Question

Help me with this math question please I'm giving away brainliests ​

Answer 1

Answer:

Step-by-step explanation:

Given function is,

$f(x)=\frac{2x^2+6x}{-4x^2+8x}$

       $=\frac{2x(x+3)}{-4x(x-2)}$

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 → $[\frac{2x^2}{-4x^2} =-\frac{1}{2}]$

Therefore, horizontal asymptote → y = $-\frac{1}{2}$

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) = $\frac{(x+3)}{-2(x-2)}$

f(0) = $\frac{(0+3)}{-2(0-2)}$

     = $\frac{3}{4}$

Therefore, hole on the graph → $(0, \frac{3}{4})$