Finding the Negative Domain of a Linear Function with Slope 1

Linear Functions with Negative Domain Identification

The slope of the function on the graph is 1.

What is the negative domain of the function?

(0, 1)(0, 1)(0, 1)xy

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:05 Let's find when the function is negative.
00:08 The function is negative when it's below the X axis.
00:13 It's positive when it’s above the X axis.
00:25 Now, let's find where it crosses the X axis.
00:30 We also need to check when it’s positive or negative.
00:34 And that's how we solve this question!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

The slope of the function on the graph is 1.

What is the negative domain of the function?

(0, 1)(0, 1)(0, 1)xy

2

Step-by-step solution

To answer the question, let's first remember what the "domain of negativity" is,

The domain of negativity: when the values of Y are less than 0.

Note that the point given to us is not the intersection point with the X-axis but with the Y-axis,

That is, at this point the function is already positive.

The point we are looking for is the second one, where the intersection with the X-axis occurs.

The function we are looking at is an increasing function, as can be seen in the diagram and the slope (a positive slope means that the function is increasing),

This means that if we want to find the point, we have to find an X that is less than 0

Now let's look at the solutions:

Option B and Option D are immediately ruled out, since in them X is greater than 0.

We are left with option A and C.

Option C describes a situation in which, as X is less than 0, the function is negative,

Remember that we know the slope is 1,

Which means that for every increase in X, Y also increases in the same proportion.

That is, if we know that when (0,1) the function is already positive, and we want to lower Y to 0,

X also decreased in the same value. If both decrease by 1, the resulting point is (0,-1)

From this we learn that option C is incorrect and option A is correct.

Whenever X is less than -1, the function is negative.

3

Final Answer

1>x -1 > x

Key Points to Remember

Essential concepts to master this topic
  • Negative Domain: Find where y-values are negative (below x-axis)
  • Technique: Use y-intercept (0,1) and slope 1 to find x-intercept at (-1,0)
  • Check: When x = -2, y = -1 which is negative, so x < -1 works ✓

Common Mistakes

Avoid these frequent errors
  • Confusing y-intercept with x-intercept
    Don't use the given point (0,1) as where the function crosses zero = wrong negative domain! This point shows where the function is already positive. Always find where y = 0 by using the slope to locate the x-intercept first.

Practice Quiz

Test your knowledge with interactive questions

Look at the linear function represented in the diagram.

When is the function positive?

–8–8–8–7–7–7–6–6–6–5–5–5–4–4–4–3–3–3–2–2–2–1–1–1111222333444555666777888–5–5–5–4–4–4–3–3–3–2–2–2–1–1–1111222333000

FAQ

Everything you need to know about this question

What does 'negative domain' actually mean?

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The negative domain is the set of x-values where the function's y-values are negative (below the x-axis). It's asking: 'For which x-values is y < 0?'

Why isn't the answer just x < 0?

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Because the function doesn't become negative until x < -1! At x = 0, y = 1 (positive). The function crosses the x-axis at x = -1, so it's only negative when x < -1.

How do I find where the line crosses the x-axis?

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Use the slope and y-intercept! Start at (0,1), then use slope = 1 to go backwards: decrease x by 1 and decrease y by 1. This gives you (-1,0) as the x-intercept.

What if the slope were different?

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The method stays the same! Use the slope to find the x-intercept from the y-intercept. For example, with slope 2 and y-intercept (0,1), you'd get x-intercept at (12,0) (-\frac{1}{2}, 0) .

How can I verify my answer is correct?

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Test a point! Pick any x-value less than -1 (like x = -2) and calculate y. If y is negative, your domain is correct. For x = -2: y = -2 + 1 = -1 (negative) ✓

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