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Chapter 1: Problem 20

In each part, graph the equation by solving for \(y\) in terms of \(x\) and graphing the resulting functions together. (a) \(x^{2} / 4+y^{2} / 9=1\) (b) \(y^{2}-x^{2}=1\)

Short Answer

Expert verified
(a) The graph is an ellipse centered at (0,0). (b) The graph is a hyperbola with vertices at (0,1) and (0,-1).

Step by step solution

01

Solve Equation (a) for y

Equation (a) is \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). First, solve for \( y^2 \) by isolating it: \[ \frac{y^2}{9} = 1 - \frac{x^2}{4} \] Multiply through by 9 to get: \[ y^2 = 9 \left( 1 - \frac{x^2}{4} \right) = 9 - \frac{9x^2}{4} \] Finally, take the square root of both sides: \[ y = \pm \sqrt{9 - \frac{9x^2}{4}} \]
02

Graph the Resulting Functions from (a)

The equation \( y = \pm \sqrt{9 - \frac{9x^2}{4}} \) describes an ellipse. Use a graphing calculator or software to plot this function. You will get an ellipse centered at the origin with a semi-minor axis along the x-axis of 2, and a semi-major axis along the y-axis of 3.
03

Solve Equation (b) for y

Equation (b) is \( y^2 - x^2 = 1 \). Rearrange this to solve for \( y^2 \): \[ y^2 = x^2 + 1 \] Then, take the square root of both sides to express \( y \) in terms of \( x \): \[ y = \pm \sqrt{x^2 + 1} \]
04

Graph the Resulting Functions from (b)

The equations \( y = \pm \sqrt{x^2 + 1} \) describe a hyperbola. When graphed, you'll see two separate curves on a graph. These curves open upwards and downwards with vertices at (0, 1) and (0, -1) and no real asymptotes intersecting them.

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Key Concepts

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

Ellipses
An ellipse is a geometric shape that appears as an elongated circle or an oval. It is defined by its equation in standard form: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively. These axes intersect at the center of the ellipse.

To graph an ellipse:
  • Determine the values of \( a \) and \( b \) to find the lengths of the axes.
  • The semi-major axis is the longest diameter, whereas the semi-minor is the shortest.
  • The center of the ellipse typically lies at the origin (0,0) unless shifted by additional terms.
  • Plot the points at the ends of the axes and sketch the curve between them.
In exercise (a), with \( x^2/4 + y^2/9 = 1 \), the ellipse has a semi-major axis of 3 along the y-axis and a semi-minor axis of 2 along the x-axis. This means the ellipse is taller than it is wide.
Hyperbolas
A hyperbola is another type of conic section that resembles two opposite-facing curves or 'branches.' These are separated and symmetrical around the coordinate axes. The standard form of a hyperbola is \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), which primarily opens vertically, or \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), which opens horizontally.

Key features of hyperbolas:
  • Two separate curves instead of a closed curve.
  • Asymptotes that the curves approach but never touch.
  • The center is where the asymptotes intersect, often at the origin.
  • Vertices are located at specific points which determine the opening direction of the curves.
In the exercise, \( y^2 - x^2 = 1 \) describes a hyperbola that opens vertically. Its vertices are located at points (0, 1) and (0, -1), with no real asymptotes explicitly given by the equation, emphasizing its opening nature.
Solving for y
Solving an equation for \( y \) involves expressing \( y \) explicitly in terms of \( x \). This is often needed to graph equations using functions that depend on \( x \).

Here鈥檚 a simple process:
  • First, isolate the term involving \( y^2 \) on one side of the equation.
  • Subtract or add terms as necessary to obtain a clear expression for \( y^2 \).
  • To solve for \( y \), take the square root of both sides, remembering it results in two curves: \( y = \pm \sqrt{\text{expression}} \).
For equation (a), by isolating \( y^2 \) as \( 9 - \frac{9x^2}{4} \), and taking the square root, \( y \) turns into the function \( \pm \sqrt{9 - \frac{9x^2}{4}} \). Similarly, equation (b) simplifies to \( y = \pm \sqrt{x^2 + 1} \). This makes graphing these functions straightforward using available graphing tools.

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Most popular questions from this chapter

Confirm that a linear model is appropriate for the relationship between \(x\) and \(y .\) Find a linear equation relating \(x\) and \(y,\) and verify that the data points lie on the graph of your equation. $$\begin{array}{|c|c|c|c|c|c|}\hline x & 0 & 1 & 2 & 4 & 6 \\\\\hline y & 2 & 3.2 & 4.4 & 6.8 & 9.2 \\\\\hline\end{array}$$

In Exercises \(17-20 .\) sketch the graph of the equation by translating, reflecting, compressing, and stretching the graph of \(y=1 / x\) appropriately, and then use a graphing utility to confirm that your sketch is correct. $$y=\frac{1}{x-3}$$

Based on your knowledge of the graphs of \(y=x\) and \(y=\sin x,\) make a sketch of the graph of \(y=x \sin x .\) Check your conclusion using a graphing utility.

Graph the equation using a graphing utility. (a) \(x=y^{2}+2 y+1\) (b) \(x=\sin y,-2 \pi \leq y \leq 2 \pi\)

Find formulas for \(f+g . f-g . f g\), and \(f / g,\) and state the domains of the functions. $$f(x)=3 x-2, g(x)=|x|$$
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