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Chapter 2: Problem 77

The table shows several points on the graph of a linear function. to see connections between the slope formula, the distance formula, the midpoint formula, and linear functions. $$\begin{array}{c|r} x & y \\ \hline 0 & -6 \\ 1 & -3 \\ 2 & 0 \\ 3 & 3 \\ 4 & 6 \\ 5 & 9 \\ 6 & 12 \end{array}$$ Use the first two points in the table to find the slope of the line.

Short Answer

Expert verified
The slope is 3.

Step by step solution

01

- Identify the Points

Identify the first two points given in the table: Point A: \( 0, -6 \) and Point B: \( 1, -3 \)
02

- Recall the Slope Formula

Recall the slope formula for a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \): \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
03

- Substitute the Coordinates

Substitute the coordinates of the first two points into the slope formula: \( x_1 = 0, y_1 = -6, x_2 = 1, y_2 = -3 \)
04

- Calculate the Slope

Calculate the slope: \[ m = \frac{-3 - (-6)}{1 - 0} = \frac{-3 + 6}{1} = \frac{3}{1} = 3 \]

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Key Concepts

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

slope formula
To understand linear functions, the slope formula is a key concept. The slope of a line indicates its steepness and direction. For two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), the slope \( m \) is calculated as: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] This formula measures the change in \( y \) (vertical) over the change in \( x \) (horizontal). By applying this, you can understand how rapidly the line rises or falls. For example, using points \( (0, -6) \) and \( (1, -3) \) from our table: \[ m = \frac{-3 - (-6)}{1 - 0} = \frac{-3 + 6}{1} = \frac{3}{1} = 3 \] So, the slope of our line is 3, meaning it rises 3 units for every 1 unit it moves horizontally.
distance formula
The distance formula helps us find the length of the segment between two points in the plane. For points \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance \( d \) is: \[ d = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \] This formula applies the Pythagorean theorem.
For instance, to find the distance between points \( (0, -6) \) and \( (1, -3) \): \[ d = \sqrt{ (1 - 0)^2 + (-3 + 6)^2} = \sqrt{ 1 + 9 } = \sqrt{10} \] This gives us the precise distance between these points.
midpoint formula
The midpoint formula finds the point exactly halfway between two given points. For points \( (x_1, y_1) \) and \( (x_2, y_2) \), the midpoint \( M \) is: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] This is handy when you need to find a central point along a line segment.
For example, using points \( (0, -6) \) and \( (1, -3) \): \[ M = \left( \frac{0 + 1}{2}, \frac{-6 - 3}{2} \right) = \left( \frac{1}{2}, \frac{-9}{2} \right) = (0.5, -4.5) \] So, the midpoint between these two points is \( (0.5, -4.5) \).
graphing linear equations
Graphing linear equations visually demonstrates their behavior. A linear equation in the form \( y = mx + c \) represents a straight line where \( m \) is the slope and \( c \) is the y-intercept. Using our slope of 3 from the table, the equation would look like \( y = 3x + c \). If we use one of our points like \( (0, -6) \) to find \( c \): \[ -6 = 3(0) + c \] \[ c = -6 \] So, our equation is \( y = 3x - 6 \). To graph, plot several points from the table and draw a line through them. You'll notice the line's rise (3) and run (1) at consistent intervals.

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Most popular questions from this chapter

For the pair of functions defined, find \((f+g)(x),(f-g)(x),(f g)(x),\) and \(\left(\frac{f}{g}\right)(x)\) Give the domain of each. $$f(x)=6-3 x, g(x)=-4 x+1$$

For certain pairs of functions \(f\) and \(g,(f \circ g)(x)=x\) and \((g \circ f)(x)=x .\) Show that this is true for each pair. $$f(x)=4 x+2, g(x)=\frac{1}{4}(x-2)$$

Find all points satisfying \(x+y=0\) that are 8 units from \((-2,3)\).

The tables give some selected ordered pairs for functions \(f\) and \(g\). $$\begin{array}{|c|c|c|c|}\hline x & 3 & 4 & 6 \\\\\hline f(x) & 1 & 3 & 9 \\\\\hline\end{array}$$ $$\begin{array}{|c|c|c|c|c|}\hline x & 2 & 7 & 1 & 9 \\\\\hline g(x) & 3 & 6 & 9 & 12 \\\\\hline\end{array}$$ Find each of the following. $$(g \circ f)(3)$$

Given functions \(f\) and \(g,\) find ( \(a\) ) \((f \circ g)(x)\) and its domain, and ( \(b\) ) \((g \circ f)(x)\) and its domain. See Examples 6 and 7 . $$f(x)=\frac{1}{x-2}, \quad g(x)=\frac{1}{x}$$
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