# Positivity and negativity - Examples, Exercises and Solutions

Positivity and Negativity of a Linear Function

The function is positive when it is above the $X$ axis when $Y<0$

The function is negative when it is below the $X$ axis as $Y>0$

When we are asked what the domains of positivity of the function are, we are actually being asked in which values of $X$ the function is positive: when it is above the $X$ axis.

In which values of $X$ does the function obtain positive $Y$ values?

When we are asked what the domain of negativity of the function is, we are actually being asked in which values of $X$ the function is negative: when it is below the $X$ axis.

In which values of $X$ does the function obtain negative $Y$ values?

## Practice Positivity and negativity

### Exercise #1

Look at the function shown in the figure.

When is the function positive?

### Step-by-Step Solution

The function we see is a decreasing function,

Because as X increases, the value of Y decreases, creating the slope of the function.

We know that this function intersects the X-axis at the point x=-4

Therefore, we can understand that up to -4, the values of Y are greater than 0, and after -4, the values of Y are less than zero.

Therefore, the function will be positive only when

X < -4

-4 > x

### Exercise #2

Given the function of the graph.

What are the areas of positivity and negativity of the function?

### Step-by-Step Solution

When we are asked what the domains of positivity of the function are, we are actually being asked at what values of X the function is positive: it is above the X-axis.

At what values of X does the function obtain positive Y values?

In the given graph, we observe that the function is above the X-axis before the point X=7, and below the line after this point. That is, the function is positive when X>7 and negative when X<7,

And this is the solution!

Positive 7 > x

Negative 7 < x

### Exercise #3

Calculate the positive domain of the function shown in the figure:

### Step-by-Step Solution

The domains of positivity and negativity are determined by the point of intersection of the function with the X-axis, so the Y values are greater or less than 0.

We are given the information of the intersection with the Y-axis, but not of the point of intersection with the X-axis,

Furthermore, there is no data about the function itself or the slope, so we do not have the ability to determine the point of intersection with the X-axis,

And so in the domains of positivity and negativity.

Not enough data

### Exercise #4

Look at the function in the figure.

What is the positive domain of the function?

### Step-by-Step Solution

Positive domain is another name for the point from which the x values are positive and not negative.

From the figure, it can be seen that the function ascends and passes through the intersection point with the X-axis (where X is equal to 0) at point 2a.

Therefore, it is possible to understand that from the moment X is greater than 2a, the function is in the domains of positivity.

Therefore, the function is positive when:

2a < x

2a < x

### Exercise #5

The slope of the function on the graph is 1.

What is the negative domain of the function?

### 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.

-1 > x

### Exercise #1

Given the function of the graph.

The slope is 1.5

What is the positive domain?

### 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!

5\frac{1}{3}>x

### Exercise #2

Which equation represents a line that is positive in domain for each value of x.

### Step-by-Step Solution

To find out if the equation intersects the x-axis, we need to substitute in the equations y=0
If the function has a solution where y=0 then the equation has an intersection point and is therefore positive.

y = 3x+8

We will substitute as instructed:

0 = 3x+8

3x = -8

x = -8/3

Although the result here is not a "nice" number, we see that we are able to arrive at a result and therefore this answer is rejected.

Let's move on to the second equation:

y = 300x+50

Here too we will substitute:

0 = 300x + 50
-50 = 300x

-50/300 = x
-1/6 = x

In this exercise too we managed to arrive at a result and therefore the answer is rejected.

Let's move on to answer C:

y = 3

We will substitute:

0 = 3

We see that here an impossible result is obtained, a "false equation", because 0 can never be equal to 3.

Therefore, we understand that the equation in answer C is the one that does not intersect the x-axis, and is in fact positive all the time.

Therefore answer D is also rejected, and only answer C is correct.

$y=3$

### Exercise #3

Look at the linear function represented in the diagram.

When is the function positive?

x>2

### Exercise #4

Given the linear function of the drawing.

What is the negative domain of the function?

### Video Solution

The always positive function

### Exercise #5

Given the function of the figure.

What are the areas of positivity and negativity of the function?

Positive x<3.5

Negative x>3.5

### Exercise #1

What is the positive domain of the function shown in the graph below?

### Video Solution

For all $x$

### Exercise #2

Given the function is negative in the domain 4 > x

Find the equation of the line given that it passes through the point $(5,9)$

### Video Solution

$y=9x-36$

### Exercise #3

Choose the equation that represents a straight line that is positive in the domain 8 > x

and passes through the point $(0,9)$.

### Video Solution

$y=-1\frac{1}{8}x+9$

### Exercise #4

Choose the equation that represents a line with a negative domain of 0 < x .

### Video Solution

$y=-2x$

### Exercise #5

Given a function that is positive from the beginning of the axes. Plus the point (7,9) on the graph of the function. Find the equation for the function.

### Video Solution

$y=1\frac{2}{7}x$