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Chapter 1: Problem 46

Sketch the line whose Cartesian equation is given. $$ x / 3+y / 9=1 $$

Short Answer

Expert verified
Sketch the line using y-intercept (0,9) and x-intercept (3,0).

Step by step solution

01

Understand the Equation

The equation given is \( \frac{x}{3} + \frac{y}{9} = 1 \). This is in the standard form of a line. Our goal is to sketch the graph of this line.
02

Convert to Slope-Intercept Form

To make it easier to sketch, convert the equation to slope-intercept form \( y = mx + b \). Multiply each term by 9 to eliminate the fractions: \( 3x + y = 9 \). Solve for \( y \): \( y = -3x + 9 \). Here, the slope \( m \) is \(-3\) and the y-intercept \( b \) is \(9\).
03

Identify the Y-Intercept

The y-intercept is where the line crosses the y-axis. From \( y = -3x + 9 \), the y-intercept \( b \) is \( 9 \). So, the point \( (0,9) \) is on the graph.
04

Find the X-Intercept

The x-intercept is where the line crosses the x-axis. Set \( y = 0 \) in the equation \( 3x + y = 9 \) to find \( x \): \( 3x = 9 \rightarrow x = 3 \). The x-intercept is \( (3,0) \).
05

Plot the Intercepts

Plot the points \((0,9)\) and \((3,0)\) on the Cartesian plane.
06

Draw the Line

Using a ruler, draw a straight line through the points \((0,9)\) and \((3,0)\) which extends infinitely in both directions. This represents the line described by the equation \( \frac{x}{3} + \frac{y}{9} = 1 \).

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form of a linear equation is a powerful and insightful way to understand and graph lines. It is typically expressed as \( y = mx + b \), where \( m \) represents the slope and \( b \) is the y-intercept. In this case, we took the original equation \( \frac{x}{3} + \frac{y}{9} = 1 \) and rearranged it to obtain \( y = -3x + 9 \). This equation is now in slope-intercept form, making it easier to identify important details about the line.

One of the main uses of the slope-intercept form is to quickly identify the slope and y-intercept:
  • Slope \( (m) \): Indicates the steepness and direction of the line. If it is positive, the line rises as it moves from left to right. If it is negative, like in our example \(-3\), the line falls as it moves from left to right.
  • Y-Intercept \( (b) \): The point where the line crosses the y-axis, which we'll dive into next.
By transforming equations into this form, you make graphing lines straightforward and intuitive.
Y-Intercept
The y-intercept of a line is a crucial concept when graphing linear equations. It is where the line crosses the y-axis. This occurs when the value of \( x \) equals zero. In slope-intercept form, \( b \) directly provides this value.

For our equation after conversion, \( y = -3x + 9 \), the y-intercept \( b \) is \( 9 \). This tells us that the line crosses the y-axis at the point \( (0,9) \).

Identifying the y-intercept is beneficial because:
  • It provides a starting point for plotting the line on a graph.
  • It gives immediate insight into the behavior of the line, especially in relation to the slope.
Always remember that the y-intercept represents a point on the graph where the entire influence of the slope hasn't yet been applied, as it is the original launch point of the line's journey across the plane.
X-Intercept
The x-intercept is another key feature of a linear graph. It is where the line crosses the x-axis, meaning the value of \( y \) is zero at this point. Finding it involves solving for \( x \) when the equation equals zero for \( y \).

For our line, the original rearranged equation was \( 3x + y = 9 \). Setting \( y = 0 \) gives us \( 3x = 9 \). Solving for \( x \), we find \( x = 3 \). Thus, the x-intercept is at the point \( (3,0) \).

This intercept helps in graph creation and understanding the equation because:
  • It offers another clear reference point to plot the line.
  • It significantly aids in checking the accuracy of the line once graphed, as both intercepts work as benchmarks.
The x-intercept, coupled with the y-intercept, completes the basic reference points needed to sketch any linear equation.
Graphing a Line
Graphing a line based on its equation is a straightforward process when utilizing both the slope-intercept form and intercepts. Once you've determined the y-intercept \( (0,9) \) and the x-intercept \( (3,0) \), plotting these on a Cartesian plane becomes the initial step.

Here's a simple approach to graphing the line for \( \frac{x}{3} + \frac{y}{9} = 1 \):
  • Start by marking the y-intercept on the graph, plotting the point \( (0,9) \).
  • Next, mark the x-intercept, \( (3,0) \), on the x-axis.
  • With both intercepts plotted, use a ruler to draw a straight line passing through these points. Extend this line evenly in both directions, representing its infinite nature.
By following these steps, the line representing the equation is accurately illustrated. Understanding graph positions and the movement based on slope allows for more complex graph interpretations and problem-solving.

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Most popular questions from this chapter

In Exercises \(87-90\), graph the curves \(\mathcal{C}\) and \(\mathcal{C}^{\prime}\) in the same viewing window. \(\mathcal{C}=\left\\{(x, y): y=\sqrt{x^{2}+2 x+2}\right\\} ; C^{\prime}\) is obtained by translating \(\mathcal{C}\) to the right by 2 units and up by 1 unit.

A function \(f: S \rightarrow T\) is specified. Determine if \(f\) is invertible. If it is, state the formula for \(f^{-1}(t) .\) Otherwise, state whether \(f\) fails to be one-to-one, onto, or both. \(S=[0,1], T=[0,1 / 2], f(s)=s /(s+1)\)

Let \(f(x)=\sqrt{2 x+5},\) and \(g(x)=x^{-1 / 3} .\) In Exercises \(19-22\) calculate the given expression. \(f^{2}(11) \cdot g^{3}(54)\)

An affine function \(f: \mathbb{R} \rightarrow \mathbb{R}\) has the form \(f(x)=a x+b\), where \(a\) and \(b\) are constants. Prove that the composition of two affine functions is affine and that the inverse of an invertible affine function is affine.

Let \(p\) be any polynomial. a. Show that i. \(q(x)=(p(x)+p(-x)) / 2\) is an even polynomial, and ii. \(r(x)=(p(x)-p(-x)) / 2\) is an odd polynomial. Because \(p=q+r,\) this shows that every polynomial can be written as the sum of an even polynomial and an odd polynomial. b. Show that \(q\) contains only even powers of \(x,\) and \(r\) contains only odd powers. c. If \(p\) is even, deduce that the coefficient of each odd power of \(x\) in \(p(x)\) is zero. If \(p\) is odd, deduce that the coefficient of each even power of \(x\) in \(p(x)\) is zero. d. If \(p\) is even, show that there is a polynomial \(s(x)\) such $$ \text { that } p(x)=s\left(x^{2}\right) $$ e. If \(p\) is odd, show that \(p(0)=0 .\) Deduce that there is an even polynomial \(t(x)\) such that \(p(x)=x \cdot t(x)\)
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