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atroni [7]
3 years ago
15

="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.

Mathematics
1 answer:
dybincka [34]3 years ago
6 0

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.

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