Answer:
When x ⇒ 4⁺ (from the rigth), f(x) ⇒+∞
When x ⇒ 4⁻ (from the left), f(x) ⇒-∞
When x ⇒ -4⁺, f(x) ⇒-∞
When x ⇒ -4⁻, f(x) ⇒+∞
Asymptotes:
x = 4
x = -4
y = 2
Step-by-step explanation:
Hi there!
Let´s write the function:
f(x) = (2x² - x) / (x² - 16)
Let´s find the x-value at which the denominator is zero:
x² - 16 = 0
x² = 16
√x² =√16
x = 4 and x = -4
Let´s evaluate the behavior of the function when x ⇒ 4 and x⇒-4 to see if x = 4 and x = -4 are vertical asymptotes:
When x ⇒ 4⁺ (from the rigth)
f(x) ⇒ 28/ 0 ⇒ +∞
when x ⇒ 4⁻ (from the left)
f(x) ⇒ 28 / 0 ⇒ -∞ because x² -16 <0 for being x²< 16
When x ⇒ -4⁺
f(x) ⇒ 36/0 ⇒ -∞ (for the same reason as when x ⇒ 4⁻)
when x ⇒ -4⁻
f(x) ⇒ 36/0 ⇒ +∞
Then, x = 4 and x = -4 are vertical asymptotes.
Let´s evaluate the end behaivor of the function:
f(x) = (2x² - x) / (x² - 16)
When x ⇒ +∞
f(x) ⇒ (2x² - x) / x²
f(x) ⇒ 2x²/x² - x/x²
f(x) ⇒ 2 - 1/x (1/x ⇒ 0 because 1 divided by a very big number is nearly zero). Then:
f(x) ⇒ 2
When x ⇒ -∞
f(x) ⇒ 2 - 1/x ⇒ 2
Then, y = 2 is a horizontal asymptote.
