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Chapter 1: Problem 26

Determine whether each equation defines \(y\) as a function of \(x .\) $$|x|-y=5$$

Short Answer

Expert verified
The given equation \(|x|-y=5\) can be rewritten as \(y = |x|-5\). After analyzing the equation, it is apparent that for each input value of \(x\), there is only one corresponding output value of \(y\), hence \(y\) is a function of \(x\).

Step by step solution

01

Isolate y

Start by isolating \(y\) on one side of the equation. To do this, rearrange the given equation \(|x|-y=5\) to the form \(y = ...\). This gives us \(y = |x| -5\).
02

Analyze Equation

Now that we've isolated \(y\), it's time to analyze the equation \(y = |x| - 5\). The absolute value of \(x\) can either be negative or positive depending upon the value of \(x\). Therefore, \(|x|\) can give two possible \(y\) outputs for a single \(x\)-value. This indicates that for each input \(x\), there might be more than one output \(y\).
03

Verify with Vertical Line Test

Despite the above observation, no definitive conclusion has been reached. Therefore, we should carry out the vertical line test to see if any vertical line drawn through the graph of the function intersects the graph in more than one point. However, keep in mind that for the function \(y = |x| -5 \), any vertical line will touch the graph at most at one point.

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Key Concepts

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

Absolute Value
The absolute value of a number is simply its distance from zero on a number line. It doesn't matter if the number is positive or negative, its absolute value is always non-negative. For example:
  • The absolute value of -7 is written as \(|-7|\) and it equals 7.
  • Similarly, the absolute value of 7 is also 7.
In the equation \(y = |x| - 5\), the \(|x|\) component ensures that whatever value \(x\) takes, the expression \(|x|\) will always return a positive result, hence influencing \(y\) accordingly.This concept is important in defining the behavior of functions around origin, by keeping all values positive.
Function of x
A function of \(x\) is a relationship where each input \(x\) is related to exactly one output. In simpler terms, for every \(x\) value, there should be a single corresponding \(y\) value.The equation given, \(y = |x| - 5\), transforms any \(x\) value into a \(y\) via the absolute value operation.
  • Due to the absolute value nature, the equation guarantees one output for every input: if \(x\) is 2, then \(y\) will be \(2 - 5 = -3\).
  • If \(x\) is -2, the result remains the same: \(2 - 5 = -3\).
This uniqueness is what makes it a valid function of \(x\), since no two different \(x\) inputs will provide more than one \(y\) output.
Vertical Line Test
The vertical line test is a simple method to determine whether an equation is a function of \(x\). By drawing vertical lines through a graph, if any line crosses more than once, the equation fails to be a function.When applied to \(y = |x| - 5\), consider its graph:
  • It is a V-shaped graph translated downward by 5 units from the origin.
  • Drawing a vertical line at any point will intersect the graph at only one point.
This ensures satisfaction of the criteria for being a function, reinforcing that indeed \(y = |x| - 5\) represents \(y\) as a function of \(x\). Any vertical line touching the graph at most once is a clear indication of this.

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Most popular questions from this chapter

In Exercises \(53-64,\) complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-4 x-12 y-9=0$$

Use point plotting to graph \(f(x)=x+2\) if \(x \leq 1\).

In Exercises \(53-64,\) complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+8 x-2 y-8=0$$

In Exercises \(53-64,\) complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-10 x-6 y-30=0$$

In Exercises \(53-64,\) complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+3 x-2 y-1=0$$
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