Question

="TexFormula1" title="f(x) = \frac{(2x+1)(x-5)}{(x-5)(x+4)^{2} }" alt="f(x) = \frac{(2x+1)(x-5)}{(x-5)(x+4)^{2} }" align="absmiddle" class="latex-formula">
Domain:
V.A:
Roots:
Y-int:
H.A:
Holes:
O.A:

Also, draw on the graph attached.

Answer 1

i) The given function is

$f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}$

The domain is

$(x-5)(x+4)^2\ne 0$

$(x-5)\ne0,(x+4)^2\ne 0$

$x\ne5,x\ne -4$

ii) For vertical asymptotes, we simplify the function to get;

$f(x)=\frac{(2x+1)}{(x+4)^2}$

The vertical asymptote occurs at

$(x+4)^2=0$

$x=-4$

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

$2x+1=0$

$2x=-1$

$x=-\frac{1}{2}$

iv) To find the y-intercept, we substitute $x=0$ into the reduced fraction.

$f(0)=\frac{(2(0)+1)}{(0+4)^2}$

$f(0)=\frac{(1)}{(4)^2}$

$f(0)=\frac{1}{16}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0$

The horizontal asymptote is $y=0$.

vi) The function has a hole at $x-5=0$.

Thus at $x=5$.

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.