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34kurt
3 years ago
9

1) Has zeros at -2 and 4, both being double roots 2) As x→∞, y→−∞

Mathematics
1 answer:
irina1246 [14]3 years ago
3 0
-(x+2)^2(x-4)^2

Not that if x= -2 and x=4 they will look like this in an equation: (x+2) and (x-4)
When (x+2) and (x-4) are set equal to zero and you solve for x, x will equal x= -2 and x=4

If they have double roots, they have a multiplicity of 2 (per root) meaning they will bounce off of the x-axis. Multiplicity can be found by using exponents.

The negative in front flips the function over the x-axis and holds true to the given limit.
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i) The given function is

f(x)=\frac{x-5}{2x^2-5x-3}

The factored form is

f(x)=\frac{x-5}{(x-3)(2x+1)}

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x\ne3,x\ne-\frac{1}{2}

ii) To find the vertical asymptotes, equate the denominator to zero.

(x-3)(2x+1)=0

(x-3)=\ne0,(2x+1)=0

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iii) To find the roots, equate the numerator to zero.

x-5=0

The root is x=5

iv) To find the y-intercept, put x=0 into the function.

f(0)=\frac{0-5}{(0-3)(2(0)+1)}

f(0)=\frac{-5}{(-3)(1)}

f(0)=\frac{5}{3}

The y-intercept is \frac{5}{3}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0

The horizontal asymptote is y=0

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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Answer:

I think the answer is 7.20.....

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