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Chapter 1: Problem 15

Work each problem related to linear functions. (a) Evaluate \(f(-2)\) and \(f(4)\) (b) Graph \(f\). How can the graph of \(f\) be used to determine the zero of \(f ?\) (c) Find the zero of \(f\) $$f(x)=0.4 x+0.15$$

Short Answer

Expert verified
(a) \(-0.65\) and \(1.75\); (b) Graph shows zero at \(-0.375\); (c) Zero is \(-0.375\).

Step by step solution

01

Step 1A - Evaluating f(-2)

To evaluate \( f(-2) \), substitute \(-2\) into the function equation. The function is \( f(x) = 0.4x + 0.15 \), so calculate \( f(-2) = 0.4(-2) + 0.15 = -0.8 + 0.15 = -0.65 \).
02

Step 1B - Evaluating f(4)

To evaluate \( f(4) \), substitute \(4\) into the function equation. The function is \( f(x) = 0.4x + 0.15 \), so calculate \( f(4) = 0.4(4) + 0.15 = 1.6 + 0.15 = 1.75 \).
03

- Graphing f(x)

To graph \( f(x) = 0.4x + 0.15 \), identify the y-intercept and slope. The y-intercept is 0.15 (the point where the line crosses the y-axis) and the slope is 0.4 (rise over run). Start by plotting the y-intercept at (0, 0.15) and use the slope to plot another point. Move up 0.4 units and right 1 unit from the y-intercept to find another point on the line, then draw the line through these points.
04

- Zero of f from Graph

The zero of \( f \) is the x-value where the graph of the function crosses the x-axis. From the graph of \( f(x) = 0.4x + 0.15 \), locate the point where the line intersects the x-axis (where y=0).
05

- Algebraically Finding the Zero of f

To find the zero algebraically, set \( f(x) = 0 \) and solve for \( x \). So, \( 0.4x + 0.15 = 0 \). Subtract 0.15 from both sides to get \( 0.4x = -0.15 \). Divide both sides by 0.4, resulting in \( x = -0.375 \). So, the zero of \( f \) is \( x = -0.375 \).

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Key Concepts

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

Function Evaluation
Function evaluation is like plugging numbers into a machine to see what comes out. If we have a function, like in our example: \[ f(x) = 0.4x + 0.15 \]we can "evaluate" the function. This means we substitute different values for \( x \) and solve the expression for each one.
For example:
  • To evaluate \( f(-2) \), substitute \(-2\) into the function: \( f(-2) = 0.4(-2) + 0.15 = -0.65 \).
  • Similarly, evaluate \( f(4) \) by substituting \( 4 \) into the function: \( f(4) = 0.4(4) + 0.15 = 1.75 \).
By doing these computations, we discover the outputs of the function when different inputs are used, which helps us understand the function's behavior.
Graphing Linear Equations
Graphing a linear equation involves plotting a straight line on a coordinate plane. Our function \[ f(x) = 0.4x + 0.15 \]is a linear function, which makes it straightforward to graph. We need two main components: the y-intercept and the slope.
  • The y-intercept is where the line crosses the y-axis. Here, it's 0.15.
  • The slope, which tells us how steep the line is, is 0.4. This means for every 1 unit we move to the right, we move 0.4 units up.
To graph:
  • Begin by marking the y-intercept at (0, 0.15).
  • Use the slope to find another point. From (0, 0.15), move right 1 unit and up 0.4 units to the point (1, 0.55).
Draw a line through these points and extend it across the graph. This visual representation helps us quickly see relationships and intersections.
Slope-Intercept Form
The slope-intercept form of a linear equation is a handy way to write and recognize equations for straight lines. It is expressed as: \[ y = mx + b \]Here, \( m \) stands for the slope of the line. It's important because it indicates the line's tilt, or how much y changes for each change in x.
In our equation, \[ f(x) = 0.4x + 0.15 \]
  • The slope \( m \) is 0.4, showing that for each increase of 1 in x, y increases by 0.4.
  • The y-intercept \( b \), which is 0.15, is the value of y when x is 0. It's where the line crosses the y-axis.
This form is perfect for quickly graphing the line and identifying its characteristics, like direction and starting point.
Finding Zero of a Function
Finding the zero of a function means determining the x-value where the function equals zero. Graphically, it's where the line crosses the x-axis, and algebraically, it's solving the equation where the function output is zero.
For \[ f(x) = 0.4x + 0.15 \]To find the zero algebraically:
  • Set the function equal to zero: \( 0.4x + 0.15 = 0 \)
  • Solve for \( x \): First subtract 0.15 from both sides, giving \( 0.4x = -0.15 \).
  • Then, divide by 0.4 to isolate \( x \): \( x = -0.375 \).
So, the zero of the function is \( x = -0.375 \). This point shows where the function has no value (or an output of zero) on the graph, marking a crucial intercept for many applications.

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