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1 answer:
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The change of base rule says
If b = 1, then we will have
which is NOT possible. We cannot divide by zero. So this is why b = 1 is NOT allowed.
If b = 1, then we will have
which is NOT possible. We cannot divide by zero. So this is why b = 1 is NOT allowed.
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)=
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)=
g(x)=
g(x)= x²-3x-18/x-2
The slant asymptote is obtained by dividing the numerator by the denominator
<u> x -1 </u>
x-2√ x²-3x-18
x² -2x
<u> - + </u>
-x-18
-x+2
<u> + - </u>
-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
