Shifting Graphs

Intuition

When you wrote a parabola in vertex form, $y = (x - h)^2 + k$ , you were really giving directions: start with the plain bowl $y = x^2$ and slide it so its lowest point lands at $(h, k)$ . This lesson makes that idea general. Take any graph you already know, and you can build a whole family of new graphs from it just by sliding it around, with no new table of values to work out. Adding a number changes the height of the graph. Changing the input changes its left-to-right position, and that is where the sign plays a trick: a graph slides sideways in the opposite direction from the sign you see inside the function. Why does that happen? This lesson shows you, so the rule stops being something to memorize and becomes something you can see for yourself.

Learning goals

Move a graph up by $k$ with $y = f(x) + k$
Shift right by $h$ with $y = f(x - h)$
Explain why subtracting inside moves the graph rightward
Combine both into $y = f(x - h) + k$
Track how a shift moves intercepts, domain and range

Sliding a graph without building a new table

A shift, also called a translation, moves every point of a graph the same distance in the same direction, without turning it, stretching it, or flipping it. The shape stays identical; only its location changes. Because a graph is the set of points $(x, f(x))$ , a shift acts on those points. We can predict exactly where each point goes by looking at how the rule changes.

There are two kinds of shift: a vertical shift moves the graph up or down, and a horizontal shift moves it left or right. We handle each on its own first, then combine them. Throughout, keep one fact from the last lesson in front of you: a point $(a, b)$ lies on the graph of $y = f(x)$ exactly when $b = f(a)$ . So the second coordinate is an output and the first is an input. Every rule below comes straight from tracking what happens to that pairing.

Vertical shifts: change the output

Adding a number to the outside of a function moves its graph straight up or down. It never changes the graph’s shape. Here is what that looks like on a graph you already know.

The dashed parabola is y = x squared. Adding 3 to every output lifts each point straight up by 3, giving the solid parabola y = x squared + 3. The vertex moves from the origin to (0, 3) and the shape is unchanged.

Worked example 1 Graph $y = x^2 + 3$ by sliding $y = x^2$ upward

The rule $y = x^2 + 3$ takes the familiar bowl $y = x^2$ and adds $3$ to every output, so the whole graph rises by $3$ . There is no need to build a new table; shift the points you already know.

The vertex of $y = x^2$ sits at $(0, 0)$ . Adding $3$ lifts it to

(0,\, 0 + 3) = (0, 3).

Take two more anchor points and raise each by $3$ . On $y = x^2$ the points $(1, 1)$ and $(2, 4)$ appear; after the shift they become

(1,\, 1 + 3) = (1, 4), \qquad (2,\, 4 + 3) = (2, 7).

Plot the lifted vertex $(0, 3)$ and the lifted points, then draw the identical bowl through them. Because the lowest output is now $3$ , the graph never dips below the line $y = 3$ , and its y-intercept is $(0, 3)$ . The parabola has not changed shape at all; it simply sits three units higher.

Look at what happened. Every output on $y = x^2$ picked up the same extra $3$ : the vertex $(0, 0)$ became $(0, 3)$ , and $(1, 1)$ became $(1, 4)$ . Adding the same number to every height is exactly what slides a shape straight up without bending it, and the same thing happens for any starting graph, not only a parabola. Written in general, adding a constant $k$ outside a function does this to every point of $y = f(x)$ :

y = f(x) + k.

Every input still passes through $f$ , but whatever comes out is then increased by $k$ . Since the output is the height of the graph, raising every output by $k$ raises the whole graph by $k$ , up when $k > 0$ and down when $k < 0$ .

Why $y = f(x) + k$ slides the graph up by $k$ #

Let $g(x) = f(x) + k$ . Pick any input $a$ . The original graph has the point $(a, f(a))$ , sitting at height $f(a)$ directly above the input $a$ . The new graph has the point $(a, g(a))$ , and

g(a) = f(a) + k,

so that point is $(a, f(a) + k)$ . The two points share the same first coordinate $a$ , so they lie on the same vertical line. The new point’s second coordinate minus the old one’s equals exactly $k$ , whatever the sign of $k$ turns out to be. That is a rise of $k$ when $k > 0$ , and a drop of $\lvert k \rvert$ when $k < 0$ . Nothing about the input changed, so the point did not move left or right at all.

Since $a$ was an arbitrary input, this happens to every point at once: each one rises by $k$ when $k > 0$ , or drops by $\lvert k \rvert$ when $k < 0$ . A rigid slide of every point by the same amount in the same direction is exactly a translation. So the graph of $y = f(x) + k$ is the graph of $y = f(x)$ moved vertically by $k$ , and its shape is untouched.

The same reasoning reads points off a graph directly. If $(a, b)$ is on $y = f(x)$ , then $(a, b + k)$ is on $y = f(x) + k$ : keep the input, add $k$ to the height.

Check your understanding

The point $(5, 2)$ lies on the graph of $y = f(x)$ . Which point must lie on the graph of $y = f(x) - 4$ ?

(5, -2)

(1, 2)

(5, 6)

(9, 2)

Horizontal shifts: the direction that looks backward

Now change the input instead of the output. Replace $x$ with $x - h$ to get

y = f(x - h).

Here is the trap. Almost everyone expects $x - h$ , with its minus sign, to move the graph left, in the negative direction. It does the opposite: for $h > 0$ the graph moves right by $h$ . Rather than memorize that, derive it, because the derivation shows exactly why the sign feels reversed.

The dashed parabola is y = x squared. Subtracting 3 inside gives y = (x minus 3) squared, the solid parabola, which sits 3 to the right, not left. The point (1, 1) reappears at (4, 1) and the vertex moves from (0, 0) to (3, 0).

Worked example 2 Relate $y = (x - 3)^2$ to $y = x^2$

The rule $y = (x - 3)^2$ has the form $f(x - h)$ with $f(x) = x^2$ and $h = 3$ . Check the direction with the vertex. The bowl $y = x^2$ bottoms out where its input is $0$ . For $y = (x - 3)^2$ , the lowest output happens where the inside is $0$ :

x - 3 = 0 \quad \Longrightarrow \quad x = 3,

so the new vertex is $(3, 0)$ , three units right of the origin, even though the rule contains a minus sign. Track one more point the same way. The point $(1, 1)$ is on $y = x^2$ ; to make the inside of $(x - 3)^2$ equal $1$ , solve $x - 3 = 1$ , giving $x = 4$ . So the matching point on the new graph is $(4, 1)$ , again shifted right by $3$ :

(1, 1) \ \longrightarrow \ (4, 1).

Every point moves the same way, so the parabola lands three units to the right, exactly as the picture shows.

Look at why that happened. To make $(x - 3)^2$ reproduce the height that $1^2$ gave on the original graph, the inside $x - 3$ had to equal $1$ , and that forced $x$ to be three more than before: $x = 4$ , not $x = -2$ . Subtracting inside asks $x$ to grow to keep the inside value the same, and that is true for every point, not just this one.

Why $y = f(x - h)$ slides the graph right by $h$ #

Let $g(x) = f(x - h)$ . Fix any input $a$ , so the original graph has the point $(a, f(a))$ at height $f(a)$ . Ask the key question: at which input $x$ does the new function $g$ produce that same height $f(a)$ ?

The function $f$ returns the value $f(a)$ when its input equals $a$ . But the input handed to $f$ inside $g$ is not $x$ ; it is $x - h$ . So $g(x)$ produces that same height $f(a)$ when the inside matches $a$ :

x - h = a \quad \Longrightarrow \quad x = a + h.

Therefore $g(a + h) = f\big((a + h) - h\big) = f(a)$ , which means the new graph has the point $(a + h, f(a))$ . Compare it with the old point $(a, f(a))$ : both sit at the same height $f(a)$ . The new point, though, is at input $a + h$ , exactly $h$ units to the right of $a$ when $h > 0$ .

Since $a$ was arbitrary, every point moves right by $h$ , so $y = f(x - h)$ is $y = f(x)$ shifted right by $h$ . The reason the minus sign feels backward is now visible: subtracting $h$ inside means the function needs a larger input $x$ to feed $f$ the same inside value. So each feature of the graph shows up $h$ units later, that is, farther to the right. Reading it the other way, $y = f(x + h)$ is $y = f(x - (-h))$ , a shift right by $-h$ , which is a shift left by $h$ .

A horizontal shift always works this way: to make the inside expression reach a value it used to reach, you move $x$ in the direction that undoes the sign in front of $h$ .

Check your understanding

Why does $y = f(x - 3)$ move the graph of $y = f(x)$ to the right, instead of to the left where the minus sign might suggest?

A Because the inside

x - 3

reaches any old input value

a

only once

x

has grown to

a + 3

, so every feature needs a bigger input to reappear. B Because subtracting a positive number always moves a graph in the negative direction. C Because

y = f(x - 3)

lowers every output by

3

, and lower outputs plot farther to the right. D Because subtracting inside first reflects the graph across the y-axis and the reflection looks like a rightward slide.

Check your understanding

How is the graph of $y = f(x + 4)$ related to the graph of $y = f(x)$ ?

A shifted left

4

B shifted right

4

C shifted up

4

D shifted down

4

Combining a horizontal and a vertical shift

The two moves do not interfere, so you can apply both at once. Starting from $y = f(x)$ , the rule

y = f(x - h) + k

shifts the graph right by $h$ (from the $x - h$ inside) and up by $k$ (from the $+ k$ outside). Track a single point to see both at work: a point $(a, f(a))$ on the original graph reappears at

\big(a + h,\ f(a) + k\big),

its input increased by $h$ and its height increased by $k$ . Negative values of $h$ or $k$ just reverse the corresponding direction.

This is where vertex form finally makes complete sense. Applying the combined shift to $y = x^2$ gives $y = (x - h)^2 + k$ , and since the vertex of $y = x^2$ is at the origin, it lands at $(h, k)$ . The numbers $h$ and $k$ in vertex form are not mysterious; they are simply how far the basic bowl was slid right and up.

Nothing in the argument used the fact that $f$ was a squaring function, so try the combined shift on a different shape: $y = \lvert x - h \rvert + k$ , a V-shaped graph with a corner instead of a smooth vertex. Start at $h = 0$ and $k = 0$ , then increase $h$ to $3$ . Watch the corner move three units to the right, even though the rule now reads $x - 3$ . Next, hold $h$ where it is and change $k$ . The corner moves straight up or down and stays at the same left-to-right position, because $k$ never touches the inside of the function. Leave the coefficient $a$ at $1$ for now; the next lesson is what that control is for.

Move the corner of $y = a\lvert x - h\rvert + k$

Stretch a Inside h Outside k

y = |x|. Its corner sits at (0, 0), not shifted at all. Its two arms point upward, rising one unit for every unit across.

An absolute value graph on a coordinate plane, with its corner placed by the inside and outside shifts and its arms set by the coefficient. The inside shift moves the graph across, the outside shift moves it up and down.

Shifting y = x squared right 2 and up 1 gives y = (x minus 2) squared + 1, the solid parabola. Its vertex is at (2, 1). This is vertex form, and (2, 1) is the point the vertex was moved to.

Worked example 3 Shift $y = x^2$ into vertex form $y = (x - 2)^2 + 1$

Read the rule $y = (x - 2)^2 + 1$ as a set of directions applied to $y = x^2$ . The inside $x - 2$ says shift right $2$ , and the $+ 1$ outside says shift up $1$ .

Move the vertex first. It starts at $(0, 0)$ , then goes right $2$ and up $1$ :

(0, 0) \ \longrightarrow \ (0 + 2,\ 0 + 1) = (2, 1).

To confirm with the rule, the lowest output happens where the squared part is $0$ , at $x - 2 = 0$ , so $x = 2$ ; there the output is $(0) + 1 = 1$ , matching the vertex $(2, 1)$ . Carry one more point along for the shape: the point $(1, 1)$ on $y = x^2$ moves right $2$ and up $1$ to $(3, 2)$ . Plot the vertex $(2, 1)$ and a few shifted points, then draw the same bowl. The graph is $y = x^2$ resting with its bottom at $(2, 1)$ , which is precisely what vertex form announces.

Check your understanding

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

(5, 7)

(-1, 7)

(5, 3)

(-1, 3)

A combined shift also relocates the intercepts. A horizontal shift slides the x-intercepts sideways by $h$ , while a vertical shift raises or lowers the graph. A vertical shift can also change how many x-intercepts there are, since it may push the curve across or away from the horizontal axis. Vertex form makes them easy to find by solving directly.

Worked example 4 Find the vertex and x-intercepts of $y = (x - 3)^2 - 4$

The rule $y = (x - 3)^2 - 4$ shifts $y = x^2$ right $3$ and down $4$ , so the vertex moves from $(0, 0)$ to

(0 + 3,\ 0 - 4) = (3, -4).

Because the vertex sits below the horizontal axis and the bowl opens upward, the graph must cross that axis twice, so expect two x-intercepts. Find them by setting the output to $0$ :

(x - 3)^2 - 4 = 0 \quad \Longrightarrow \quad (x - 3)^2 = 4.

A number whose square is $4$ is $2$ or $-2$ , so $x - 3 = 2$ or $x - 3 = -2$ , giving

x = 5 \quad \text{or} \quad x = 1.

The x-intercepts are $(1, 0)$ and $(5, 0)$ , sitting symmetrically on either side of the vertex’s input $x = 3$ . The down- $4$ shift is what dropped the vertex below the axis and created two crossings; without it, $y = (x - 3)^2$ would only touch the axis at its vertex.

Check your understanding

A graph of $y = f(x)$ crosses the x-axis at $x = -2$ and $x = 5$ . What are the x-intercepts of $y = f(x - 3)$ ?

x = 1

and

x = 8

x = -5

and

x = 2

x = -2

and

x = 5

x = 1

and

x = 2

Worked example 5 Shift a general function and track its domain and range

Suppose $f$ has domain $0 \le x \le 6$ and range $-1 \le y \le 5$ , and the point $(4, -1)$ lies on its graph. Describe the graph of $y = f(x - 2) + 3$ .

The rule shifts the graph right $2$ and up $3$ , so apply that to the point first:

(4, -1) \ \longrightarrow \ (4 + 2,\ -1 + 3) = (6, 2).

The domain is the horizontal spread of the graph, so a shift right by $2$ slides both ends of the domain right by $2$ , adding $2$ to each:

0 \le x \le 6 \quad \longrightarrow \quad 2 \le x \le 8.

The range is the vertical spread, so a shift up by $3$ raises both ends of the range by $3$ :

-1 \le y \le 5 \quad \longrightarrow \quad 2 \le y \le 8.

The horizontal shift touches only the inputs (the domain) and the vertical shift touches only the outputs (the range). The two moves can therefore be handled separately even when they happen together.

Check your understanding

A function $h$ has domain $-4 \le x \le 2$ and range $1 \le y \le 9$ . What are the domain and range of $y = h(x - 1) - 3$ ?

A domain

-3 \le x \le 3

, range

-2 \le y \le 6

B domain

-5 \le x \le 1

, range

4 \le y \le 12

C domain

-4 \le x \le 2

, range

-2 \le y \le 6

D domain

-3 \le x \le 3

, range

1 \le y \le 9

The four basic translations are worth keeping in one place. In each row, $h$ and $k$ stand for positive numbers.

Rule	Effect on the graph of $y = f(x)$
$y = f(x) + k$	up $k$
$y = f(x) - k$	down $k$
$y = f(x - h)$	right $h$
$y = f(x + h)$	left $h$

Check your understanding

What is the vertex of the parabola $y = (x - 5)^2 + 2$ ?

(5, 2)

(-5, 2)

(5, -2)

(2, 5)

Common mistakes

Shifting horizontally the wrong way. $y = f(x - h)$ moves right by $h$ , and $y = f(x + h)$ moves left. The direction is opposite the sign inside. Check it with one point: the inside $x - h$ reaches an old value only when $x$ grows by $h$ , which is a move to the right.
Swapping which change moves which way. A number added outside the function, $f(x) + k$ , is a vertical shift (it changes the height). A number combined with $x$ inside the function, $f(x - h)$ , is a horizontal shift (it changes the position). Watch where the number sits.
Moving a point in both coordinates for a single shift. A vertical shift changes only the output, so $(a, b) \to (a, b + k)$ ; the input is untouched. A horizontal shift changes only the input, so $(a, b) \to (a + h, b)$ . Do not shift both coordinates unless both shifts are present.
Misreading vertex form. In $y = (x - h)^2 + k$ the vertex is $(h, k)$ , so $y = (x + 3)^2 - 5$ has vertex $(-3, -5)$ , taking the sign from $x - (-3)$ . Read $h$ as the value that makes the inside zero.
Expecting the shape to change. A translation slides the graph; it never stretches, squeezes, or flips it. If your new graph looks wider, narrower, or upside down, you have done something other than a shift.

Key takeaways

A shift (translation) moves every point of a graph the same distance in the same direction; the shape is unchanged.
Vertical: $y = f(x) + k$ moves the graph up by $k$ (down when $k < 0$ ), because adding $k$ to every output raises every point. Points map by $(a, b) \to (a, b + k)$ .
Horizontal: $y = f(x - h)$ moves the graph right by $h$ (left when $h < 0$ ), because $x - h$ reaches an old input value only when $x$ grows by $h$ . Points map by $(a, b) \to (a + h, b)$ .
The horizontal direction looks backward on purpose: subtracting inside makes the graph appear farther to the right, and adding inside makes it appear to the left.
Combined: $y = f(x - h) + k$ shifts right $h$ and up $k$ , sending $(a, b) \to (a + h, b + k)$ . For a parabola this is vertex form $y = (x - h)^2 + k$ , with vertex $(h, k)$ .
A horizontal shift moves the domain and the x-intercepts sideways by $h$ . A vertical shift moves the range up or down by $k$ and can change the number of x-intercepts.

Shifting Graphs

Learning goals

Sliding a graph without building a new table

Vertical shifts: change the output

Horizontal shifts: the direction that looks backward

Combining a horizontal and a vertical shift

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice