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Chapter 2: Problem 4

Find the \(x\) - and \(y\) -intercepts of the graph of the equation. $$ x^{2}+y^{2}=25 $$

Short Answer

Expert verified
x-intercepts: (5, 0), (-5, 0); y-intercepts: (0, 5), (0, -5)

Step by step solution

01

- Understand the Problem

To find the intercepts of the equation given, which is the equation of a circle, identify the points where the graph crosses the x-axis (y=0) and y-axis (x=0).
02

- Find the x-intercepts

Set y = 0 in the equation and solve for x: \( x^2 + 0^2 = 25 \) \( x^2 = 25 \) \( x = \ \text{{positive or negative}} 5 \). The x-intercepts are at (5, 0) and (-5, 0).
03

- Find the y-intercepts

Set x = 0 in the equation and solve for y: \( 0^2 + y^2 = 25 \) \( y^2 = 25 \) \( y = \ \text{{positive or negative}} 5 \). The y-intercepts are at (0, 5) and (0, -5).

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Key Concepts

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

x-intercepts
Finding the x-intercepts of a circle involves solving for the points where the circle crosses the x-axis. This happens when the value of y is 0. Given the equation of the circle: \[ x^{2} + y^{2} = 25 \],we set y = 0 and substitute this into the equation:\[ x^{2} + 0^{2} = 25 \]which simplifies to:\[ x^{2} = 25 \]Next, solve for x:\[ x = \text{positive or negative} \sqrt{25} \]which gives us:\[ x = 5 \text{ or } x = -5 \]Thus, the x-intercepts are at the points (5, 0) and (-5, 0). These are the coordinates where the circle touches the x-axis.
y-intercepts
Finding the y-intercepts of a circle means determining where the circle crosses the y-axis. This occurs when x is 0. Using the same circle equation: \[ x^{2} + y^{2} = 25 \],we set x = 0 and substitute this into the equation:\[ 0^{2} + y^{2} = 25 \]which simplifies to:\[ y^{2} = 25 \]Now, solve for y:\[ y = \text{positive or negative} \sqrt{25} \]This gives us:\[ y = 5 \text{ or } y = -5 \]So, the y-intercepts are at the points (0, 5) and (0, -5). These are the points where the circle touches the y-axis.
graphing circles
Graphing a circle involves plotting all the points that satisfy the equation of the circle. For the equation \( x^{2} + y^{2} = 25 \), we note that the circle is centered at the origin (0, 0) with a radius of 5. To graph this circle:
  • Start by plotting the center, which is the point (0, 0).
  • Measure a distance of 5 units in all directions: up, down, left, and right from the center.
  • Plot the points (5, 0), (-5, 0), (0, 5), and (0, -5).
  • Draw a smooth curve connecting these points to form the circle.
This will give you a visual representation of the circle described by the equation. Remember that the radius is the distance from the center to any point on the circle.
algebraic solutions
Solving algebraically for intercepts and other properties of circles involves substituting specific values and simplifying the equation. Here, we looked at how to find the intercepts:
  • For x-intercepts: Set y = 0 and solve for x.
  • For y-intercepts: Set x = 0 and solve for y.
These steps transform our equation into simpler forms that are solvable using basic algebra. Another important concept is understanding the standard form of a circle's equation: \[ (x - h)^{2} + (y - k)^{2} = r^{2} \],where (h, k) is the center, and r is the radius. This form is particularly useful for identifying key properties of the circle from the equation.

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Most popular questions from this chapter

Graph the function. $$ s(x)=\left\\{\begin{array}{ll} -x-1 & \text { for } x \leq-1 \\ \sqrt{x+1} & \text { for } x>-1 \end{array}\right. $$

Refer to the functions \(f\) and \(g\) and evaluate the functions for the given values of \(x\). \(f=\\{(2,4),(6,-1),(4,-2),(0,3),(-1,6)\\} \quad\) and \(\quad g=\\{(4,3),(0,6),(5,7),(6,0)\\}\) $$(g \circ g)(6)$$

Evaluate the step function defined by \(f(x)=[x]\) for the given values of \(x\). $$ f(-5) $$

a. Graph \(a(x)=x\) for \(x<1\). b. Graph \(b(x)=\sqrt{x-1}\) for \(x \geq 1\). c. Graph \(c(x)=\left\\{\begin{array}{ll}x & \text { for } x<1 \\ \sqrt{x-1} & \text { for } x \geq 1\end{array}\right.\)

Graph the function. $$ g(x)=\left\\{\begin{aligned} x+2 & \text { for } x<-1 \\ -x+2 & \text { for } x \geq-1 \end{aligned}\right. $$
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