Answer:
The domain of the function is the set of all real numbers except 2.
When x ⇒ 2⁻, f(x) ⇒ +∞
When x ⇒ 2⁺, f(x) ⇒ -∞
x = 2 is a vertical asymptote.
When x ⇒ +∞, f(x) ⇒ -∞
When x ⇒ -∞, f(x) ⇒ +∞
At x = 0 and x = 3/2 the function intersect the x-axis.
Explanation:
Hi there!
First, let´s write the fucntion:
f(x) = (3x - 2x²) / (x - 2)
Let´s find the value of x for which the denominator is zero. That value will not be included in the domain of the function:
x - 2 = 0
x = 2
Then, the domain of the function is the set of all real numbers except 2.
If the function tends to infinity (∞) when "x" tends to 2, then x = 2 is a vertical asymptote. So let´s evaluate the behavior of the function when we near to 2 from left (x tends to 2 from the left = x ⇒2⁻) and from the right (x tends to 2 from the right = x ⇒2⁺).
When x ⇒ 2⁻
f(x) ⇒ 3 · 2 - 2 · 2²/2 - 2 ⇒6 - 8 /0 ⇒ -2/0 (since x tends to 2 from the left, x<2, then, x - 2<0).
Then, a number divided by a very small number (almost zero) gives infinty. In this case, both numbers are negative, therefore:
When x ⇒ 2⁻, f(x) ⇒ +∞
The same procedure should be done to evaluate the behavior of the function when x nears 2 from the right. In this case, x>2, then x-2>0. The numerator will be the same, -2. Therefore:
When x ⇒ 2⁺, f(x) ⇒ -∞
Then, x = 2 is a vertical asymptote.
Now let´s find if the function tends to a certain value when it tends to infinty. In that case, that value will be a horizontal asymptote of f(x). So:
When x ⇒ +∞
f(x) ⇒ (3x - 2x²) / x (x>>>2, then it can be considered that x - 2 ≈ x)
Let´s apply distributive property:
f(x) ⇒ 3x/x - 2x²/x
f(x) ⇒ 3 - 2x ⇒ -2x ⇒ -∞
Then, when x ⇒ +∞, f(x) ⇒ -∞, there is no asymptote.
In the same way, when x ⇒ -∞
f(x) ⇒ -2x ⇒ +∞
Now, let´s find the zeros of the function:
f(x) = (3x - 2x²) / (x - 2)
0 = (3x - 2x²) / (x - 2)
0 = 3x - 2x²
0 = x(3- 2x) (x = 0)
0 = 3-2x
-3 = -2x
-3/-2 = x
x = 3/2
Then at x = 0 and x = 3/2 the function intersect the x-axis.
Attached, you can find the graph of the function.