Answer:
Horizontal Asymptote: None (degree of numerator is larger than denominator)
Vertical Asymptote: None (domain of function is [tex]\mathbb{R}[/tex])
x-intercept: (0, 0)
y-intercept: (0, 0)
Hole: (-2, 1)
Step-by-step explanation:
HA: Since the degree of the numerator (2) is larger than the degree of the denominator (1), no HA exists.
VA: Since the domain of the function of is [tex]\mathbb{R}[/tex] (all real numbers), no VA exists. *no real value of x will make the denominator equal to 0 and numerator [tex]\neq[/tex] 0
x-intercept:
The x-intercept occurs at [tex]y=0[/tex].
Solving for x:
[tex]0=\frac{-2x^2-4x}{4x+8}[/tex]:
[tex]0=-2x^2-4x,\\x=0[/tex]. Therefore, the x-intercept is (0, 0).
y-intercept:
The y-intercept occurs when [tex]x=0[/tex].
Solving for y:
[tex]\frac{-2x^2-4x}{4x+8}, x=0,\\\frac{-2(0)^2-4(0)}{4(0)+8}=0[/tex]. Therefore, the y-intercept is (0, 0).
Hole:
When looking for a hole, we are looking for a factor that occurs in both the numerator and denominator:
[tex]\frac{-2x^2-4x}{4x+8}=\frac{-2x(x+2)}{4(x+2)}[/tex]
Rewriting our function here, we see that the term [tex](x+2)[/tex] occurs in both the numerator and denominator. Thus, when this term is set to 0, it will make both the numerator and denominator 0, thus rendering an undefined value (since we cannot divide by 0). We call this a hole. *Note: Usually when an x-value makes the denominator zero, this will create a vertical asymptote. However, if the x-value also makes the numerator zero, it's considered a hole, not an asymptote. The x-value of this hole will be the value of when the common term is equal to 0:
[tex]x+2=0,\\x=-2[/tex]
To find the y-value of the hole, remove the common term and plug in your x-value:
[tex]\frac{-2x^2-4x}{4x+8}=\frac{-2x(x+2)}{4(x+2)}=\frac{-2x}{4},\\\frac{-2(-2)}{4}=\frac{4}{4}=1[/tex]
Therefore, the hole of this function is at (-2, 1).
