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Magazine Chapter 2: Problem 33
Given each set of information, find a linear equation satisfying the conditions, if possible $$ f(-5)=-4, \text { and } f(5)=2 $$
Short Answer
Expert verified
The linear equation is \(y = \frac{3}{5}x - 1\).
Step by step solution
01
Identify Given Points
The problem provides two points from the function: - The first point is \((-5, -4)\) since \(f(-5) = -4\).- The second point is \((5, 2)\) since \(f(5) = 2\).These represent points on the linear graph of the function.
02
Calculate Slope
The formula to find the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]Substituting the given points \((-5, -4)\) and \((5, 2)\) into the formula:\[ m = \frac{2 - (-4)}{5 - (-5)} = \frac{6}{10} = \frac{3}{5} \] So the slope of the line is \(\frac{3}{5}\).
03
Use Point-Slope Form
With the slope \(m = \frac{3}{5}\) and the point \((-5, -4)\), use the point-slope form of a linear equation:\[ y - y_1 = m(x - x_1) \]Substitute the point and slope:\[ y - (-4) = \frac{3}{5}(x - (-5)) \]Simplify to write:\[ y + 4 = \frac{3}{5}(x + 5) \]
04
Convert to Slope-Intercept Form
Now, simplify the equation into slope-intercept form (\(y = mx + b\)):\[ y + 4 = \frac{3}{5}(x + 5) \]Distribute \(\frac{3}{5}\) on the right side:\[ y + 4 = \frac{3}{5}x + \frac{15}{5} \]\[ y + 4 = \frac{3}{5}x + 3 \]Subtract 4 from both sides to solve for \(y\):\[ y = \frac{3}{5}x + 3 - 4 \]\[ y = \frac{3}{5}x - 1 \]
05
Final Check
Verify by plugging the original points into the equation \(y = \frac{3}{5}x - 1\):- For \(x = -5\): \[ y = \frac{3}{5}(-5) - 1 = -3 - 1 = -4 \] - For \(x = 5\): \[ y = \frac{3}{5}(5) - 1 = 3 - 1 = 2 \] Both points satisfy the equation, confirming its correctness.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope Calculation
Slope calculation is the process of determining the steepness or incline of a line, which can be found using two points on the line. The formula for calculating the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
- \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
- \( m = \frac{2 - (-4)}{5 - (-5)} = \frac{6}{10} = \frac{3}{5} \)
Point-Slope Form
The point-slope form is a standard way to express the equation of a line when you know the slope and one point on the line. Its general formula is:
- \( y - y_1 = m(x - x_1) \)
- \( y - (-4) = \frac{3}{5}(x - (-5)) \)
- \( y + 4 = \frac{3}{5}(x + 5) \)
Slope-Intercept Form
The slope-intercept form of a linear equation is expressed as \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. It provides a clear representation of the line's slope and where it crosses the y-axis.
Converting from point-slope form to slope-intercept form involves simplifying the expression. Starting from:
Converting from point-slope form to slope-intercept form involves simplifying the expression. Starting from:
- \( y + 4 = \frac{3}{5}(x + 5) \)
- \( y + 4 = \frac{3}{5}x + 3 \)
- \( y = \frac{3}{5}x - 1 \)
Linear Graph
A linear graph represents the set of solutions to a linear equation, forming a straight line on a coordinate plane. Each point on the line is a solution to the equation. The variables and constants of the equation govern the line's direction, slope, and position.
In our problem, the linear equation is \(y = \frac{3}{5}x - 1\). This gives us a visual tool to plot the function. The slope \(\frac{3}{5}\) tells us for each unit increase in \(x\), \(y\) increases by \(\frac{3}{5}\) units, and the y-intercept \(-1\) tells us where the line crosses the y-axis.
In our problem, the linear equation is \(y = \frac{3}{5}x - 1\). This gives us a visual tool to plot the function. The slope \(\frac{3}{5}\) tells us for each unit increase in \(x\), \(y\) increases by \(\frac{3}{5}\) units, and the y-intercept \(-1\) tells us where the line crosses the y-axis.
- The line rises from left to right, indicating a positive slope.
- Intersecting the y-axis at \(-1\) gives an initial point of plotting.
Function Verification
Function verification involves checking whether a given equation correctly represents the relationship described by the initial conditions. In our problem, we verify by substituting original points into the final equation.
Using the equation \(y = \frac{3}{5}x - 1\):
This step ensures there's no mistake in our calculations and that the equation accurately models the linear function.
Using the equation \(y = \frac{3}{5}x - 1\):
- Plug in \(x = -5\):\[ y = \frac{3}{5}(-5) - 1 = -3 - 1 = -4 \]
- Plug in \(x = 5\):\[ y = \frac{3}{5}(5) - 1 = 3 - 1 = 2 \]
This step ensures there's no mistake in our calculations and that the equation accurately models the linear function.
