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Chapter 1: Problem 10

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$x=y^{2}-1$$

Short Answer

Expert verified
The graph has x-intercept \((-1, 0)\), y-intercepts \((0, 1)\) and \((0, -1)\), with symmetry about the x-axis.

Step by step solution

01

Identify the equation format

The given equation is in the form of \( x = y^2 - 1 \), which represents a parabola opening horizontally.
02

Find the y-intercept

The y-intercept is found by setting \( x = 0 \) and solving for \( y \). The equation becomes \( 0 = y^2 - 1 \), which leads to \( y^2 = 1 \). Solving for \( y \), we get \( y = \pm 1 \). Thus, the y-intercepts are \((0, 1)\) and \((0, -1)\).
03

Find the x-intercept

The x-intercept is found by setting \( y = 0 \) and solving for \( x \). Substituting \( y = 0 \) into the equation, we get \( x = 0^2 - 1 = -1 \). Thus, the x-intercept is \((-1, 0)\).
04

Determine symmetry

To check symmetry about the x-axis, replace \( y \) with \( -y \), resulting in \( x = (-y)^2 - 1 = y^2 - 1 \), which is the same equation, indicating symmetry about the x-axis. To check symmetry about the y-axis, replace \( x \) with \( -x \), resulting in \( -x = y^2 - 1 \), which is not the same equation, indicating no symmetry about the y-axis. For symmetry about the origin, replace \( x \) with \( -x \) and \( y \) with \( -y \), leading to \( -x = (-y)^2 - 1 = y^2 - 1 \), which is also not the same as the original, indicating no symmetry about the origin.
05

Draw the graph

Using the intercepts, plot the points \((0, 1), (0, -1)\), and \((-1, 0)\) on the coordinate system. Since the graph is a parabola opening sideways, sketch the parabola opening to the right from the vertex at \((-1, 0)\). The parabolic shape should pass through the y-intercepts and exhibit symmetry about the x-axis.

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Key Concepts

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

x-intercepts
The x-intercepts of a graph are the points where the graph crosses the x-axis. This means that at these points, the y-value is zero. To find the x-intercepts:
  • Set the y variable to zero in the equation.
  • Solve the resulting equation for the x values.
In the provided exercise, we set the y value to zero in the equation: \[ x = y^2 - 1 \] This simplifies to: \[ x = 0^2 - 1, \] resulting in \[ x = -1. \] Therefore, the x-intercept for this parabola is located at the point \((-1, 0)\). This point indicates where the curve intersects the x-axis, crucial for sketching the graph.
y-intercepts
The y-intercepts of a graph are where the graph crosses the y-axis. At these points, the x-value is always zero. To find the y-intercepts:
  • Set the x variable to zero.
  • Solve the equation for y-values.
In our equation, \[ x = y^2 - 1, \] setting \( x = 0 \) gives:\[ 0 = y^2 - 1, \]which simplifies to \[ y^2 = 1. \] Solving for y, we find \( y = 1 \) and \( y = -1 \). Thus, the y-intercepts are the points \((0,1)\) and \((0,-1)\). These points highlight where the graph intersects the y-axis, and you can imagine them as the starting and stopping points when plotting.
symmetry
Symmetry in a graph refers to how a shape can be mirrored or replicated across an axis or a point. The three main types of symmetry in graphing are:
  • Symmetry about the x-axis
  • Symmetry about the y-axis
  • Symmetry about the origin
To check for x-axis symmetry, replace y with \(-y\) in the equation. If the equation remains unchanged, the graph is symmetric about the x-axis. In our case: \[ x = (-y)^2 - 1 = y^2 - 1, \] indicating x-axis symmetry. For y-axis symmetry, replace x with \(-x\). If the equation still holds, then there is y-axis symmetry. Attempting this, the equation\(-x = y^2 - 1 \) differs from the original, so no y-axis symmetry exists here. Finally, origin symmetry requires both variables to be replaced: \(-x = (-y)^2 - 1 = y^2 - 1, \) which isn't equivalent to the original. Thus, the graph is not symmetric about the origin.
parabolas
Parabolas are U-shaped curves on a graph, typically represented by quadratic equations like \( y = ax^2 + bx + c \). However, the exercise given is formatted differently: \( x = y^2 - 1 \), indicating a parabola that opens sideways rather than upward or downward. This is an important distinction. For sideways-opening parabolas like this one, the vertex forms a turning point of the curve. Here, the vertex is at \((-1, 0)\). This vertex allows the parabola to extend rightward from it, creating a mirror image along the x-axis. Understanding the orientation of your parabola is vital when graphing; for instance, with this equation, the sideways opening results in the graph appearing horizontally symmetric. So, remember:
  • The direction of the opening depends on the variable with the squared term.
  • The vertex acts as the focal point from which the graph extends.
  • This distinctive formation defines the parabola's shape.
These characteristics help us accurately draw and interpret the graph.

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Most popular questions from this chapter

Find the standard equation of the circle tangent to the \(y\) -axis and with center (3,5)

Rewrite each statement using absolute value notation, as in Example 5.The distance between \(x\) and 1 is \(1 / 2\).

Rewrite each statement using absolute value notation, as in Example 5. The distance between \(x\) and 1 is at least \(1 / 2\).

Find an equation for the line that is described, and sketch the graph. Write the final answer in the form \(y=m x+b ;\) (a) Passes through (-7,-2) and (0,0) (b) Passes through (6,-3) and has \(y\) -intercept 8 (c) Passes through (0,-1) and has the same slope as the line \(3 x+4 y=12\) (d) Passes through (6,2) and has the same \(x\) -intercept as the line \(-2 x+y=1\) (e) Has \(x\) -intercept -6 and \(y\) -intercept \(\sqrt{2}\)

This exercise outlines a proof of the fact that two nonvertical lines with slopes \(m_{1}\) and \(m_{2}\) are perpendicular if and only if \(m_{1} m_{2}=-1 .\) In the following figure, we've assumed that our two nonvertical lines \(y=m_{1} x\) and \(y=m_{2} x\) intersect at the origin. [If they did not intersect there, we could just as well work with lines parallel to these that do intersect at \((0,0),\) recalling that parallel lines have the same slope.] The proof relies on the following geometric fact: \(\overline{O A} \perp \overline{O B} \quad\) if and only if \(\quad(O A)^{2}+(O B)^{2}=(A B)^{2}\). (a) Verify that the coordinates of \(A\) and \(B\) are \(A\left(1, m_{1}\right)\) and \(B\left(1, m_{2}\right)\) (b) Show that $$\begin{aligned}O A^{2} &=1+m_{1}^{2} \\\O B^{2} &=1+m_{2}^{2} \\\A B^{2} &=m_{1}^{2}-2 m_{1} m_{2}+m_{2}^{2} \end{aligned}$$ (c) Use part (b) to show that the equation $$O A^{2}+O B^{2}=A B^{2}$$ is equivalent to \(m_{1} m_{2}=-1\) (GRAPH CAN'T COPY)
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