Answer:
Step-by-step explanation:
Given function in the question is,
[tex]f(x)=\frac{-2x^2-4x}{4x+8}[/tex]
[tex]=\frac{-2x(x+2)}{4(x+2)}[/tex]
[tex]=-\frac{1}{2}x[/tex] If [tex]x\neq -2[/tex]
Given function is not defined at x = -2
At x = -2,
[tex]f(-2)=\frac{1}{2}(2)[/tex]
[tex]=1[/tex]
Therefore, there is a hole in the graph at (-2, 1).
Graph of the function 'f' will be a straight line with a hole at (-2, 1).
Horizontal asymptote → None
Vertical asymptote → None
x-intercept → No x-intercept [Line passes through origin (0,0)]
y-intercept → No y-intercept [Line passes through origin (0,0)]
Hole → (-2, 1)
