Stretching and Reflecting Graphs

Intuition

The last lesson slid graphs around without changing their shape. This one changes the shape itself. Take a graph you already know and you can make it taller or shorter, flip it over an axis, or squeeze it narrower. All of that comes from editing the rule rather than from replotting a table. Multiplying the output by a number stretches the graph up and down; putting a minus sign on the output turns it upside down. The genuinely surprising move is horizontal scaling: when you multiply the input by a number bigger than $1$ , the graph gets narrower, not wider. This lesson derives why that direction looks backward, the same way the shift lesson explained why $x - h$ moves a graph to the right. By the end you will read a rule like $y = -2f(x)$ or $y = f(3x)$ and know exactly what it does to the picture.

Learning goals

Scale heights with $y = a f(x)$ , stretching when $a > 1$
Flip across the $x$ -axis with $y = -f(x)$
Compress toward the $y$ -axis with $y = f(bx)$ when $b > 1$
Mirror across the $y$ -axis with $y = f(-x)$
Read $y = a f(x - h) + k$ as vertex form for a parabola
Say which of domain and range each kind of scaling reshapes

Vertical stretches and compressions: scale the output

Start with the graph of $y = f(x)$ and multiply every output by a positive constant $a$ :

y = a\,f(x).

Each input still passes through $f$ , but whatever comes out is then multiplied by $a$ . Because the output is the height of the graph, scaling every output by $a$ scales the height of every point by $a$ . When $a > 1$ that makes the graph taller and steeper (a vertical stretch); when $0 < a < 1$ it makes the graph shorter and flatter (a vertical compression).

Why $y = a\,f(x)$ scales every height by $a$ #

Let $g(x) = a\,f(x)$ with $a > 0$ , and pick any input $x$ in the domain. The original graph has the point $(x, f(x))$ , sitting at height $f(x)$ directly above the input. The new graph has the point $(x, g(x))$ , and since $g(x) = a\,f(x)$ , that point is $(x,\, a\,f(x))$ . The two share the same first coordinate, so the new point lies on the same vertical line as the old one, at $a$ times its height.

Multiplying a height by $a > 1$ pushes the point farther from the horizontal axis, and multiplying by an $a$ between $0$ and $1$ pulls it closer. Both follow because scaling a distance by $a$ enlarges it when $a > 1$ and shrinks it when $a < 1$ . A point that already sits on the horizontal axis has height $0$ , and $a \times 0 = 0$ , so it does not move at all.

Since $x$ was arbitrary, every height is scaled by the same factor $a$ : the graph stretches vertically when $a > 1$ and compresses vertically when $0 < a < 1$ . In particular, every point on the horizontal axis, including all x-intercepts, stays pinned exactly where it was.

Reading points off the graph is immediate. If $(x_0, y_0)$ lies on $y = f(x)$ , then $(x_0,\, a y_0)$ lies on $y = a\,f(x)$ : keep the input, multiply the height by $a$ .

The dashed parabola is y = x squared. Multiplying every output by 2 gives the solid parabola y = 2x squared, which rises twice as high above each input, so it looks narrower and steeper. The point (1, 1) rises to (1, 2), while the vertex (0, 0) stays fixed because twice a height of 0 is still 0.

Worked example 1 Stretch and compress $y = x^2$

Both new rules keep $f(x) = x^2$ and only rescale its outputs, so shift the points you already know instead of building a fresh table.

For $y = 2x^2$ , multiply every height by $2$ . The vertex $(0, 0)$ stays at $(0, 0)$ because $2 \times 0 = 0$ , while the anchor points $(1, 1)$ and $(2, 4)$ climb to

(1,\, 2 \times 1) = (1, 2), \qquad (2,\, 2 \times 4) = (2, 8).

The bowl passes through those higher points, so it looks narrower and steeper, though it still opens upward from the same vertex.

For $y = \tfrac{1}{2}x^2$ , multiply every height by $\tfrac{1}{2}$ instead. The same anchor points sink to

\left(1,\, \tfrac{1}{2}\right), \qquad (2, 2),

so this bowl is wider and flatter. In both cases the vertex and the single x-intercept at the origin never budge. Scaling a height of $0$ leaves it at $0$ , so only the nonzero heights move.

Check your understanding

The point $(3, 4)$ lies on the graph of $y = f(x)$ . Which point must lie on the graph of $y = 5f(x)$ ?

(3, 20)

(15, 4)

(3, 9)

(8, 4)

Reflecting across the x-axis: negate the output

The case $a = -1$ deserves its own name. The rule

y = -f(x)

negates every output, sending a point at height $y_0$ to height $-y_0$ . A point above the horizontal axis drops to the mirror-image spot the same distance below it, and a point below rises to the matching spot above. That is exactly a reflection across the x-axis: the graph flips over that axis like a card turning face down. Points sitting on the axis have height $0$ , and $-0 = 0$ , so they stay fixed and act as the hinge of the flip.

A negative scale factor with size other than $1$ just does both jobs at once. Writing $a = -\lvert a \rvert$ , the rule $y = a\,f(x)$ scales every height by $\lvert a \rvert$ and then negates it, so it stretches or compresses vertically and reflects across the x-axis together.

The dashed parabola y = x squared opens upward with a lowest point at the origin. Negating every output gives the solid parabola y = negative x squared, the same bowl reflected across the x-axis so it opens downward with a highest point at the origin. The point (1, 1) flips to (1, negative 1).

Worked example 2 Reflect $y = x^2$ across the x-axis

The rule $y = -x^2$ is $y = -f(x)$ with $f(x) = x^2$ , so it takes the familiar upward bowl and negates every output. Track a few points by flipping the sign of each height:

(1, 1) \longrightarrow (1, -1), \qquad (2, 4) \longrightarrow (2, -4).

The vertex $(0, 0)$ stays put, since $-0 = 0$ . Plotting the flipped points gives the same parabola turned upside down, so it now opens downward. The origin that used to be its lowest point is now its highest. Nothing about the width changed, because reflecting only reverses the sign of each height without rescaling it. What was a minimum has become a maximum, purely from the minus sign.

Horizontal stretches and compressions: the direction that looks backward again

Now change the input instead of the output. Replace $x$ with $bx$ for a positive constant $b$ :

y = f(bx).

Here is the trap, and it mirrors the one from the shift lesson. You might expect multiplying the input by a large $b$ to stretch the graph wider. It does the opposite: for $b > 1$ the graph is squeezed narrower, pulled in toward the vertical axis. Rather than memorize that, derive it, because the derivation shows precisely why the direction feels reversed.

Why $y = f(bx)$ scales every width by $\tfrac{1}{b}$ #

Let $g(x) = f(bx)$ with $b > 0$ . 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 $bx$ . So $g(x)$ produces that same height when the inside matches $a$ :

bx = a \quad \Longrightarrow \quad x = \frac{a}{b}.

Therefore $g\!\left(\tfrac{a}{b}\right) = f\!\left(b \cdot \tfrac{a}{b}\right) = f(a)$ , which means the new graph has the point $\left(\tfrac{a}{b},\, f(a)\right)$ . Compare it with the old point $(a, f(a))$ : both sit at the same height. The new point, though, is at input $\tfrac{a}{b}$ , the old input scaled by the factor $\tfrac{1}{b}$ .

When $b > 1$ the factor $\tfrac{1}{b}$ is less than $1$ , so $\tfrac{a}{b}$ is closer to $0$ than $a$ was. Every point is therefore pulled toward the vertical axis, a horizontal compression. When $0 < b < 1$ the factor $\tfrac{1}{b}$ is greater than $1$ , so $\tfrac{a}{b}$ is farther from $0$ : every point is pushed away, a horizontal stretch. A point on the vertical axis has $a = 0$ , and $\tfrac{0}{b} = 0$ , so it never moves.

The direction feels backward for exactly the reason it did with shifts. To make the inside $bx$ reach a value it used to reach at input $a$ , and since $b$ multiplies $x$ , you need a smaller $x$ when $b > 1$ . So each feature of the graph shows up at a smaller input, nearer the axis.

Replacing x with 2x pulls the whole curve in toward the y-axis by a factor of one half. The dashed curve y = f(x) has its peak at input 2; on the solid curve y = f(2x) that peak reappears at input 1, because the inside 2x equals 2 exactly when x = 1. Every input is halved, so b = 2 compresses the graph rather than stretching it.

Worked example 3 Locate points and intercepts on $y = f(2x)$ and $y = f\left(\tfrac{x}{2}\right)$

Suppose $(8, 5)$ lies on $y = f(x)$ , and suppose $f$ has x-intercepts at $x = 2$ and $x = 6$ . Handle the two scalings by asking, in each case, what input makes the inside equal an old input.

For $y = f(2x)$ , a point reappears where $2x$ equals the old input. To get the height $5$ back, set the inside equal to $8$ :

2x = 8 \quad \Longrightarrow \quad x = 4,

so $(8, 5)$ moves to $(4, 5)$ , half the input. The x-intercepts move the same way: $2x = 2$ gives $x = 1$ and $2x = 6$ gives $x = 3$ , so the zeros at $2$ and $6$ are pulled in to $1$ and $3$ . With $b = 2 > 1$ the whole graph is compressed toward the vertical axis by a factor of $\tfrac{1}{2}$ .

For $y = f\!\left(\tfrac{x}{2}\right)$ , the inside is $\tfrac{x}{2}$ , so $b = \tfrac{1}{2}$ . Now $\tfrac{x}{2} = 8$ gives $x = 16$ , so $(8, 5)$ moves out to $(16, 5)$ . The zeros stretch out too, to $x = 4$ and $x = 12$ . Because $0 < b < 1$ , the graph is stretched away from the vertical axis by a factor of $\tfrac{1}{b} = 2$ .

Check your understanding

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

A horizontally compressed toward the y-axis by a factor of

\tfrac{1}{4}

B horizontally stretched away from the y-axis by a factor of

4

C vertically stretched by a factor of

4

D shifted left

4

units

Reflecting across the y-axis: negate the input

The case $b = -1$ is the horizontal partner of the x-axis reflection. The rule

y = f(-x)

feeds $f$ the opposite input. Following the derivation above, a point $(a, f(a))$ reappears where the inside $-x$ equals $a$ , that is, at $x = -a$ , so it moves to $(-a, f(a))$ . Every point keeps its height but lands at the opposite horizontal position, which is a reflection across the y-axis: the graph flips left to right. A point already on the vertical axis has $a = 0$ , and $-0 = 0$ , so it stays fixed and serves as the hinge.

The dashed curve y = f(x) leans to the right, with its peak above input 2. Replacing x with negative x gives the solid curve y = f(negative x), the same shape flipped left to right across the y-axis, with its peak above input negative 2. Each point keeps its height and swaps to the opposite side of the vertical axis.

Worked example 4 Reflect a graph across the y-axis

Suppose $f$ passes through $(3, 5)$ and $(-1, 2)$ , and its only x-intercept is at $x = 4$ . Describe the graph of $y = f(-x)$ .

Reflection across the y-axis keeps every height and flips the sign of every input, so map each feature by negating its first coordinate:

(3, 5) \longrightarrow (-3, 5), \qquad (-1, 2) \longrightarrow (1, 2).

The x-intercept at $x = 4$ moves to $x = -4$ , since setting the inside $-x = 4$ gives $x = -4$ . Any point that happened to sit on the y-axis, where the input is $0$ , would not move at all. So $y = f(-x)$ is the mirror image of $y = f(x)$ across the vertical axis, with the whole picture swapped left for right while every height is preserved.

Check your understanding

The point $(-6, 2)$ lies on the graph of $y = f(x)$ . Which point must lie on the graph of $y = f(-x)$ ?

(6, 2)

(-6, -2)

(6, -2)

(-6, 2)

Combining a scaling with a shift

Scalings and shifts stack, so a rule can do several things at once. Reading it correctly means separating what happens to the output from what happens to the input. For a vertical scaling followed by a shift,

y = a\,f(x - h) + k,

the inside $x - h$ moves the graph right by $h$ , while the outside multiplies each height by $a$ and then adds $k$ . Tracking a single point makes the order clear: a point $(p, q)$ on $y = f(x)$ reappears at

\big(p + h,\ a q + k\big).

The input picks up the shift $h$ , and the output is first scaled by $a$ (which also flips it when $a$ is negative) and then raised by $k$ . The horizontal part touches only the first coordinate and the vertical part only the second, exactly as with pure shifts.

For a parabola this is the vertex form you already know, now read as a transformation. The rule $y = a(x - h)^2 + k$ takes $y = x^2$ , scales it vertically by $a$ (reflecting it downward when $a < 0$ ), then moves the vertex to $(h, k)$ . The number $a$ controls how narrow the bowl is and which way it opens; $h$ and $k$ say where its vertex sits.

The dashed parabola y = x squared opens upward from the origin. The rule y = negative 2x squared multiplies every output by negative 2, so it reflects across the x-axis and stretches to twice the height, giving the solid downward parabola. The point (1, 1) lands at (1, negative 2), and the shared vertex at the origin is now a maximum instead of a minimum.

Worked example 5 Graph $y = -2x^2$

Read the rule $y = -2x^2$ as $y = a\,x^2$ with $a = -2$ , a single number that does two jobs. Its size, $\lvert a \rvert = 2$ , stretches the bowl vertically to twice its height, and its sign, negative, reflects it across the x-axis so it opens downward.

Apply both to the anchor points of $y = x^2$ . Multiply each height by $-2$ :

(1, 1) \longrightarrow (1, -2), \qquad (2, 4) \longrightarrow (2, -8).

The vertex $(0, 0)$ holds still, since $-2 \times 0 = 0$ . Plotting the results gives a downward parabola, narrower than $y = x^2$ , whose highest point is the origin. What was the minimum of the plain bowl has become a maximum, and the graph is twice as steep, all encoded in the single coefficient $-2$ .

A scaling also reshapes the domain and range, and each type touches only one of them. A vertical scaling by $a$ multiplies every output by $a$ , so it scales the range. A horizontal scaling by $b$ divides every input by $b$ , so it scales the domain. A reflection swaps the endpoints of whichever interval it acts on.

Worked example 6 Track the domain and range through a scaling

Suppose $f$ has domain $0 \le x \le 8$ and range $-2 \le y \le 6$ . Describe the domain and range of two transformed graphs.

For $y = -2f(x)$ , only the outputs change, so the domain stays $0 \le x \le 8$ . Each output is multiplied by $-2$ , which scales the range endpoints and swaps them because the factor is negative:

-2 \times (-2) = 4, \qquad -2 \times 6 = -12,

so the new range runs from the lower value up, $-12 \le y \le 4$ . The reflection is what turned the top of the old range into the bottom of the new one.

For $y = f(2x)$ , only the inputs change, so the range stays $-2 \le y \le 6$ . Each input is divided by $2$ (the inside $2x$ reaches the old inputs at half the $x$ ), so the domain endpoints halve:

0 \le x \le 8 \quad \longrightarrow \quad 0 \le x \le 4.

The graph is squeezed into half the horizontal room while covering the same heights.

The scalings and reflections are worth keeping in one place. In each row, $a > 1$ and $b > 1$ .

Rule	Effect on the graph of $y = f(x)$
$y = a\,f(x)$	vertical stretch by $a$ (taller)
$y = \tfrac{1}{a}f(x)$	vertical compression by $a$ (shorter)
$y = -f(x)$	reflection across the x-axis
$y = f(bx)$	horizontal compression by $b$ (narrower)
$y = f\!\left(\tfrac{x}{b}\right)$	horizontal stretch by $b$ (wider)
$y = f(-x)$	reflection across the y-axis

Check your understanding

The parabola $y = -3x^2$ transforms $y = x^2$ . Which description is correct?

A reflected across the x-axis and stretched vertically by

3

B reflected across the y-axis and stretched vertically by

3

C compressed vertically by

3

and opens upward D shifted down

3

units

Common mistakes

Scaling horizontally the wrong way. $y = f(bx)$ with $b > 1$ compresses the graph toward the vertical axis; it does not stretch it. The width scales by $\tfrac{1}{b}$ , so a larger $b$ means a narrower graph. Check it with one point: the inside $bx$ reaches an old input at $x = \tfrac{a}{b}$ , which is closer to $0$ .
Mixing up which change affects which direction. A number multiplied outside the function, $a\,f(x)$ , is a vertical scaling (it changes heights). A number multiplied with $x$ inside the function, $f(bx)$ , is a horizontal scaling (it changes positions). Watch where the factor sits, just as with shifts.
Adding instead of multiplying. A stretch multiplies a coordinate; a shift adds to it. For $y = 3f(x)$ a point $(2, 5)$ goes to $(2, 15)$ , not $(2, 8)$ . Multiply the height, do not add.
Moving a point in the wrong coordinate. A vertical scaling changes only the output, so $(x_0, y_0) \to (x_0,\, a y_0)$ . A horizontal scaling changes only the input, so $(x_0, y_0) \to \left(\tfrac{x_0}{b},\, y_0\right)$ . Do not rescale both coordinates for a single factor.
Thinking a reflection changes the shape. Reflecting across either axis flips the graph but never stretches or compresses it. Only a scale factor whose size differs from $1$ changes how tall or wide the graph is.
Losing the fixed points. A vertical scaling pins every x-intercept in place (a height of $0$ scales to $0$ ); a horizontal scaling pins the y-intercept (an input of $0$ scales to $0$ ). If those anchor points drifted, recheck the work.

Key takeaways

Vertical scaling: $y = a\,f(x)$ multiplies every height by $a$ , stretching the graph when $a > 1$ and compressing it when $0 < a < 1$ . Points map by $(x_0, y_0) \to (x_0,\, a y_0)$ , and every x-intercept stays fixed.
Reflection across the x-axis: $y = -f(x)$ negates every output, flipping the graph over the x-axis (the case $a = -1$ ). A negative $a$ scales and flips at once.
Horizontal scaling: $y = f(bx)$ divides every input by $b$ , so $b > 1$ compresses the graph toward the y-axis and $0 < b < 1$ stretches it away. Points map by $(x_0, y_0) \to \left(\tfrac{x_0}{b},\, y_0\right)$ , and the direction looks backward for the same reason shifts do.
Reflection across the y-axis: $y = f(-x)$ negates every input, flipping the graph over the y-axis (the case $b = -1$ ), sending $(x_0, y_0) \to (-x_0,\, y_0)$ .
Combined: $y = a\,f(x - h) + k$ sends $(p, q) \to (p + h,\ a q + k)$ ; for a parabola this is vertex form $y = a(x - h)^2 + k$ , where $a$ sets the width and opening and $(h, k)$ is the vertex.
A vertical scaling reshapes the range and a horizontal scaling reshapes the domain; a reflection swaps the endpoints of the interval it acts on.

Stretching and Reflecting Graphs

Learning goals

Vertical stretches and compressions: scale the output

Reflecting across the x-axis: negate the output

Horizontal stretches and compressions: the direction that looks backward again

Reflecting across the y-axis: negate the input

Combining a scaling with a shift

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)