Rom

$\text{The possible rational roots are of the form }\dfrac p q \\ \text{where }p \text{ is a factor of the constant term, in this case -4}\\ \text{and }q \text{ is a factor of the highest order term, i.e. 2}\\ \text{so possible roots are}\\ \text{factors of -4 are} \pm 1,~\pm 2,~\pm 4\\ \text{factors of 2 are }\pm 1,~\pm 2\\ \text{so our possible roots are}\\ x=\pm \dfrac 1 2,~\pm 1,~\pm 2,~\pm 4$

$\text{we plug these into }f(x) \text{ to determine which are actually roots}\\ \text{and see that}\ x = -2,~-\dfrac 1 2,~2\\ \text{are the only actual roots, and as }f(x) \text{ is degree 3 that is all 3 of them}\\ \text{so }f(x) = c(x+2)\left(x+\dfrac 1 2\right)(x-2), \text{ for some }c \in \mathbb{Q} \\ c \text{ is the coefficient of the }x^3 \text{ term so}\\ f(x) = 2(x+2)\left(x+\dfrac 1 2\right)(x-2)$

$6 = \log\left(\dfrac{I_1}{I_s}\right) \\ 8=\log\left(\dfrac{I_2}{I_s}\right) \\ 10^6=\dfrac{I_1}{I_s}\\ 10^8=\dfrac{I_2}{I_s} \\ \dfrac{I_2}{I_1} = \dfrac{10^8}{10^6} = 10^2 = 100$

$\text{suppose she has to make }n \text{ consecutive throws}\\ \dfrac{12+n}{15+n} = 0.85 \\ 12+n = 12.75 + 0.85n\\ 0.15n =0.75 \\ n=5$

1)

$\dfrac{5}{\frac 1 6 + \frac{1}{x+1}} =\\ \dfrac{5}{\frac{x+1+6}{6(x+1)}} =\\ \dfrac{5\cdot 6(x+1)}{x+7} = \\ \dfrac{30(x+1)}{x+7}$

for (1) ... what is f(0)?  With that just look at the y intercept of the graphs

(2) same as (1)

(3) as x goes to +infinity g(x) goes off to infinity so that's not an asymptote

as x goes to -infinity 3^x goes to 0 and so g(x) goes to 6 which is an asymptote

$\dfrac 8 q + 2 = \dfrac{q+4}{q-1} \\ \dfrac{8+2q}{q} = \dfrac{q+4}{q-1} \\ (8+2q)(q-1) = q(q+4) \\ 2q^2+6q-8 = q^2 + 4q \\ q^2 + 2q-8=0 \\ (q+4)(q-2) = 0 \\ q= -4, ~2$

$200 = R + P + C = \\ \\ R + (3R+4) + C =\\ \\ 4(R+1)+C\\ \\ C = 196-4R$

$E[C] =\\ E[196-4R]=\\ 196-4E[R] = \\ 196-4(2) = 188$

$Var[C] = Var[196-4R] = \\ 4^2 Var[R] = \\ 16Var[R] =\\ 16(1.2) = 19.2$

$E[a X + b] = a E[X] + b\\ \\ Var[a X + b]= a^2 Var[X]\\ \\ E[Y] = E[100X-50] = 100E[X]-50\\ \\ Var[Y] = Var[100X-50] = 10000Var[X]$

I bet you have E[X] and Var[X] from some other part of the problem so plug those in.

$Var[a X + b] = a^2 Var[X]$

$Var[3Y+2] = 9 Var[Y] = 11.61\\ \\ Var[1-2Y] = 4 Var[Y] = 5.16$

$E[aX + b] = aE[X] + b\\ V[a X + b] = a^2V[X]$

$\tilde{R} = 1-3R\\ E[\tilde{R}] = E[1-3R] = \\ 1 - 3E[R] =\\ 1-3\left(\dfrac{24}{11}\right)=\\ 1-\dfrac{72}{11} = -\dfrac{61}{11}$

$V[\tilde{R}] = V[1-3R] = \\ (3)^2 V[R] = 9\left(\dfrac{95}{121}\right) = \dfrac{855}{11}$

$\text{Let }N \text{ be the number of games in the season}\\ \text{and let }p \text{ be the percentage of games of the last 20% of the season that must be won}\\ \text{the total wins equals the total losses}\\ (0.8)(0.4)N + (0.2)pN = (0.8)(0.6)N + (0.2)(1-p)N \\ 0.32 + (0.2)p = 0.48 + 0.2 - (0.2)p \\ (0.4)p = 0.36 \\ p = 0.9 = 90\%$

You can see that the N wasn't really needed as it immediately cancels out 

but I think it makes things a bit clearer.