NJSLA Algebra II Building Functions. Practice it free.

Work from the outside in. Since $p(5) = 1$, we need $q(r(x)) = 5$. Since $q(4) = 5$, we need $r(x) = 4$. From the table, $r(3) = 4$, so $x = 3$.

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

The common difference is $d = 3$. With $a_1 = 1$, the $16$th term is $a_{16} = 1 + 15(3) = 46$.

Replacing $x$ with $3x$ means outputs occur at one-third the original input values, so the graph is compressed horizontally by a factor of $\dfrac{1}{3}$.

To shift left, $x$ gets smaller but the height stays the same, so $4$ is added inside $f$. That is why you see $(x + 4)$.

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

A downward shift of $3$ replaces $y$ with $y + 3$. At the $x$-intercept, $y = 0$, so $3x + 2(3) = 12$. Then $3x + 6 = 12$, $3x = 6$, and $x = 2$.

Terms: $8, 4, 2$. Sum $= 8 + 4 + 2 = 14$.