Shifts and Reflections of Graphs, precalculus lesson

Slide a drawing of a parabola two squares to the right on graph paper and its shape stays exactly the same. Only its formula changes. Once you know how each change in a formula moves the graph, a handful of basic shapes lets you sketch a whole family of functions and read off features such as a vertex without plotting a single point.

Key terms

Parent function

Definition A family of functions is a collection of functions whose graphs all come from one basic graph by shifts, reflections and stretches. The parent function of the family is the function with that basic graph, usually the simplest formula in the family. The common parent functions are listed below.

Parent	Formula	Domain	Points to track
identity	$x$	$\mathbb{R}$	$(0, 0)$ , $(1, 1)$
square	$x^2$	$\mathbb{R}$	$(-1, 1)$ , $(0, 0)$ , $(1, 1)$
cube	$x^3$	$\mathbb{R}$	$(-1, -1)$ , $(0, 0)$ , $(1, 1)$
absolute value	$\lvert x \rvert$	$\mathbb{R}$	$(-1, 1)$ , $(0, 0)$ , $(1, 1)$
square root	$\sqrt{x}$	$[0, \infty)$	$(0, 0)$ , $(1, 1)$ , $(4, 2)$
reciprocal	$1/x$	all $x \ne 0$	$(-1, -1)$ , $(1, 1)$

In plain words A parent function is the plain version of a shape. Know its graph and a few of its points, and every relative of it becomes a moved, flipped or stretched copy.

Example The function $g(x) = (x - 3)^2 + 1$ belongs to the family with parent $x^2$ . Its graph is the parabola $y = x^2$ moved $3$ units right and $1$ unit up, so the vertex $(0, 0)$ goes to $(3, 1)$ and the point $(1, 1)$ goes to $(4, 2)$ . The formula agrees, since $g(3) = 0 + 1 = 1$ and $g(4) = 1 + 1 = 2$ .

Rigid transformation

Definition A rigid transformation of the plane moves each point of the plane to a point of the plane so that distances are preserved. In other words, if points $P$ and $Q$ move to $P'$ and $Q'$ , then the distance from $P'$ to $Q'$ equals the distance from $P$ to $Q$ . The rigid transformations in this lesson are the shifts and the reflections across the coordinate axes,

\begin{gathered} (x, y) \mapsto (x + h, y + k), \\[0.4em] (x, y) \mapsto (x, -y), \\[0.4em] (x, y) \mapsto (-x, y). \end{gathered}

A rigid transformation does not change the size or shape of a graph, only its position or orientation. Stretches and compressions are not rigid.

In plain words Sliding or flipping a drawing does not bend or resize it. Only its place on the page changes.

Example The points $P = (0, 0)$ and $Q = (2, 4)$ lie on $y = x^2$ , at distance $\sqrt{2^2 + 4^2} = \sqrt{20}$ . Compare where three different moves send them.

Move	New points	Distance
shift by $3$ right and $1$ up	$(3, 1)$ and $(5, 5)$	$\sqrt{2^2 + 4^2} = \sqrt{20}$
reflection across the $x$ -axis	$(0, 0)$ and $(2, -4)$	$\sqrt{2^2 + 4^2} = \sqrt{20}$
vertical stretch by $2$ , doubling each $y$	$(0, 0)$ and $(2, 8)$	$\sqrt{2^2 + 8^2} = \sqrt{68}$

The shift and the reflection keep the distance, as rigid transformations must. The stretch changes it, so it is not rigid.

Vertical shift

Definition Let $f$ be a function and $k$ a real number. The function $g(x) = f(x) + k$ has the same domain as $f$ , and each point $(x, y)$ of the graph of $f$ corresponds to the point $(x, y + k)$ of the graph of $g$ . So the graph of $g$ is the graph of $f$ shifted up by $k$ units when $k > 0$ and down by $\lvert k \rvert$ units when $k < 0$ . Every output, and with it the whole range, moves by $k$ .

In plain words Adding a number outside the function changes every output by that number. The graph slides straight up or down.

Example Take $f(x) = \sqrt{x}$ and $g(x) = \sqrt{x} - 2$ , so $k = -2$ .

$x$	$0$	$1$	$4$	$9$
$f(x)$	$0$	$1$	$2$	$3$
$g(x)$	$-2$	$-1$	$0$	$1$

Each point drops by $2$ . The domain stays $[0, \infty)$ , the range moves from $[0, \infty)$ to $[-2, \infty)$ , and the graph now crosses the $x$ -axis at $(4, 0)$ .

Horizontal shift

Definition Let $f$ be a function and $h$ a real number. The function $g(x) = f(x - h)$ is defined at $x$ exactly when $x - h$ is in the domain of $f$ . Since

g(x + h) = f(x),

each point $(x, y)$ of the graph of $f$ corresponds to the point $(x + h, y)$ of the graph of $g$ . So the graph of $g$ is the graph of $f$ shifted right by $h$ units when $h > 0$ and left by $\lvert h \rvert$ units when $h < 0$ . The domain moves by $h$ and the range stays the same.

In plain words A number added inside the function changes the input, and it moves the graph the opposite way from what the sign suggests. In $f(x - 3)$ the input has to be $3$ larger to produce the same output as before, so the graph moves $3$ units to the right.

Example The function $g(x) = \sqrt{x + 3}$ is $f(x - h)$ with $f(x) = \sqrt{x}$ and $h = -3$ , so its graph is the graph of $\sqrt{x}$ shifted $3$ units left.

$x$	$-3$	$-2$	$1$	$6$
$x + 3$	$0$	$1$	$4$	$9$
$g(x)$	$0$	$1$	$2$	$3$

The starting point moves from $(0, 0)$ to $(-3, 0)$ , and the domain becomes $[-3, \infty)$ .

Reflection across the x-axis

Definition Let $f$ be a function. The function $g(x) = -f(x)$ has the same domain as $f$ , and each point $(x, y)$ of the graph of $f$ corresponds to the point $(x, -y)$ of the graph of $g$ . So the graph of $g$ is the mirror image of the graph of $f$ across the $x$ -axis. Points on the $x$ -axis stay where they are, and the range is replaced by the negatives of its values.

In plain words A minus sign outside the function changes the sign of every output. What was above the $x$ -axis ends up below it, at the same distance.

Example Let $f(x) = x^2 - 2x$ , a parabola opening upward with vertex $(1, -1)$ . Then $g(x) = -f(x) = 2x - x^2$ .

$x$	$-1$	$0$	$1$	$2$	$3$
$f(x)$	$3$	$0$	$-1$	$0$	$3$
$g(x)$	$-3$	$0$	$1$	$0$	$-3$

The zeros $0$ and $2$ do not move. The vertex goes to $(1, 1)$ , the parabola now opens downward, and the range $[-1, \infty)$ becomes $(-\infty, 1]$ .

Reflection across the y-axis

Definition Let $f$ be a function. The function $g(x) = f(-x)$ is defined at $x$ exactly when $-x$ is in the domain of $f$ . Since $g(-x) = f(x)$ , each point $(x, y)$ of the graph of $f$ corresponds to the point $(-x, y)$ of the graph of $g$ . So the graph of $g$ is the mirror image of the graph of $f$ across the $y$ -axis. Points on the $y$ -axis stay where they are, and the domain is replaced by the negatives of its values. For an even function this reflection leaves the graph unchanged, and for an odd function it gives the same graph as the reflection across the $x$ -axis.

In plain words A minus sign on the input swaps left and right. The outputs stay the same, but each one now happens at the opposite input.

Example Reflecting $f(x) = \sqrt{x}$ across the $y$ -axis gives $g(x) = \sqrt{-x}$ , which is defined when $-x \ge 0$ , that is, for $x \le 0$ .

$x$	$0$	$-1$	$-4$
$g(x)$	$0$	$1$	$2$

The points $(1, 1)$ and $(4, 2)$ of $y = \sqrt{x}$ become $(-1, 1)$ and $(-4, 2)$ , and the domain $[0, \infty)$ becomes $(-\infty, 0]$ .

Combining shifts and reflections

Definition Let $a$ and $b$ each be $1$ or $-1$ , and let $h$ and $k$ be real numbers. The graph of

g(x) = a f\bigl(b(x - h)\bigr) + k

is obtained from the graph of $f$ by moving each point $(x, y)$ to $(bx + h, ay + k)$ . Read this as a reflection followed by a shift, separately for the input and for the output. Every combination of shifts and reflections across the axes can be written in this form. The order of the steps matters, because reflecting and then shifting is in general not the same as shifting and then reflecting.

In plain words Follow a few points instead of the whole curve. Each point is reflected first and then shifted, sideways according to the inside of the formula and up or down according to the outside.

Example Build $g(x) = -(x - 2)^2 + 3$ from $x^2$ in three steps and follow the vertex $(0, 0)$ and the point $(1, 1)$ .

Step	Formula	Vertex	Image of $(1, 1)$
start	$x^2$	$(0, 0)$	$(1, 1)$
shift right $2$	$(x - 2)^2$	$(2, 0)$	$(3, 1)$
reflect across the $x$ -axis	$-(x - 2)^2$	$(2, 0)$	$(3, -1)$
shift up $3$	$-(x - 2)^2 + 3$	$(2, 3)$	$(3, 2)$

Indeed $g(3) = -1 + 3 = 2$ . Doing the last two steps in the other order gives

-\bigl((x - 2)^2 + 3\bigr) = -(x - 2)^2 - 3,

a parabola with vertex $(2, -3)$ , which is a different graph.

Check yourself

How is the graph of $g(x) = \lvert x + 4 \rvert - 1$ related to the graph of $y = \lvert x \rvert$ , and where is its vertex?

Show the answer

Since $x + 4 = x - (-4)$ , the graph is the graph of $y = \lvert x \rvert$ shifted $4$ units left and $1$ unit down. The vertex moves from $(0, 0)$ to $(-4, -1)$ , and indeed $g(-4) = 0 - 1 = -1$ .

The point $(3, 5)$ lies on the graph of $f$ . Which point must lie on the graph of $g(x) = -f(x - 1) + 2$ ?

Show the answer

Here $a = -1$ , $b = 1$ , $h = 1$ and $k = 2$ , so $(3, 5)$ moves to $(3 + 1, -5 + 2) = (4, -3)$ . As a check, $g(4) = -f(3) + 2 = -5 + 2 = -3$ .

Reflect the graph of $y = \sqrt{x}$ across the $y$ -axis and then shift it $3$ units right. Find the new formula and its domain.

Show the answer

The reflection gives $y = \sqrt{-x}$ . Shifting right by $3$ replaces $x$ with $x - 3$ , which gives

y = \sqrt{-(x - 3)} = \sqrt{3 - x}.

It is defined when $3 - x \ge 0$ , so the domain is $(-\infty, 3]$ . The point $(4, 2)$ of $y = \sqrt{x}$ goes to $(-4, 2)$ and then to $(-1, 2)$ , and indeed $\sqrt{3 - (-1)} = 2$ .

Shifts and Reflections of Graphs

Key terms

Parent function

Rigid transformation

Vertical shift

Horizontal shift

Reflection across the x-axis

Reflection across the y-axis

Combining shifts and reflections

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus

Resources for this lesson

Function transformation explorer