Finding the Positive Domain of a Linear Function with Slope 1.5

Linear Functions with Positive Domain Analysis

Given the function of the graph.

The slope is 1.5

What is the positive domain?

–6–4–2246810121416–12–10–8–6–4–20(0, -8)

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find the negative domain of the function
00:07 Draw the function continuation until intersection with X-axis
00:19 Use the line equation to find intersection point with X-axis
00:24 Substitute appropriate values to find intersection point with X-axis
00:41 Isolate X
01:01 This is the intersection point with X-axis
01:08 And negative when the function is below X-axis
01:13 The function is positive when it's above X-axis
01:21 Identify when the function is positive and when negative
01:25 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Given the function of the graph.

The slope is 1.5

What is the positive domain?

–6–4–2246810121416–12–10–8–6–4–20(0, -8)

2

Step-by-step solution

To find the domain of positivity, we need to find the point of intersection of the equation with the x-axis.

For this, we need to find the formula of the equation.

We know that a linear equation is constructed as follows:

Y=MX+B

m represents the slope of the line, which is given to us: 1.5

b represents the point of intersection of the line with the Y-axis, which can be extracted from the existing point on the graph, -8.

And therefore:

Y=1.5X-8

Now, we replace:

Y=0, since we are trying to find the point of intersection with the X-axis.

0=1.5X-8
8=1.5X
5.3333 = X

We reveal that the point of intersection with the X-axis is five and one third (5.333)

Now, as we know that the slope is positive and the function is increasing, we can conclude that the domain of positivity is when the x values are less than five and one third.

That is:

5.333>X

And this is the solution!

3

Final Answer

513>x 5\frac{1}{3}>x

Key Points to Remember

Essential concepts to master this topic
  • Linear Function Form: Use y = mx + b with slope m = 1.5
  • X-Intercept Method: Set y = 0, solve: 0 = 1.5x - 8 gives x = 5⅓
  • Domain Check: Since slope is positive, function is positive when x < 5⅓ ✓

Common Mistakes

Avoid these frequent errors
  • Confusing positive domain with function being above x-axis
    Don't think positive domain means where y > 0 = wrong interpretation! The question asks for positive domain of the function itself. Always find where the function intersects the x-axis first, then determine based on the slope direction.

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 exactly is a 'positive domain'?

+

The positive domain refers to the x-values where the function output (y-values) is positive. You need to find where the line is above the x-axis.

Why do I set y = 0 to find the positive domain?

+

Setting y = 0 finds the x-intercept - the boundary point where the function changes from positive to negative (or vice versa). This is your reference point!

How do I know which side of the x-intercept is positive?

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Look at the slope! If slope is positive (like 1.5), the line goes up from left to right. So the function is positive on the left side of the x-intercept.

Why is the answer 513>x5\frac{1}{3} > x instead of x<513x < 5\frac{1}{3}?

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Both expressions mean the same thing! 513>x5\frac{1}{3} > x reads as "5⅓ is greater than x" which is equivalent to "x is less than 5⅓".

What if I got x=163x = \frac{16}{3} instead of 5135\frac{1}{3}?

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That's correct too! 163=513=5.333...\frac{16}{3} = 5\frac{1}{3} = 5.333... These are all the same value written in different forms.

How can I verify my answer is correct?

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Test a point! Pick an x-value less than 5⅓ (like x = 0) and substitute: y = 1.5(0) - 8 = -8. Wait, that's negative! Check your work - the function is actually negative when x < 5⅓.

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