Stretching and Reflecting Graphs: Free Response

Multiply each height by the vertical scale factor $3$, and leave every input alone.

$(-4,\ 3\times 2) = (-4, 6), \qquad (0,\ 3\times(-1)) = (0, -3), \qquad (4,\ 3\times 5) = (4, 15).$

Each input is untouched because $y=3f(x)$ only changes what comes OUT of $f$.

Divide each input by the horizontal scale factor $2$, and leave every height alone: the inside $2x$ reaches an old input $a$ exactly when $x=\tfrac{a}{2}$.

$\left(\tfrac{-4}{2}, 2\right) = (-2, 2), \qquad \left(\tfrac{0}{2}, -1\right) = (0, -1), \qquad \left(\tfrac{4}{2}, 5\right) = (2, 5).$

Because $2 > 1$, every input moved closer to $0$: this is a compression, not a stretch.

$y=3f(x)$ multiplies every output and leaves every input fixed, so only the graph's HEIGHT changed, and with $3>1$ that makes it taller. $y=f(2x)$ divides every input and leaves every output fixed, so only the graph's WIDTH changed, and with $2>1$ the inputs moved toward $0$, which narrows it rather than widening it.

$(x_0,y_0)\to(x_0,\ 3y_0), \qquad (x_0,y_0)\to\left(\tfrac{x_0}{2},\ y_0\right).$

The rule $y=-f(x)$ negates the OUTPUT and leaves the input alone, since the minus sign sits outside $f$.

$(4,\ -(-5)) = (4, 5).$

The rule $y=f(-x)$ negates the INPUT and leaves the output alone, since the minus sign sits inside $f$.

$\big(-(4),\ -5\big) = (-4, -5).$

In $y=-f(x)$, the minus sign is applied AFTER $f$ runs, to whatever comes out, so it can only touch the height. In $y=f(-x)$, the minus sign is applied to $x$ BEFORE $f$ ever sees it, so it can only touch the input; $f$ itself never learns the sign changed until it is handed $-x$.

$y=-f(x):\ (x_0,y_0)\to(x_0,-y_0), \qquad y=f(-x):\ (x_0,y_0)\to(-x_0,y_0).$

A single transformation touches whichever side of the point the negative sign sits on, never both.

Fix $x=p+h$ and check what $y=a\,f(x-h)+k$ produces there. The inside of $f$ becomes

$x-h = (p+h)-h = p,$

so $f(x-h)=f(p)=q$ at that input. Substituting into the whole rule,

$y = a\,f(x-h)+k = a\,f(p)+k = aq+k.$

So at $x=p+h$ the rule outputs $aq+k$, meaning $(p+h,\ aq+k)$ lies on the graph, for ANY $p$, $q$, $a$, $h$, $k$ satisfying $f(p)=q$, not merely for one example.

Apply the part A formula with $p=6$, $q=4$, $a=5$, $h=2$, $k=-3$.

$\big(6+2,\ \ 5(4)-3\big) = (8,\ 20-3) = (8, 17).$

The height is scaled by $5$ FIRST, giving $20$, and only then does the $-3$ get added.

The misremembered version scales $q+k$ together, as though the shift happened before $f$ was even evaluated. But part A showed the input's shift, $x=p+h$, has nothing to do with $k$ at all: $k$ never enters until AFTER $a$ has already multiplied $f(p)=q$.

$a(q+k) \;\neq\; aq+k \quad \text{whenever } k\neq 0 \text{ and } a\neq 1.$

Folding $k$ inside the multiplication would give $a(q+k)$ instead of the correct $aq+k$; on the part B numbers, $5(4+(-3))=5$, nowhere near the correct height.

A vertical scaling changes only the output, so the domain is untouched: $-3 \le x \le 9$. Each range endpoint is multiplied by $-3$, and a negative factor swaps which end is smaller.

$-3\times(-4) = 12, \qquad -3\times 2 = -6.$

Ordering the results from least to greatest gives the new range $-6 \le y \le 12$.

A horizontal scaling changes only the input, so the range is untouched: $-4 \le y \le 2$. Each domain endpoint is divided by $3$, since the inside $3x$ reaches an old input $a$ at $x=\tfrac{a}{3}$.

$\frac{-3}{3} = -1, \qquad \frac{9}{3} = 3.$

The new domain is $-1 \le x \le 3$.

$y=-3g(x)$ left the domain exactly as given, and $y=g(3x)$ left the range exactly as given, matching parts A and B. This is not tied to these particular numbers: the range is the set of OUTPUTS, and a factor multiplied with the input, as in $g(3x)$, is applied before $g$ ever runs, so it can only relocate WHICH input produces a given output, never change what that output IS.

$g(3x)\Big|_{x=\,x_0/3} \;=\; g(x_0) \;=\; y_0.$

The input moved from $x_0$ to $x_0/3$, but the output $y_0$ it reaches is exactly the same, so the set of outputs a function can produce is identical whether the input was scaled or not.

Line 1 restates the given point correctly, and Line 2 correctly identifies the condition $4x=6$ from the definition of $y=f(4x)$; both are sound. Line 3 is the first line that fails: solving $4x=6$ for $x$ requires DIVIDING both sides by $4$, not multiplying by it.

$4x = 6 \quad\Longrightarrow\quad x = \frac{6}{4} = 1.5,$

not $x=4\times 6$. Line 4 inherits this error directly; line 5 then states the direction rule the wrong way round as well, but line 3 is the first line that fails.

Using the corrected input, the point is $(1.5, 5)$. Checking it against the general rule with $b=4$ and $(x_0,y_0)=(6,5)$:

$\frac{x_0}{b} = \frac{6}{4} = 1.5,$

which matches. The two routes, solving $4x=6$ directly and applying the general formula, agree once the equation is solved correctly.

The equation $bx=x_0$ names $x$ as a factor multiplied by $b$; undoing a multiplication by $b$ means dividing by $b$, the same operation that undoes $4x=6$ in any other linear equation. Multiplying by $b$ instead does not undo anything: it produces a new input $b$ TIMES too far from $0$, not $\tfrac{1}{b}$ as far.

$x = \frac{x_0}{b} \quad\text{(correct)}, \qquad x = b\,x_0 \quad\text{(the slip)}.$

That is exactly the shape of Line 3's mistake, and it is also exactly why the graph looked like a stretch by a factor of $4$: an input $b$ times too far out is precisely what a genuine stretch would look like, so the arithmetic error manufactures the appearance of the wrong transformation.