Question 1136720
<br>
I will assume you mean<br>
{{{(x^2+4x+3)/(2x+4)}}}<br>
and not<br>
{{{x^2+4x+3/2x+4}}}<br>
If you are working a problem like this, you should know that parentheses are required....<br>
(1) y-intercept: set x=0 --> 3/4<br>
(2) x-intercepts: where the function value is 0 -- i.e., where the numerator is 0.
{{{x^2+4x+3 = (x+1)(x+3) = 0}}}
so x-intercepts at -3 and -1<br>
(3) horizontal asymptote: none (the degree of the numerator is 1 more than the degree of the denominator; the function has a (linear) oblique asymptote.<br>
(4) vertical asymptote: where the denominator is 0 -- at x=-2.<br>
Perform polynomial division to find the oblique asymptote:
{{{(x^2+4x+3)/(2x+4) = ((x^2+4x+4)-1)/(2x+4) = ((x+2)^2-1)/(2(x+2)) = (x+2)/2-1/(2(x+2)) = (1/2)x+1 - 1/(2x+4)}}}<br>
The oblique asymptote is y = (1/2)x+1.<br>
A graph of the function and the oblique asymptote:<br>
{{{graph(400,400,-5,3,-5,5,(x^2+4x+3)/(2x+4),(1/2)x+1)}}}