Question

A polynomial f (x) has the given zeros of 6, –1, and –3.

Answer 1

Using the factor theorem the given zeros can be written in the form

f(x) = (x-6)(x+1)(x+3)   where (x-6)=0 ;(x+1)=0; (x+3) =0  

or

f(x) =  (x-6) (x²+4x+3)

f(x) =  x³-6x² +4x²-24x+3x-18

f(x) =  x³-2x²-21x-18

g(x)= $\frac{(x-6)(x+3) }{(x-2)}$

The equation of the slant asymptote is y= x-1

The polynomial g(x)= x²-3x-18/x-2 has infinite discontinuity at x=2

And x ∈ R: x≠ 2   where R= set of real numbers

Part B:

f (x)= (x² – x – 2)

can be written as

f (x) = (x² – 2x+x – 2)

or

f (x) = (x+1) ( x – 2)

Dividing the original polynomial with the second polynomial to get g(x)

g(x)= $\frac{(x-6)(x+1)(x+3) }{(x+1)(x-2)}$

g(x)= $\frac{(x-6)(x+3) }{(x-2)}$

g(x)= x²-3x-18/x-2

The slant asymptote is obtained by dividing the numerator by the denominator

         x -1                    

x-2√ x²-3x-18

        x² -2x

        -    +    

               -x-18

               -x+2

               +  -  

                   -20

The equation of the slant asymptote is y= x-1

The polynomial g(x)= x²-3x-18/x-2 has infinite discontinuity at x=2

When  the denominator becomes zero (2-2) then it cannot be defined therefore it is discontinuous.

And x ∈ R: x≠ 2   where R= set of real numbers

The graph shows that it is infinitely discontinuous at point 2

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