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Chapter 3: Problem 80

Find a function whose graph is the given curve. The line segment joining the points \((-3,-2)\) and \((6,3)\)

Short Answer

Expert verified
The function is \( y = \frac{5}{9}x - \frac{1}{3} \).

Step by step solution

01

Determine the Slope

To find the slope of the line (denoted as \( m \)), use the formula: \( m = \frac{y_2-y_1}{x_2-x_1} \). For the points \( (-3,-2) \) and \( (6,3) \), substitute the coordinates: \( m = \frac{3 - (-2)}{6 - (-3)} = \frac{5}{9} \). So, the slope of the line is \( \frac{5}{9} \).
02

Use the Point-Slope Form Equation

The equation of a line using the point-slope formula is \( y - y_1 = m(x - x_1) \). Choose one of the given points, let's use \( (-3, -2) \). Substituting the values, we get: \( y + 2 = \frac{5}{9}(x + 3) \).
03

Simplify the Equation

To express the equation in the slope-intercept form (\( y = mx + b \)), simplify \( y + 2 = \frac{5}{9}(x + 3) \). Distribute \( \frac{5}{9} \): \( y + 2 = \frac{5}{9}x + \frac{15}{9} \). Simplifying further: \( y = \frac{5}{9}x + \frac{15}{9} - 2 \).
04

Express in Simplified Form

Convert \( -2 \) to a fraction with a denominator of 9: \( -2 = \frac{-18}{9} \). Now combine terms: \( y = \frac{5}{9}x + \frac{15}{9} - \frac{18}{9} \). This simplifies to \( y = \frac{5}{9}x - \frac{3}{9} \), or \( y = \frac{5}{9}x - \frac{1}{3} \).

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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 is a straightforward way to write the equation of a straight line. It allows you to see the slope and the y-intercept at a glance. This form is given by the equation \( y = mx + b \), where:
  • \( m \) represents the slope of the line.
  • \( b \) is the y-intercept, which is the value of y when x equals zero.
Once you have the slope and y-intercept of a line, you can easily formulate its equation using this form. In our case, after simplifying the point-slope equation, we arrived at \( y = \frac{5}{9}x - \frac{1}{3} \). Here, \( \frac{5}{9} \) is the slope, and \( -\frac{1}{3} \) is the y-intercept. This form makes it very clear how the line behaves and where it crosses the y-axis.
Point-Slope Formula
The point-slope formula is a powerful tool when you know a point on a line and the line's slope. Its equation is: \( y - y_1 = m(x - x_1) \), where:
  • \( (x_1, y_1) \) is a known point on the line.
  • \( m \) is the slope of the line.
You start by choosing any point on the line, though either could be used. For our line between points \((-3, -2)\) and \((6, 3)\), we opted for \((-3, -2)\). By knowing the slope as \( \frac{5}{9} \), we can substitute these values into the formula, leading to \( y + 2 = \frac{5}{9}(x + 3) \). This step bridges the gap between the raw data given (points) and the neat equation in slope-intercept form.
Slope Calculation
Calculating the slope is the essential first step in forming the equation of a line. The slope tells us how steep the line is, or how much y increases as x increases. The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
  • \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
For our specific points \((-3, -2)\) and \((6, 3)\), the slope calculation becomes \( m = \frac{3 - (-2)}{6 - (-3)} = \frac{5}{9} \). Think of the slope as the 'rise' over 'run' metric, illustrating how much y 'rises' for each unit x 'runs' to the right. This important value is central to deriving further iterations of line equations, such as in point-slope or slope-intercept forms.

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Most popular questions from this chapter

Sums of Even and Odd Functions If \(f\) and \(g\) are both even functions, is \(f+g\) necessarily even? If both are odd, is their sum necessarily odd? What can you say about the sum if one is odd and one is even? In each case, prove your answer.

\(29-40\) Find the functions \(f \circ g, g \circ f, f \circ f,\) and \(g \circ g\) and their domains. $$ f(x)=x^{3}+2, \quad g(x)=\sqrt[3]{x} $$

Inflating a Balloon A spherical balloon is being inflated. The radius of the balloon is increasing at the rate of 1 \(\mathrm{cm} / \mathrm{s}\) . (a) Find a function \(f\) that models the radius as a function of time. (b) Find a function \(g\) that models the volume as a function of the radius. (c) Find \(g \circ f .\) What does this function represent?

\(17-22=\) Use \(f(x)=3 x-5\) and \(g(x)=2-x^{2}\) to evaluate the expression. $$ \begin{array}{ll}{\text { (a) }(f \circ f)(x)} & {\text { (b) }(g \circ g)(x)}\end{array} $$

\(51-54\) Express the function in the form \(f \circ g \circ h\) $$ F(x)=\sqrt[3]{\sqrt{x}-1} $$
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