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Chapter 2: Problem 76

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 equation of the line is \( y = \frac{5}{9}x - \frac{1}{3} \).

Step by step solution

01

Identify the form of the function

Lines on a graph can be represented using the linear equation formula, which is given by: \( y = mx + c \), where \( m \) is the slope, and \( c \) is the y-intercept.
02

Calculate the slope \( m \)

The slope \( m \) of a line between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substitute the given points \((-3, -2)\) and \((6, 3)\) into the formula: \[ m = \frac{3 - (-2)}{6 - (-3)} = \frac{3 + 2}{6 + 3} = \frac{5}{9} \]
03

Use the point-slope form to find the equation

We use the point-slope form of the line equation to find \( c \). The point-slope form is \( y - y_1 = m(x - x_1) \). Let's choose point \((-3, -2)\) for our calculations:\[ y - (-2) = \frac{5}{9}(x - (-3)) \]\[ y + 2 = \frac{5}{9}(x + 3) \]
04

Simplify the equation

Expand and simplify the equation from Step 3:\[ y + 2 = \frac{5}{9}x + \frac{5}{3} \]Subtract 2 from both sides:\[ y = \frac{5}{9}x + \frac{5}{3} - 2 \]Convert -2 into fractional form to combine:\[ y = \frac{5}{9}x + \frac{5}{3} - \frac{6}{3} \]\[ y = \frac{5}{9}x - \frac{1}{3} \]
05

Verify the function with both points

To confirm that our function works, plug in both points to verify:For \((-3, -2)\):\( y = \frac{5}{9}(-3) - \frac{1}{3} = -\frac{15}{9} - \frac{1}{3} = -\frac{20}{9} = -2 \)For \((6, 3)\):\( y = \frac{5}{9}(6) - \frac{1}{3} = \frac{30}{9} - \frac{1}{3} = 3 \)Both points satisfy the function

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Key Concepts

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

Slope of a Line
The slope of a line is a measure of its steepness and direction. It is represented by the letter \( m \) in the linear equation \( y = mx + c \). Calculating the slope involves taking two points on the line, such as \((-3, -2)\) and \((6, 3)\), and using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). By substituting the coordinates into this formula, we find the slope:
  • Calculate the difference in \( y \)-coordinates: \( 3 - (-2) = 3 + 2 = 5 \)
  • Calculate the difference in \( x \)-coordinates: \( 6 - (-3) = 6 + 3 = 9 \)
  • Divide the differences: \( \frac{5}{9} \)
This tells us the line rises \( 5 \) units for every \( 9 \) units it moves to the right. Understanding slope helps determine how quickly a line ascends or descends.
Point-Slope Form
The point-slope form is a handy equation used to easily find the equation of a line when you have a point on the line and its slope. It is written as \( y - y_1 = m(x - x_1) \). This can help simplify finding the line鈥檚 equation without first finding its y-intercept. For example, using the point \((-3, -2)\) and slope \( \frac{5}{9} \), the setup is:
  • Plug into formula: \( y - (-2) = \frac{5}{9}(x - (-3)) \)
  • Simplify to \( y + 2 = \frac{5}{9}(x + 3) \)
This expression is the starting point to simplify into the standard form \( y = mx + c \). It鈥檚 ideal to immediately capture a line鈥檚 characteristics with given data points.
Y-Intercept
The y-intercept is the point where the line crosses the y-axis, denoted as \( c \) in the equation \( y = mx + c \). To find it, you adjust the point-slope form by removing any constants on the left side through algebraic manipulation.
  • Start with: \( y + 2 = \frac{5}{9}x + \frac{5}{3} \)
  • Subtract \( 2 \): \( y = \frac{5}{9}x + \frac{5}{3} - 2 \)
  • Convert to common fractions: \( y = \frac{5}{9}x + \frac{5/3} - \frac{6/3} \)
  • Result: \( y = \frac{5}{9}x - \frac{1}{3} \)
Here, the y-intercept is \( -\frac{1}{3} \). It鈥檚 crucial for interpreting where the line will cross the y-axis, providing an anchor point for graphing and understanding line behavior.
Graph of a Line
Graphing a line starts with plotting its equation, allowing us to see its trajectory and intersection points. With the equation \( y = \frac{5}{9}x - \frac{1}{3} \), observe its key features:
  • Slope \( \frac{5}{9} \) means a gentle upward incline鈥擻( 5 \) units up for \( 9 \) units along the x-axis.
  • The y-intercept, where it crosses the y-axis, is at \( -\frac{1}{3} \).
Steps to graph:
  • Begin at the y-intercept \((-\frac{1}{3})\) on the y-axis.
  • From the intercept, use the slope to determine points along the line. Move right \( 9 \) units, then up \( 5 \) units to plot another point.
  • Draw a straight line through the points to extend it in both directions.
This visualization aids in comprehending how lines are situated on the coordinate plane, demonstrating their relationship through calculated points.

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Most popular questions from this chapter

Temperature readings \(T\) (in 'F) were recorded every 2 hours from midnight to noon in Atlanta, Georgia, on March 18, 1996. The time \(t\) was measured in hours from midnight. Sketch a rough graph of \(T\) as a function of \(t\) \begin{array}{|c|c|c|c|c|c|c|c|} \hline t & 0 & 2 & 4 & 6 & 8 & 10 & 12 \\ \hline T & 58 & 57 & 53 & 50 & 51 & 57 & 61 \\ \hline \end{array}

The population \(P\) (in thousands) of San Jose, California, from 1988 to 2000 is shown in the table. (Midycar estimates are given.) Draw a rough graph of \(P\) as a function of time \(t\) $$\begin{array}{|c|c|c|c|c|c|c|c|} \hline t & 1988 & 1990 & 1992 & 1994 & 1996 & 1998 & 2000 \\ \hline P & 733 & 782 & 800 & 817 & 838 & 861 & 895 \\ \hline \end{array}$$

A function is given. (a) Find all the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. $$g(x)=x^{4}-2 x^{3}-11 x^{2}$$

Complete the table. $$f(x)=2(x-1)^{2}$$ $$\begin{array}{|r|r|} \hline x & f(x) \\ \hline-1 & \\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ \hline \end{array}$$

Find the domain of the function. $$f(t)=\sqrt[3]{t}-1$$
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