at the origin) on the right side, which is not a function.
1Step 1: Identify the shape of the graph
The equation given is \(x^2 + y^2 = 9\). This is the equation of a circle centered at the origin with a radius of 3.
2Step 2: Apply the constraint for \(x\)
The constraint \(x \geq 0\) indicates that only the right half of the circle (including the boundary along the y-axis) should be considered. Thus, the circle is only plotted for \(x \geq 0\).
3Step 3: Sketch the graph
To sketch the graph, draw a semicircle of radius 3 centered at the origin that spans the positive x-axis and both the positive and negative y-axes. This semicircle represents the portion of the circle for \(x \geq 0\).
4Step 4: Determine if it is a function
A graph represents a function if each \(x\) value corresponds to exactly one \(y\) value. However, for this semicircle, there are values of \(x\) where both positive and negative \(y\) values satisfy the equation. Thus, this graph is not a function.
Key Concepts
Graphing CirclesFunctions and RelationsEquation of a CircleDetermining Functions from Graphs
Graphing Circles
When it comes to graphing circles, the process starts with understanding the equation that represents the circle. The standard form of a circle's equation is \[(x - h)^2 + (y - k)^2 = r^2 \]where
However, given the constraint \( x \geq 0 \), we only plot the right half of this circle, which is called a semicircle.
- \( (h, k) \) is the center of the circle
- \( r \) is the radius of the circle.
However, given the constraint \( x \geq 0 \), we only plot the right half of this circle, which is called a semicircle.
Functions and Relations
To better grasp functions and relations, it's important to see the key difference:
Thus, while the equation suggests a relation, it does not fulfill the criteria to be a function.
- A function is a special type of relation where each input (usually represented by \( x \)) maps to exactly one output (\( y \)).
- A relation is more general - it's a set of ordered pairs without the restriction of mapping one input to just one output.
Thus, while the equation suggests a relation, it does not fulfill the criteria to be a function.
Equation of a Circle
Understanding the equation of a circle is essential when engaging in problems involving the geometric figure. The basic form is \[(x - h)^2 + (y - k)^2 = r^2\]and it precisely defines a circle in terms of its center and radius.
Such equations are instrumental in drawing circles graphically, furnishing a clear method to map points equidistant from the center point.
- Here, \( (h, k) \) indicates the center's coordinates.
- \( r \) is the radius.
Such equations are instrumental in drawing circles graphically, furnishing a clear method to map points equidistant from the center point.
Determining Functions from Graphs
Determining whether a graph represents a function involves a handy test called the vertical line test. This test helps quickly assess the function status:
Such an evaluation is crucial as it informs us about how inputs relate to outputs and the uniqueness of this correspondence in mathematical functions.
- If any vertical line drawn through the graph intersects it at more than one point, the graph is not a function.
- If a vertical line only crosses the graph once no matter where it's drawn, then it represents a valid function.
Such an evaluation is crucial as it informs us about how inputs relate to outputs and the uniqueness of this correspondence in mathematical functions.
Other exercises in this chapter
Problem 36
Solve the inequality. $$ \frac{t^{2}-2 t-3}{t^{2}-8 t+15}>0 $$
View solution Problem 36
Find the domain of the function. $$ f(x)=\sqrt{4-\sqrt{1+9 x^{2}}} $$
View solution Problem 36
Find \(g\) if \(f(x)=\left(x^{2}-4\right) /(x+3)\) and \((f g)(x)=1\), for \(x \neq 2\), \(-2\), and \(-3 .\)
View solution Problem 37
Show that \(\ln \left(x+\sqrt{x^{2}-1}\right)=-\ln \left(x-\sqrt{x^{2}-1}\right)\) for \(x \geq 1\).
View solution