f(x) = (x-5)(x+4)(x-3)

Question

Find the y-intercept(s), the x-intercept(s), the values of x where f(x)>0, and the values of x where f(x)<0.

What are the values of x where f(x)>0?

Answer (in interval notation):

 

What are the values of x where f(x)<0?

Answer (in interval notation):

Answer 1

Let's start with the y-intercept. The y-intercept is the value of y when x = 0 (since that's when the function will cross the y-axis), so plug in x = 0 and solve for y (or, in this case, f(0)):

 

f(0) = (0-5)(0+4)(0-3)

f(0) = (-5)(4)(-3)

f(0) = 60

 

So, the y-intercept is (0, 60).

 

 

Now, let's work on the x-intercepts. Those are all the values of x when y = 0 (since that's when the function will cross the x-axis). So, plug in f(x) = 0:

 

0 = (x-5)(x+4)(x-3)

 

The right-hand side is zero if and only if one of its factors is zero. So, the right-hand side is zero when x - 5 = 0, when x + 4 = 0, and when x - 3 = 0. In other words, when x = 5, when x = -4, and when x = 3. That means the x-intercepts are (5, 0), (3, 0), and (-4, 0).

 

 

Using those points, you can sketch out a rough graph of the function. f(x) > 0 whenever the function is above the x-axis and f(x) < 0 whenever the function is below the x-axis. The function crosses the x-axis at 5, 3, and -4 (since those are its x-intercepts) and is above the x-axis at 0 (since it includes the point (0, 60)).

 

At 0, the function is above the x-axis, so it must be above the x-axis until it crosses in either direction, at 3 and -4. Beyond -4, the function remains below the x-axis, since it never crosses again. Beyond 3, the function remains below the x-axis until it crosses again at 5, then remains above the x-axis since it never crosses again. The pattern of the function is:

 

 

<---(neg)---|(-4)|------(pos)------|(3)|-----(neg)------|(5)|------(pos)----->

 

 

So, f(x) > 0 when -4 < x < 3 and when x > 5:

 

(-4, 3)∪(5, infinity)

 

And, f(x) < 0 when x < -4 and when 3 < x < 5:

 

(-infinity, -4)∪(3, 5)

 

 

By the way, this line of reasoning only works because all of the function's zeros appear only once. If any zero appeared twice, the function would bounce off the x-axis instead of crossing it, and two regions next to each other might be both above the x-axis or both below it. To be safe, you should test points from each region:

 

1. Plug x = -5 into the initial equation to determine whether the function is positive or negative below -4

2. Since x = 0 makes f(x) positive, you know the function is positive between -4 and 3

3. Plug x = 4 into the initial equation to determine whether the function is positive or negative between 3 and 5

4. Plug x = 6 into the initial equation to determine whether the function is positive or negative above 5