PARCC Algebra II Building Functions. Practice it free.

The common difference is $d = 6$. With $a_1 = 8$, the $18$th term is $a_{18} = 8 + 17(6) = 110$.

Work from the outside in. Since $j(4) = 5$, we need $k(l(x)) = 4$. Since $k(1) = 4$, we need $l(x) = 1$. From the table, $l(5) = 1$, so $x = 5$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $9 \cdot 3^{0}$
2nd term, or $n = 2\textsf{:}$ $9 \cdot 3^{1}$
3rd term, or $n = 3\textsf{:}$ $9 \cdot 3^{2}$
4th term, or $n = 4\textsf{:}$ $9 \cdot 3^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $a_n = 9 \cdot 3^{n-1}$.

Start with the inner function: $4 \geq 0$, so $f(4)=1-2(4)=-7$. Then $-7 \leq 1$, so $g(-7)=3(-7)=-21\textsf{.}$

The common difference is $14 - 20 = -6\textsf{.}$ From the sixth term to the ninth term there are $3$ steps, so the ninth term is $14 + (-6) \times 3 = -4\textsf{.}$

Set $y = 1 - 3x\textsf{.}$ Swap: $x = 1 - 3y\textsf{.}$ Then $3y = 1 - x\textsf{,}$ so $y = \dfrac{1 - x}{3}\textsf{.}$ So $f^{-1}(x) = \dfrac{1 - x}{3}\textsf{.}$

Replacing $x$ with $x + 5$ is equivalent to $f(x - (-5))$, so the input must be $5$ smaller to produce the same output and the graph shifts $5$ units left.

To shift right, $x$ gets larger but the height stays the same, so $6$ is subtracted inside $f$. Adding $1$ outside $f$ is a separate move up.