Mathematics Problems (Math Problem Sample)

Quiz 3
Name
Institution
Date
QUESTION 1
Part a
Apply log rule : loga(1/x)=-logax……………….(1)
Simplify loga(a3)
Apply log rule :loga(xb)=b logax…………….(2) assuming x≥0
=3loga(a)
Apply log rule :loga(a)=1………………………(3)
3loga(a)=3
This implies that loga(1/a3)=-3 from………….(1)
Hence answer is -3
Part b
Rewrite 3 in power-base form
3=√9
Log9(3)=log9√9
Apply log rule :logaxb=b logax
Log9√9=1/2log9(9)
Apply log rule :logaa=1
Log99=1
This implies that 1/2log9(9)=1/2
Hence final answer is ½
QUESTION 2
Part a
For rational functions ,the vertical asymptotes are the undefined point ,also known as the zeros of the denominator ,of the simplified function.
Take the denominator of (6x2-x-2)/(x2+3x-28) and compare to zero
X2+3x-28=0
For a quadratic equation of the form ax2+bx+c=0
X1,2=(-b±√b2-4ac)/2a
For a =1,b=3 and c=-28
X1,2=(-3±√32-4*1*-28)/2*1
X1=4 and x2=-7
The vertical asymptotes are x=-7 and x=4
Part b
X intercept is a point on the graph where y=0
(6x2-x-2)/(x2+3x-28)=0
f(x)/g(x)=0 .This implies that f(x)=0
6x2-x-2=0
a=6,b=-1 and c=-2
X1,2=(-b±√b2-4ac)/2a
X1,2=(-(-1)±√(-1)2-4*6*-2)/(2*6)
X1=2/3 and x2=-1/2
This implies that x intercepts are (2/3,0) and (-1/2,0)
The y intercept is the point on the graph where x=0
Y=(6*02)-0-2/02+3*0-28
=-2/-28
=1/14
This implies that y intercept is (0,1/14)
Part c
If denominator’s degree>numerator’s degree,then the horizontal asymptote is the x axis(y=0)
If numerator’s degree =1+denominator’s degree,then the asymptote is a slant asymptote of the form y=mx+b
If the degrees are equal,the asymptote is y=(numerator’s leading coefficient)/(denominator’s leading coefficient)
If the numerator’s degree>1+denominator’s degree,there is no horizontal asymptote
The degree of the numerator=2.The degree of the denominator =2.
The degrees are equal.The asymptote is y=(numerator’s leading coefficient)/(denominator’s leading coefficient)
Numerator’s leading coefficient=6
Denominator’s leading coefficient =1
Therefore ,y=6/1
The horizontal asymptote is y=6
Part d
Part e
Use the degree and the leading coefficient to determine the behavior.It rises to the left and rises to the right.
Part f
Factor x2+3x-28
=(x2-4x)+(7x-28)
X2-4x=x(x-4)
7x-28=7(x-4)
=(x2-4x)+(7x-28)=x(x-4)+7(x-4)
=(x-4)(x+7)
Factor 6x2-x-2
=(6x2+3x)+(-4x-2)
6x2+3x=3x(2x+1)
-4x-2=-2(2x+1)
=(6x2+3x)+(-4x-2)
=3x(2x+1)-2(2x+1)
=(2x+1)(3x-2)
((2x+1)(3x-2))/((x-4)(x+7))≥0
Compute the sign of the factors of ((2x+1)(3x-2))/((x-4)(x+7))
2x+1 is zero for x=-1/2
2x+1 is negative for x<-1/2
2x+1 is positive for x>-1/2
3x-2 is zero for x=2/3
3x-2 is negative for x<2/3
3x-2 is positive for x>2/3
x-4 is zero for x=4
x-4 is negative for x<4
x-4 is positive for x>4
x+7 is zero for x=-7
x+7 is negative for x<-7
x+7 is positive for x>-7
Choosing ranges that satisfy the required conditions :x≥0
X<-7 or -1/2≤x≤2/3 or x>4
QUESTION 3
We can take g(x)=(x-1)3 and f(x)=1/x
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