Question

https://drive.google.com/file/d/1pTlataLgJOPDtv7E5gmfG4QnekPcoK5E/view

I have tried all of these questions and I can't come up with the right answers. Need some help.

Answer 1

I'll help you with a few so that you can get an idea

Q1)
\[ \text{What is the simplified form of} \frac{4z^2+16z+15}{2z^2-z-15}?\]
\[\frac{4z^2+16z+15}{2z^2-z-15}\Rightarrow\frac{(2z+3)(2z+5)}{(2z+5)(z-3)} \Rightarrow \frac{{(2z+3)}\cancel{(2z+5)}}{\cancel{(2z+5)}(z-3)} \Rightarrow \frac{2z+3}{z-3}\]

You should factor the numerator and the denominator. You will get a common factor that you can cancel.
In case you don't know how to factor, check this out --> https://www.dummies.com/education/math/pre-calculus/how-to-factor-a-polynomial-expression/

Answer 2

With Q2 you are just going to multiply and employ the same method as described above (i.e., factor and cancel). For Q3, since there's a division sign, there's one extra step you need to (take the reciprocal of one of the terms and then multiply).

Answer 3

Q7
\[\text{What are the discontinuities of the function}~f(x) = \frac{x^2-16}{6x-24}?\]
One way you can find the discontinuity is to find a value of x that sets the denominator to 0
6x-24=0
x=4

The more appropriate way is to apply the conditions of continuity, although it might not be required at your level..
\[1) \lim_{x\to c^+} f(x)=\lim_{x\to\ c^-} f(x)\]
\[2) f(c)~\text{exists}\]
\[3) \lim_{x\to c} f(x)=f(c)\]

\[\lim_{x\to 4}~x^2-16=0\]
\[\lim_{x\to 4}~6x-24=0\]
Since the limit yields an indeterminate form of \(\frac{0}{0}\), we can apply L'Hopital's rule..
\[\lim_{x\to 4}~2x=8\]
\[\lim_{x\to 4}~6=6\]
\[\lim_{x\to 4} \frac{x^2-16}{6x-24}=\frac{8}{6} = \frac{4}{3}\]
We can conclude that the limit exists.
However, f(x) at x=4 does not exist.
Thus we can conclude that \(\lim_{x\to 4} f(x)≠f(4)\)...a removable discontinuity at x=4

Answer 4

For question 8) you have to find the discontinuity again by prescribing the similar method above. You also need to see whether f is increasing or decreasing on the given interval...you could just look at the sign assuming you don't know the first derivative test
for 9, oblique asymptotes can be found by performing long division. Let me know if you need help on that.
Q10 is the same idea as 8

Answer 5

Q4 and Q5 is very similar to adding/subtracting fractions with numbers instead of variables. You must find the LCD so that you can add, and you do that by evaluating common denominators or multiplying the two. They already gave the LCD for Q4, but you need to do that for Q5. See this if you don't understand --> https://opentextbc.ca/elementaryalgebraopenstax/chapter/add-and-subtract-rational-expressions-with-unlike-denominators/