Shifting Graphs: Free Response

A flat surcharge is added to whatever the company already charges, so it changes the OUTPUT of $C$, not the input. Adding a fixed amount to every output is exactly a vertical shift.

$\text{new cost}(x) = C(x) + 7.$

The weight $x$ itself never appears inside this rule in a new way, so nothing about which weights the company accepts has been touched yet.

A vertical shift changes only the output, so the set of accepted weights, the domain, is untouched:

$2 \le x \le 18 \quad \text{(unchanged)}.$

The range is exactly the old range with $7$ added to each end, since every cost the graph can produce is now $7$ dollars higher:

$9 + 7 = 16, \qquad 41 + 7 = 48 \quad\Longrightarrow\quad 16 \le C \le 48.$

A vertical shift, by definition, changes only the output of a function; the input is whatever it always was. Here the input is the package's weight, and the surcharge sits entirely on the output side, added to $C$.

$2 \le x \le 18 \quad \text{(same domain, before and after)}.$

So the customer's fear, that some weight near $18$ pounds might now be refused, describes something a HORIZONTAL shift could cause, not this vertical one. Every weight the company shipped before, it still ships, just at a higher price.

Test the claim with the definition rather than with how the sign looks. A horizontal shift asks: at what input does the INSIDE expression reach a value $f$ already produces? For $y = f(x-6)$, the inside is $x - 6$, and it reaches an old input $a$ only when $x = a + 6$, a LARGER number.

$x - 6 = a \quad\Longrightarrow\quad x = a + 6.$

So every feature of the graph appears at a larger $x$, a move to the right, not the left the classmate claimed. The classmate's error is treating $x-6$ as if subtracting made $x$ itself smaller, rather than asking what value of $x$ is needed to reproduce an old output.

Take one concrete case: a point sitting at input $7$ on $y = f(x)$, at height $f(7)$. On $y = f(x-6)$, that same height reappears where the inside equals $7$:

$x - 6 = 7 \quad\Longrightarrow\quad x = 13.$

The height $f(7)$ now shows up at input $13$, which is $6$ units to the right of $7$, not $6$ units to the left. This single check settles the disagreement: the graph moves right $6$, confirming part A and directly contradicting the classmate's conclusion.

Apply the two definitions in the stated order. Shifting $f$ right by $h$ means building the rule $q(x) = f(x-h)$. Shifting the RESULT $q$ up by $k$ means adding $k$ to its output:

$q(x) + k = f(x-h) + k.$

So doing the horizontal shift first and the vertical shift second gives $f(x-h)+k$.

Reverse the order. Shifting $f$ up by $k$ first means building the rule $p(x) = f(x) + k$. Shifting the RESULT $p$ right by $h$ second means replacing $x$ with $x - h$ inside $p$'s rule, not inside $f$ directly:

$p(x - h) = f(x-h) + k.$

So doing the vertical shift first and the horizontal shift second also gives $f(x-h)+k$.

Parts A and B computed the two orders and landed on the identical expression

$f(x-h) + k,$

with no restriction anywhere on which function $f$ was used or which real numbers $h$ and $k$ were chosen, so the agreement holds in general, not just for one case that happened to work out.

The reason is structural. The horizontal shift only ever changes what gets FED INTO $f$, and the vertical shift only ever changes what happens to the number that comes back OUT of $f$. Those are two different places in the expression, so performing one first can never disturb what the other does; the two operations simply do not touch the same thing.

To evaluate $g$ at a given input, find what the INSIDE expression, $x+3$, equals there, since that is the input $g$ actually feeds to $f$.

$g(-7) = f(-7+3) - 1 = f(-4) - 1.$

The marked point gives $f(-4) = -2$, so $g(-7) = -2 - 1 = -3$. The same idea gives the other two:

$g(-4) = f(-4+3)-1 = f(-1)-1 = 3-1 = 2, \qquad g(-1) = f(-1+3)-1 = f(2)-1 = -2-1 = -3.$

A shift only relocates the domain; it never changes its width. The rule $g(x) = f(x+3)-1$ has $x+3$ inside, which is $x - (-3)$, a horizontal shift, so every endpoint of $f$'s domain moves by the same amount used inside $g$:

$-4 - 3 = -7, \qquad 2 - 3 = -1 \quad\Longrightarrow\quad -7 \le x \le -1.$

Both moves used to build $g$ are translations. The horizontal one only decides WHICH input of $f$ a given input of $g$ reaches; the vertical one only ADDS a fixed amount to whatever height comes back. Neither one multiplies a length or a height by anything other than $1$, and only a factor different from $1$ can stretch, compress, or flip a graph.

$g(x) = f(x+3) - 1 \; \text{moves every point the same distance in the same direction; it scales nothing.}$

So although each value of $g$ traces back to a different input than the matching value of $f$, and sits at a different height, the whole family of points moves together, rigidly, which is exactly what keeps the shape identical.

The first coordinate only ever changes because of $h$, and the second only ever changes because of $k$, so read the two off separately.

$h = 9 - 2 = 7, \qquad k = -1 - 3 = -4.$

Apply the $h$ and $k$ from part A to the point $(5,-2)$: add $h$ to the input and $k$ to the output.

$(5 + 7,\; -2 - 4) = (12, -6).$

That is exactly the given correspondence, so the same shift explains the second point too.

Apply the same $h$ and $k$ to $(0,8)$: add $h$ to the input and $k$ to the output.

$(0 + 7,\; 8 - 4) = (7, 4).$

Two unknowns, $h$ and $k$, need two pieces of information to pin down, and a single correspondence already supplies exactly two: the change in the first coordinate fixes $h$, and the change in the second fixes $k$.

$h = 9 - 2, \qquad k = -1 - 3$

So the second correspondence was never needed to solve for anything; both unknowns were already determined by the first one alone. What it does instead is confirm that the very same $h$ and $k$ also explain a second, independent point, which is what shows the two graphs are related by one consistent translation, rather than by two point-coincidences that only happen to match a translation at a single spot.