Shifting and Reflecting Graphs, foundations for calculus lesson

Once you know the graph of $y = x^2$ well, you do not need a new table for $y = (x - 2)^2 - 3$ . The new graph is the old one picked up and moved. The trick is to follow a few marked points and ask one honest question about each of them, which input now produces the output it used to have.

Key terms

Parent graph

Definition A parent graph is the graph of a basic function $f$ whose shape and key points you already know, such as $y = x^2$ , $y = \sqrt{x}$ , $y = \lvert x \rvert$ , $y = \sqrt[3]{x}$ or $y = \frac{1}{x}$ . A new function built from $f$ by changing its input or its output is compared with this graph point by point.

In plain words The parent graph is the familiar picture you start from. A few easy points on it, called anchor points, are what you carry along when you move it.

Example Useful anchor points for $y = x^2$ are $(-1, 1)$ , $(0, 0)$ , $(1, 1)$ and $(2, 4)$ . For $y = \sqrt{x}$ , with domain $[0, \infty)$ , good choices are $(0, 0)$ , $(1, 1)$ and $(4, 2)$ , because their square roots are whole numbers.

Image of a point

Definition If a new function $g$ is built from $f$ , the image of a point $(u, f(u))$ on the graph of $f$ is the point on the graph of $g$ that comes from it. To find it, solve for the input $x$ at which the inside of $g$ equals $u$ , then compute the output of $g$ there. For

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

the inside equals $u$ when $x = u + h$ , so $(u, f(u))$ has image $(u + h, f(u) + k)$ .

In plain words Ask which new input feeds the old input into $f$ . That tells you where the point moves sideways, and whatever is done to the output tells you where it moves up or down.

Example Let $g(x) = (x - 2)^2 - 3$ , built from $f(x) = x^2$ with $h = 2$ and $k = -3$ . The point $(1, 1)$ goes to $(3, -2)$ , and indeed $g(3) = 1 - 3 = -2$ . The vertex $(0, 0)$ goes to $(2, -3)$ .

Horizontal shift

Definition For a real number $h$ , the graph of $g(x) = f(x - h)$ is the graph of $f$ moved $h$ units horizontally, to the right when $h > 0$ and to the left when $h < 0$ . Each point $(u, f(u))$ becomes $(u + h, f(u))$ , because

g(u + h) = f\bigl((u + h) - h\bigr) = f(u).

In plain words A shift sideways changes which input produces each output. Do not guess the direction from a plus or minus sign. Solve for the input that gives the old value.

Example Take $g(x) = (x + 3)^2$ . The inside $x + 3$ equals $u$ when $x = u - 3$ , so every point moves $3$ to the left. The vertex $(0, 0)$ goes to $(-3, 0)$ and the point $(1, 1)$ goes to $(-2, 1)$ . Check that $g(-2) = 1^2 = 1$ .

Vertical shift

Definition For a real number $k$ , the graph of $g(x) = f(x) + k$ is the graph of $f$ moved $k$ units vertically, up when $k > 0$ and down when $k < 0$ . Each point $(u, f(u))$ becomes $(u, f(u) + k)$ . The domain is unchanged, and every output moves by $k$ , so the whole range does too.

In plain words Adding a number after $f$ has done its work lifts or lowers every point by that same amount.

Example The graph of $g(x) = \sqrt{x} + 2$ sends $(0, 0)$ to $(0, 2)$ and $(4, 2)$ to $(4, 4)$ . The domain stays $[0, \infty)$ and the range becomes $[2, \infty)$ .

Reflection in the x-axis

Definition The graph of $g(x) = -f(x)$ is the reflection of the graph of $f$ in the $x$ -axis. Each point $(u, f(u))$ becomes $(u, -f(u))$ . The domain is unchanged and the range is the set of negatives of the old outputs.

In plain words Every output changes sign, so the graph flips upside down across the $x$ -axis. Points on the $x$ -axis stay where they are.

Example For $g(x) = -\sqrt{x}$ the point $(4, 2)$ becomes $(4, -2)$ and $(1, 1)$ becomes $(1, -1)$ . The domain is still $[0, \infty)$ and the range is $(-\infty, 0]$ .

Reflection in the y-axis

Definition The graph of $g(x) = f(-x)$ is the reflection of the graph of $f$ in the $y$ -axis. Each point $(u, f(u))$ becomes $(-u, f(u))$ , since $g(-u) = f(u)$ . The domain of $g$ is the set of negatives of the inputs in the domain of $f$ , and the range is unchanged.

In plain words To get the old output you now have to feed in the opposite input, so the graph flips left to right across the $y$ -axis.

Example The function $g(x) = \sqrt{-x}$ is defined when $-x \ge 0$ , that is for $x \le 0$ . The point $(4, 2)$ on $y = \sqrt{x}$ becomes $(-4, 2)$ , and $g(-4) = \sqrt{4} = 2$ confirms it.

New domain

Definition If $g$ is built from $f$ by replacing the input $x$ with an expression $w(x)$ , then $x$ is in the domain of $g$ exactly when $w(x)$ is in the domain of $f$ . For $g(x) = f(x - h) + k$ this means

\operatorname{dom} g = \{\, x : x - h \in \operatorname{dom} f \,\}.

A vertical shift or a reflection in the $x$ -axis does not change the domain.

In plain words The new function works only where its inside is something the old function accepts. Write that condition down and solve it.

Example Let $g(x) = \sqrt{3 - x} - 1$ . The condition $3 - x \ge 0$ gives the domain $(-\infty, 3]$ . To find images, solve $3 - x = u$ , which gives $x = 3 - u$ .

Point on $y = \sqrt{x}$	Input with $3 - x = u$	Image on $y = g(x)$
$(0, 0)$	$x = 3$	$(3, -1)$
$(1, 1)$	$x = 2$	$(2, 0)$
$(4, 2)$	$x = -1$	$(-1, 1)$

The graph opens to the left from $(3, -1)$ . Reading only the minus sign in $3 - x$ would wrongly suggest a simple shift to the right. In fact $g(x) = f\bigl(-(x - 3)\bigr) - 1$ with $f(x) = \sqrt{x}$ , so the graph of $\sqrt{x}$ is reflected in the $y$ -axis, then shifted $3$ to the right and $1$ down. The table confirms it, since $(4, 2)$ goes to $(-4, 2)$ , then to $(-1, 2)$ , then to $(-1, 1)$ .

Check yourself

Starting from $y = x^2$ , describe $g(x) = -(x + 1)^2 + 4$ . Find its vertex, the image of $(2, 4)$ and its range.

Show the answer

The graph is moved $1$ left, flipped in the $x$ -axis and moved $4$ up. The vertex $(0, 0)$ goes to $(-1, 4)$ . For $(2, 4)$ solve $x + 1 = 2$ to get $x = 1$ , and the output is $-4 + 4 = 0$ , so the image is $(1, 0)$ . The range is $(-\infty, 4]$ .

Find the domain and range of $h(x) = -\sqrt{x + 4}$ and the image of the point $(4, 2)$ on $y = \sqrt{x}$ .

Show the answer

The function needs $x + 4 \ge 0$ , so the domain is $[-4, \infty)$ . Solving $x + 4 = 4$ gives $x = 0$ , and the output is $-2$ , so the image is $(0, -2)$ . Every output is $0$ or negative, so the range is $(-\infty, 0]$ .

The point $(5, -2)$ lies on the graph of $f$ . Name a point on the graph of $y = f(-x)$ and one on the graph of $y = -f(x - 1)$ .

Show the answer

For $y = f(-x)$ the input $x = -5$ gives $f(5) = -2$ , so $(-5, -2)$ is on the graph. For $y = -f(x - 1)$ the input $x = 6$ gives $-f(5) = 2$ , so $(6, 2)$ is on the graph.

Shifting and Reflecting Graphs

Key terms

Parent graph

Image of a point

Horizontal shift

Vertical shift

Reflection in the x-axis

Reflection in the y-axis

New domain

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus

Resources for this lesson

Function transformation explorer