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Chapter 1: Problem 43

Find the \(x\) - and \(y\) -intercepts of the graph of the equation. $$(x-2)^{2}+y^{2}=9$$

Short Answer

Expert verified
Answer: The x-intercepts are (5, 0) and (-1, 0), and the y-intercepts are (0, 鈭�5) and (0, -鈭�5).

Step by step solution

01

Find the x-intercepts

To find the x-intercepts, set \(y\) to \(0\) and solve for \(x\). $$(x-2)^{2}+0^{2}=9$$ Simplify and solve for \(x\).
02

Solve the equation for x

Expanding and solving the equation from step 1: $$(x-2)^{2}=9$$ $$x-2=\pm\sqrt{9}$$ $$x-2=\pm3$$ $$x=2\pm3$$ So, we have two x-intercepts, \(x=5\) and \(x=-1\).
03

Find the y-intercepts

To find the y-intercepts, set \(x\) to \(0\) and solve for \(y\). $$(0-2)^{2}+y^{2}=9$$ Simplify and solve for \(y\).
04

Solve the equation for y

Expanding and solving the equation from step 3: $$(-2)^{2}+y^{2}=9$$ $$4+y^{2}=9$$ $$y^{2}=5$$ $$y=\pm\sqrt{5}$$ So, we have two y-intercepts, \(y=\sqrt{5}\) and \(y=-\sqrt{5}\).
05

Write the coordinates of the x- and y-intercepts

Now, we can write down the coordinates of the x- and y-intercepts: x-intercepts: \((5, 0)\) and \((-1, 0)\) y-intercepts: \((0, \sqrt{5})\) and \((0, -\sqrt{5})\)

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Key Concepts

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

Understanding X-Intercepts
The x-intercept of a graph is where the graph crosses the x-axis. At this point, the y-coordinate is always 0. To find the x-intercepts of an equation, we set the y-value to 0 and then solve the equation for x.
For the equation \[(x-2)^2 + y^2 = 9,\]we substitute y with 0, simplifying to \[(x-2)^2 = 9.\]Once simplified, solve for x by taking the square root of both sides, giving:
  • \(x - 2 = 3\), which gives \(x = 5\)
  • \(x - 2 = -3\), which gives \(x = -1\)
Thus, the x-intercepts are the points where the graph intersects the x-axis: \((5, 0)\) and \((-1, 0)\).
Determining Y-Intercepts
To find a graph's y-intercepts, identify where it crosses the y-axis, which means x is 0 at these points. For the given equation, \[(x-2)^2 + y^2 = 9,\]set x to 0 and solve for y.
Substitute to form: \[(0-2)^2 + y^2 = 9,\] which simplifies and is solved for y by:
  • Calculating \((-2)^2 =4\), converting the equation to \(4 + y^2 = 9\).
  • Solving for \(y^2\) gives \(y^2 = 5\).
  • Thus, \(y = \pm \sqrt{5}\).
This calculation shows the y-intercepts are \((0, \sqrt{5})\) and \((0, -\sqrt{5})\).
Mastering Equation Solving
Solving equations is a critical skill in finding intercepts on a graph. Here, solving \((x-2)^2 = 9\)required taking square roots. However, understanding each step:
  • Simplifies complex equations into simpler forms like \( (x-2)^2\).
  • Identifies both positive and negative solutions (\(x - 2 = \pm 3\)).
  • Subsequently leads to x-values \(5\) and \(-1\).
Through consistent practice, reading equations like narratives becomes easy, thereby leading to quick and error-free intercept calculations.
Exploring Coordinate Geometry
Coordinate geometry is the study of geometry using a coordinate system, allowing geometrical problems to transform into algebraic equations. In this context, equations like \((x-2)^2 + y^2 = 9\)represent circles centered at (2,0) with a radius of 3.
This exercise involves placing theoretical points on the coordinate plane:
  • x-intercepts like \((5, 0)\) and\((-1, 0)\) show horizontal crossings of the circle from its center (2,0).
  • y-intercepts such as \((0, \sqrt{5})\) and \((0, -\sqrt{5})\) indicate vertical touches, balanced along the y-axis.
Thus, visualizing intercepts provides a clear reflection of the circle's position and scope on the coordinate grid.

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Most popular questions from this chapter

Determine whether the line through \(P\) and \(Q\) is parallel or perpendicular to the line through \(P=(0,3 / 2), Q=(1,1)\) and \(R=(2,7), S=(3,9)\)

One diagonal of a square has endpoints (-3,1) and \((2,-4) .\) Find the endpoints of the other diagonal.

Sketch the graph of the equation. Label the \(x\) - and y-intercepts. $$(x+1)^{2}+(y-3)^{2}=9$$

Simplify the expression without using a calculator. Your answer should not have any radicals in it. \(\sqrt{3 x} \sqrt{75 x^{3}}(x \geq 0)\)

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