Transformations and piecewise functions

Changes outside move the graph up or down

Adding a number to a function's output moves its whole graph up. If $g(x) = f(x) + k$, every output of $g$ is $k$ more than the output of $f$ at the same input. A negative $k$ moves the graph down.

The shift $k$ is measured in the output's units. If $f(t)$ is a temperature in °C at hour $t$, then $f(t) + 5$ is the same day's pattern, 5 °C warmer.

Changes inside move it sideways

Changing the input moves the graph sideways, and the direction surprises most people. $g(x) = f(x - h)$ moves the graph $h$ units to the right.

Here is why. At input $x$, $g$ gives what $f$ gave at $x - h$, which is $h$ units further left. So every feature of $f$ shows up $h$ units later in $g$.

With the temperature, $f(t - 2)$ is the same pattern 2 hours later.

Take $f(x) = x^2$ and $g(x) = (x - 3)^2 - 4$, which is $f(x - 3) - 4$. A table shows the move:

Each value of $f$ turns up in $g$ three columns later and 4 lower: $g(3) = f(0) - 4 = -4$ and $g(5) = f(2) - 4 = 0$.

Algebra says the same. $x^2$ is smallest at $x = 0$, and $(x - 3)^2$ is smallest where $x - 3 = 0$, at $x = 3$. There $g$ is $-4$, so the lowest point moves from $(0, 0)$ to $(3, -4)$.

The zeros of $g$ come from $(x - 3)^2 = 4$, so $x - 3 = 2$ or $-2$: $x = 5$ or $x = 1$.

The domain moves with the graph. The function $\sqrt{x - 3}$ from lesson 1.1.1 is $\sqrt{x}$ moved 3 right, which is why its domain starts at 3 instead of 0.

Flips and stretches

A minus sign flips a graph, and a multiplier stretches it. Where you put them decides which way.

• $-f(x)$ turns every output upside down: a flip over the $x$-axis.
• $f(-x)$ swaps left and right: a flip over the $y$-axis.
• $a\,f(x)$ with $a > 1$ stretches the graph away from the $x$-axis; with $0 < a < 1$ it squashes it towards the axis. A negative $a$ stretches and flips at once.

Follow the point $(4, 2)$ of $\sqrt{x}$ through each one:

The flip $\sqrt{-x}$ needs $-x \ge 0$, so its domain is $x \le 0$: the domain flips too. It gives 0, 1, 2 and 3 at $x = 0, -1, -4, -9$.

Three cases catch people out.

• Zeros stay put under a stretch or a flip over the $x$-axis: $2 \times 0$ and $-0$ are still 0.
• A symmetric graph ignores $f(-x)$. For $x^2$, $(-x)^2 = x^2$, so the flip over the $y$-axis changes nothing.
• $-x^2$ is not $(-x)^2$. By the order of operations (lesson 0.3.1), $-x^2$ means $-(x^2)$, the parabola opening downwards. $(-x)^2$ is $x^2$ again.

Multiplying the input acts sideways, again the opposite way. $f(2x)$ reaches each height at half the input, so the graph is squeezed towards the $y$-axis. $x^2$ reaches 4 at $x = 2$, but $(2x)^2$ reaches it at $x = 1$.

Several moves at once

Most formulas combine moves: $a\,f(x - h) + k$ shifts right by $h$, stretches by $a$ (flipping if $a$ is negative), then shifts up by $k$. Work from the inside out, the order you would evaluate the formula in (lesson 0.3.1).

The order matters. With $f(x) = x^2$, $2f(x) + 1$ stretches and then lifts by 1: at $x = 3$ it gives 19. But $2(f(x) + 1)$ lifts first, and the stretch doubles the lift too: at $x = 3$ it gives 20.

A second example: $-2(x - 1)^2 + 3$ is $x^2$ moved 1 right, stretched by 2 and flipped, then lifted 3. Its highest point is $(1, 3)$. At $x = 0, 1, 2, 3$ it gives 1, 3, 1 and $-5$.

Lesson 1.2.2 writes every quadratic in this form to find its lowest or highest point.

Try the moves yourself. Set $h = 3$ and $k = -4$ to get the graph of the first figure, then make $a$ negative and watch the lowest point become the highest.

Piecewise functions

A piecewise function uses a different rule on different parts of its domain. You met one in lesson 0.3.3: the absolute value is $x$ when $x \ge 0$ and $-x$ when $x < 0$. Here is another:

To evaluate it, first find which piece the input belongs to. $p(3) = 3 + 1 = 4$, but $p(-2) = 2 \times (-2) = -4$. At the boundary, $x = 1$ falls in the second piece, so $p(1) = 2$.

Here the left piece also gives $2 \times 1 = 2$ at $x = 1$, so the two pieces meet and the graph has no break. They need not meet. Take

Just left of 2 the output is 1, as in $j(1.99) = 1$. At 2 it is $j(2) = 2$. The graph jumps up by 1 at $x = 2$.

Two more cases matter. If two pieces both claimed $x = 1$ with different outputs, the rule would not be a function at all. If no piece covers an input, that input is not in the domain.

Lesson 2.1.2 calls a graph with no break continuous, and studies jumps like this one.

The ramp

The most useful piecewise function in this course is the ramp, $\max(x, 0)$: "the larger of $x$ and 0". It is 0 for negative inputs and $x$ itself for positive ones. Its two pieces meet at a corner, $(0, 0)$.

Every move of this lesson applies. $\max(x - 3, 0)$ is flat at 0 up to 3, then climbs with slope 1. The flip over the $y$-axis gives $\max(-x, 0)$, and moved to 3 it is $\max(3 - x, 0)$: it falls to 0 at 3 and stays there.

A ramp and its mirror image add up to the absolute value: $\max(x, 0) + \max(-x, 0) = |x|$. For any $x$, one of the two is 0 and the other is the distance from 0. Case study 1.1.5 builds every option payoff from these ramps.

Common mistakes

1. The sideways shift goes against the sign. $f(x - 3)$ moves the graph right, not left.
2. $-x^2$ is not $(-x)^2$. The first is never positive; the second is never negative.
3. The order of moves matters. $2f(x) + 1$ is not $2(f(x) + 1)$: the stretch also doubles any lift made before it.
4. An inside multiplier squeezes. $f(2x)$ is narrower than $f(x)$, not wider.
5. The domain moves too. $\sqrt{x - 3}$ starts at 3, and $\sqrt{-x}$ lives on $x \le 0$.
6. Check which piece applies, including at the boundary: a closed dot is on the graph, an open dot is not.
7. Use np.maximum on arrays, not Python's max.