Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.[tex]g(x) = \frac{(2x+1)(x-5)}{(x-5)(x+4)^{2} }[/tex]

Accepted Solution

Answer:1. Horizontal Asymptote is y = 02. Vertical Asymptote is x = -43. No Slant Asymptote4. Hole at (5, 0.14)5. x-intercepts:x-intercept [tex](-\frac{1}{2},0)[/tex]y-intercepts:y-intercept Β [tex](0,\frac{1}{16})[/tex]6.Domain is [tex]{x|x\neq -4,5}[/tex]Step-by-step explanation:1. Horizontal Asymptotes* If the degree of the numerator is less than the degree of the denominator (this is our case since multiplying will give the numerator a degree of 2 and denominator a degree of 3), then y = 0 is the only horizontal asymptoteHorizontal Asymptote is y = 02. Vertical Asymptotes* To get VA (vertical asymptote), we set the denominator equal to zero.Before doing this, we see that we can cancel out (x-5) from both numerator and denominator so the denominator becomes (x+4)^2. Now we find VA:[tex](x+4)^2=0\\x+4=0\\x=-4[/tex]Vertical Asymptote is x = -43. Oblique asymptotes* If the degree of numerator is less than the degree of the denominator (this is our case as explained above), then there is no slant asymptote.No Slant Asymptote4. HolesThere is hole in a rational function if there is the same factor in both numerator and denominator (before simplifying, only after factoring). Set that equal to 0 and solve. Then, cross out the common factor and put the x-value into the function and get the y-value of the hole.We can see that there is a factor of (x-5) in both the numerator and denominator. We set it equal to 0 and solve for x:[tex]x-5=0\\x=5[/tex]Putting x = 5, we get:Y value of hole = [tex]g(x)=\frac{2x+1}{(x+4)^2}\\g(5)=\frac{2(5)+1}{(5+4)^2}\\g(5)=0.14[/tex]Hole at (5, 0.14)5. InterceptsTo get x-intercepts, we set y = 0 (g(x) = 0) and for y-intercepts we set x = 0.x-intercepts:[tex]0=\frac{2x+1}{(x+4)^2}\\2x+1=0\\2x=-1\\x=-\frac{1}{2}[/tex]x-intercept [tex](-\frac{1}{2},0)[/tex]y-intercepts:[tex]y=\frac{2x+1}{(x+4)^2}\\y=\frac{2(0)+1}{(0+4)^2}\\y=\frac{1}{16}[/tex]y-intercept Β [tex](0,\frac{1}{16})[/tex]6. DomainThis is the set of allowed x-values of the function. We simply disregard any value that would make the denominator equal to 0.So we have:x - 5 = 0, x = 5and(x+4)^2 = 0, x = -4Domain is the set of all real numbers x EXCEPT x = -4 and x = 5Domain is [tex]{x|x\neq -4,5}[/tex]