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Chapter 3: Problem 48

Exer. 47-48: If a line \(l\) has nonzero \(x\) - and \(y\)-intercepts \(a\) and \(b\), respectively, then its intercept form is $$ \frac{x}{a}+\frac{y}{b}=1 . $$ Find the intercept form for the given line. $$ x-3 y=-2 $$

Short Answer

Expert verified
The intercept form is \(\frac{x}{-2} - \frac{y}{3} = 1\).

Step by step solution

01

Identify coefficients of x and y

From the equation \(x - 3y = -2\), identify the coefficients of \(x\) and \(y\). The coefficient of \(x\) is 1 and for \(y\) is -3.
02

Convert to slope-intercept form

Rearrange the equation to the slope-intercept form \(y = mx + c\). Starting with \(x - 3y = -2\): \[-3y = -x - 2\] which simplifies to \[y = \frac{1}{3}x + \frac{2}{3}\].
03

Use intercept form formula

The formula for the intercept form of a line is \(\frac{x}{a} + \frac{y}{b} = 1\). We can directly rearrange the original equation \(x - 3y = -2\) to make it similar to this form: \[\frac{x}{-2} + \frac{y}{\frac{2}{3}} = 1\].
04

Rearrange to standard intercept form

For neatness, multiply through by 3 to eliminate fractions: \[3\left(\frac{x}{-2} + \frac{y}{\frac{2}{3}}\right) = 3 \times 1\] which simplifies to \[\frac{3x}{-2} + y = 1\]. Convert \(\frac{3x}{-2}\) to \(-\frac{3}{2}x\) before stating the intercept form.
05

Present intercept form

After simplification, the intercept form of the line is \(\frac{x}{-2} - \frac{y}{3} = 1\) or equivalent variants based on standardization.

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Key Concepts

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

Equation of a Line
The equation of a line is a mathematical expression that represents all the points along a straight line in a coordinate plane. Depending on the context, there are different forms to express this equation.

Common forms include:
  • Slope-intercept form: \( y = mx + c \)
  • Point-slope form: \( y - y_1 = m(x - x_1) \)
  • Intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \)
Each form gives different insights about the line's properties, like slope, y-intercept, and positioning. Understanding the equation of a line allows you to graph it easily and analyze linear relationships. For the specific example in the exercise, the equation \( x - 3y = -2 \) is offered in standard form, but can be converted to other forms for further insight.
x-intercept
An x-intercept is where a line crosses the x-axis on a graph. This point has coordinates where the y-value is zero. To find the x-intercept of a line, you set \( y = 0 \) in the line's equation and solve for \( x \).

For example, in the equation \( x - 3y = -2 \), substitute \( y = 0 \):
  • \( x - 3(0) = -2 \) or simply \( x = -2 \)
This tells us the x-intercept at (-2, 0). On the graph, the line touches the x-axis at this point. Knowing the x-intercept provides a starting or endpoint for graphing or understanding a line's behavior.
y-intercept
Similar to the x-intercept, the y-intercept is where a line crosses the y-axis. At this point, the x-value is zero. To determine the y-intercept, you set \( x = 0 \) and solve for \( y \).

Using the slope-intercept form can make this easier to understand, yet let's use the equation \( x - 3y = -2 \) to find it:
  • Set \( x = 0 \): \( 0 - 3y = -2 \)
  • Then solve for \( y \) to get: \( y = \frac{2}{3} \)
Thus, the y-intercept here is at (0, \( \frac{2}{3} \)). Graphically, the line will cross the y-axis at this point. The y-intercept gives insight into the vertical position of the line.
Slope-Intercept Form
The slope-intercept form of a line is an equation that features both slope and y-intercept. It takes the form \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. This form is very useful for quickly understanding a line鈥檚 rise over run (slope) and where it crosses the y-axis.

Let's use the equation \( x - 3y = -2 \) and convert it into slope-intercept form:
  • Rearrange: \(-3y = -x - 2 \)
  • Simplify: \( y = \frac{1}{3}x + \frac{2}{3} \)
Here, the slope \( m \) is \( \frac{1}{3} \) and the y-intercept \( c \) is \( \frac{2}{3} \). This provides a simple way to plot and interpret the line since you understand both its incline and starting position on the y-axis.

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Most popular questions from this chapter

Exer. 53-60: Find a composite function form for \(y\). $$ y=\frac{1}{(x-3)^{4}} $$

Exer. 3-12: Determine whether \(f\) is even, odd, or neither even nor odd. f(x)=5 x^{3}+2 x

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{3 x+5}{2}, \quad g(x)=\frac{2 x-5}{3} $$

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=c x^{3} ; \quad c=-\frac{1}{3}, 1,2 $$

Exer. 3-12: Determine whether \(f\) is even, odd, or neither even nor odd. $$ f(x)=x^{3}-\frac{1}{x} $$
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