Answer:
Step-by-step explanation:
Given function is,
[tex]f(x)=\frac{2x^2+6x}{-4x^2+8x}[/tex]
[tex]=\frac{2x(x+3)}{-4x(x-2)}[/tex]
This function is not defined at "x(x - 2) = 0"
Therefore, for the values of x = 0 and 2, given function is not defined.
Vertical asymptote → x = 2 [Denominator = 0]
For Horizontal asymptote → [tex][\frac{2x^2}{-4x^2} =-\frac{1}{2}][/tex]
Therefore, horizontal asymptote → y = [tex]-\frac{1}{2}[/tex]
x-intercept,
For f(x) = 0,
Numerator of the function = 0
2x² + 6x = 0
2x(x + 3) = 0
x = -3
Therefore, x - intercept → (-3, 0)
For y-intercept,
x = 0
f(0) = 0
Therefore, graph has no y-intercept.
For holes on the graph,
Since, function is not defined at x = 0,
f(x) = [tex]\frac{(x+3)}{-2(x-2)}[/tex]
f(0) = [tex]\frac{(0+3)}{-2(0-2)}[/tex]
= [tex]\frac{3}{4}[/tex]
Therefore, hole on the graph → [tex](0, \frac{3}{4})[/tex]
