Transformations of Graphs: Free Response

The height $8$ belongs to the input $1$ of $f$, so on the new graph it reappears at whichever $x$ makes the inside equal $1$.

$x + 5 = 1 \ \Longrightarrow\ x = -4$

The outside $5$ is added to whatever height $f$ returns, so that height becomes $8 + 5 = 13$. The single point the given fact determines is $(-4, 13)$.

Read what the rule actually computes at the input $1$.

$f(1 + 5) + 5 = f(6) + 5$

The inside carries the input $1$ to the input $6$ of $f$, and nothing at all is known about $f$ there. The given fact is about the input $1$ of $f$, which the new rule reaches from a different place, so that fact settles the height somewhere else rather than here. One further value, $f(6)$, would settle this one.

The two $5$s are applied at different moments. The outside $5$ meets a number $f$ has already produced, so it does exactly what it says: every height gains $5$.

The inside $5$ meets $x$ before $f$ has run, so it changes the question rather than the answer. To find where an old input $u$ reappears, ask which $x$ makes the inside equal $u$, and answering that means solving for $x$.

$x + 5 = u \ \Longrightarrow\ x = u - 5$

The solving step subtracts what the rule added, which is why an inside plus produces a move to the left. The outside never has to be solved, so it never reverses.

$C$ expects a number of minutes, and $t$ hours is $60t$ minutes, so the conversion has to reach the time before $C$ is applied to it.

$H(t) = C(60t)$

That is an inside change with $b = 60$, so every time coordinate is divided by $60$. The logged reading reappears where the inside equals $45$.

$60t = 45 \ \Longrightarrow\ t = \tfrac{3}{4}$

Nothing acted on the temperature, so the height is still $80$ degrees Celsius and the reading sits at $\left(\tfrac{3}{4}, 80\right)$, three quarters of an hour after switch-on.

The Fahrenheit conversion is applied to a temperature $C$ has already produced, so it sits outside $C$, with $a = \tfrac{9}{5}$ and $k = 32$.

$F(m) = \tfrac{9}{5}\,C(m) + 32$

Nothing acted on the time, so the logged reading keeps its $45$ minutes, and its height converts with the multiplication carried out before the addition.

$\tfrac{9}{5}(80) + 32 = 144 + 32 = 176$

The reading sits at $(45, 176)$, that is $176$ degrees Fahrenheit at $45$ minutes.

A constant can only act on whatever is in front of it at the moment it is applied. The Fahrenheit factor and the $32$ meet a temperature $C$ has already returned, so they can reach the second coordinate and nothing else. The $60$ meets a time $C$ has not yet seen, so it can reach the first coordinate and nothing else.

That is also why the two multipliers pull in opposite directions. The outside factor multiplies each height by $\tfrac{9}{5}$, while the inside factor divides each time by $60$, because the new time is whatever solves the inside.

$60t = 45 \ \Longrightarrow\ t = \tfrac{3}{4}, \qquad \text{whereas } 45 \times 60 = 2700$

Multiplying the logged time by $60$ would place the reading $2700$ hours after switch-on, roughly four months later, rather than at the same instant counted in a larger unit.

Factor the input multiplier out of both terms, not only out of the term carrying $x$.

$4x + 10 = 4\left(x + \tfrac{5}{2}\right) = 4\left(x - \left(-\tfrac{5}{2}\right)\right)$

Matching this against $b(x - h)$ gives $b = 4$ and $h = -\tfrac{5}{2}$. Every input is therefore divided by $4$, a horizontal compression by $4$, and the graph is then shifted by $h$, which is $\tfrac{5}{2}$ to the left.

The height $-6$ reappears where the inside equals the old input $2$.

$4x + 10 = 2 \ \Longrightarrow\ 4x = -8 \ \Longrightarrow\ x = -2$

Nothing outside $f$ was changed, so the height stays $-6$ and the landing point is $(-2, -6)$. The factored reading has to agree, because it is the same equation solved in two named steps, divide by $b$ and then add $h$.

$\frac{2}{4} + \left(-\tfrac{5}{2}\right) = \tfrac{1}{2} - \tfrac{5}{2} = -2$

To shift left $10$ after the compression, the $10$ has to sit in the $h$ slot of $b(x - h)$, so that it is the shift rather than a distance the compression has yet to shrink.

$y = f\big(4(x + 10)\big) = f(4x + 40)$

That is not the rule in the stem, so the graph of $y = f(4x + 10)$ is not the compressed graph moved $10$ to the left.

Something in the student's reading is real, though. Shift the graph of $f$ left $10$ first, building $f(x + 10)$, and then compress that result by $4$ by replacing $x$ with $4x$:

$f(x + 10) \ \longrightarrow\ f(4x + 10)$

So the rule does come from a shift of $10$, but from one performed before the compression. The compression that follows squeezes every horizontal distance by a factor of $4$, that displacement included, which leaves $\tfrac{10}{4} = \tfrac{5}{2}$ once the two moves are read in the order the finished graph shows them, scaling first and shifting second. That is exactly the $h$ the factoring produced.

Reflecting the input builds the rule $g(x) = f(-x)$. Shifting that graph right $4$ replaces $x$ by $x - 4$ in the rule of $g$, not back in the rule of $f$.

$g(x - 4) = f\big(-(x - 4)\big) = f(4 - x)$

Shifting first builds $p(x) = f(x - 4)$. Reflecting that graph negates the input of the rule of $p$, so the $-4$ is already inside whatever gets negated.

$p(-x) = f(-x - 4) = f\big(-(x + 4)\big)$

In each order, the height $f(6)$ appears where that order's inside equals $6$.

$-(x - 4) = 6 \ \Longrightarrow\ x = -2, \qquad -(x + 4) = 6 \ \Longrightarrow\ x = -10$

The same point of $f$ is sent to two different places, $8$ apart. Nothing about the input $6$ was special: for a general old input $u$ the first order lands at $4 - u$ and the second at $-u - 4$, and those two differ by $8$ whatever $u$ is. So the two finished graphs are congruent, each being the other slid $8$ along the horizontal axis.

Redo both orders with the shift left as a letter. Reflecting first and then shifting right $h$ gives $f\big(-(x - h)\big)$, while shifting right $h$ first and then reflecting gives $f\big(-(x + h)\big)$. Ask where an old input $u$ lands in each.

$-(x - h) = u \ \Longrightarrow\ x = h - u, \qquad -(x + h) = u \ \Longrightarrow\ x = -u - h$

Those two landings agree for every $u$ exactly when $h = -h$, since the $-u$ cancels from both sides.

$h - u = -u - h \ \Longleftrightarrow\ 2h = 0 \ \Longleftrightarrow\ h = 0$

When $h = 0$, then, both orders carry every point of $f$ to the same place, $-u$, so the finished graphs are identical, whatever $f$ is.

That settles one direction only. Two different landings need not build two different graphs: a constant $f$ gives the same horizontal line under both orders for every $h$, and a periodic one can be carried onto itself as well. So for $h \ne 0$ test the two orders on a function that cannot absorb the difference, $f(x) = x$.

$f\big(-(x - h)\big) = h - x, \qquad f\big(-(x + h)\big) = -h - x$

These two lines differ by the constant $2h$ at every input, which is not zero unless $h$ is, so every $h \ne 0$ is already refuted by this one function. The only shift that this reflection can be interchanged with for every function $f$ is therefore the one that moves nothing. For any other $h$ the landings $h - u$ and $-u - h$ put the first order's graph $2h$ to the right of the second's, whatever $f$ is.

Build each rule by applying the second move to the rule the first move produced. The first student stretches the heights and then raises them:

$5f(x) + 20 = 5x + 20$

The second student shifts the input left $4$, building $f(x + 4)$, and then stretches the heights of that graph by $5$:

$5f(x + 4) = 5(x + 4) = 5x + 20$

The two students used different moves on different sides of $f$, and their rules simplify to the identical formula.

Everything turns on one identity, available because this $f$ hands back its own input.

$f(x + c) = x + c = f(x) + c$

So on this graph a shift left by $c$ and a shift up by $c$ produce the same set of points, and either description of the move is correct. The second student's shift left $4$ is a shift up $4$ in disguise, and the stretch by $5$ that follows multiplies that $4$ along with every other height, which is where the first student's $20$ comes from.

$5f(x + 4) = 5\big(f(x) + 4\big) = 5f(x) + 20$

Take $f(x) = x^2$ and run both sequences on it. The first student's moves give

$5f(x) + 20 = 5x^2 + 20,$

and the second student's moves give

$5f(x + 4) = 5(x + 4)^2.$

At the input $0$ these are $20$ and $5 \times 16 = 80$, so the two graphs are not the same one. The identity used in part B fails here, because squaring does not hand back the input: $f(x + 4)$ is $x^2 + 8x + 16$, which is not $f(x) + 4$.

Both sequences produce the line exactly, as part A showed, so the line is consistent with either one, and with any other pair of moves that finishes in the same place.

$\text{for this } f, \quad 5f(x) + 20 = 5f(x + 4) = 5x + 20$

What a graph carries is a set of points; the moves that put them there leave no mark on it. So the finished line settles where every point is, and it settles that both students are describing it correctly, but it cannot single one of them out. Calling any of these 'the' transformation behind the line claims something the picture does not contain.