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How do stretches and reflections change a function graph?

Quick answer

For g(x)=a f(x), the multiplier changes every output: points (x,y) become (x,ay). If |a|>1, the graph stretches vertically; if 0<|a|<1, it compresses vertically. A negative a also reflects the graph across the x-axis. Changing the input instead, as in f(bx), changes horizontal positions; a negative b reflects across the y-axis. [S1]

LET’S WORK THROUGH IT

The question, the details, the next step.

Question

What does multiplying a function by a number do to its graph? I keep mixing up which direction the stretch goes.

Geez

Check where the multiplier is placed. In g(x)=a f(x), it multiplies the output, so it changes the graph vertically: each point (x,y) on f becomes (x,ay) on g. The x-coordinate stays put. If a is greater than 1, points move farther from the x-axis and the graph is stretched vertically. If 0<a<1, they move toward the x-axis and the graph is compressed vertically. [S1]

Use the parent function f(x)=√x. It has points (0,0), (1,1), and (4,2). For g(x)=2√x, those points become (0,0), (1,2), and (4,4). The graph is taller, but its x-values have not changed.

Follow-up question

How can I tell a reflection from a stretch, especially when the multiplier is negative?

Geez

For g(x)=a f(x), look at both the size and sign of a. A negative multiplier reverses the sign of every output, reflecting the graph across the x-axis. Its absolute value determines any additional vertical scaling. For example, g(x)=−2√x sends (1,1) to (1,−2) and (4,2) to (4,−4): the points are below the x-axis and twice as far from it as before. [S1]

A useful coordinate check is: original point (x,y) becomes (x,ay). A point on the x-axis, where y=0, stays there. This example shows the rule for the square-root function; it does not mean every graph has the same shape or domain.

Follow-up question

Then what changes in f(2x)? Is that also a vertical stretch?

Geez

No. In f(2x), the 2 is inside the input, so it changes horizontal positions, not output heights. To match an original point (u,f(u)), solve 2x=u: the corresponding point is (u/2,f(u)). For √x, the original point (4,2) therefore appears at (2,2) on √(2x). The graph is compressed horizontally toward the y-axis. In general, a positive input factor greater than 1 compresses horizontally; a negative input factor also reflects across the y-axis. [S1]

Compare the two formulas at that same original point: 2√x moves (4,2) to (4,4), changing height; √(2x) moves it to (2,2), changing horizontal position. This point-tracking method helps avoid relying on whether a curve merely looks wider or taller.

Follow-up question

Can you give me a quick way to work these out on homework?

Geez

First identify whether the number multiplies the whole function or is inside its input. Then track one or two known points before sketching. For example, starting with f(x)=√x and point (1,1):

  • 3f(x): (1,1) becomes (1,3), a vertical stretch.
  • −f(x): (1,1) becomes (1,−1), an x-axis reflection.
  • f(3x): (1,1) corresponds to (1/3,1), a horizontal compression.
  • f(−x): the input reversal reflects across the y-axis; for √x, the real-valued domain changes, so do not assume the whole original curve simply appears on both sides.

The general transformation rules describe how coordinates relate where the functions are defined. A quick sketch should also respect the original function's domain. [S1]

A practical checklist

  • Locate the multiplier: outside f(x) affects outputs; inside the input affects horizontal positions.

  • For a·f(x), change each point (x,y) to (x,ay); check whether a is negative.

  • For f(bx), solve bx=u for the new x-coordinate of an original point (u,f(u)).

  • Track at least two known points and keep the original function's domain in mind.

  • Label the reflection axis and whether the graph is stretched or compressed.

A few more useful answers

Does a negative vertical multiplier change the domain?

Multiplying the outputs changes their values and may reflect them across the x-axis; it does not by itself change which inputs the original function accepts.

Why does f(2x) compress horizontally instead of stretching?

An original input u is reached when 2x=u, so its new horizontal position is u/2. The graph's corresponding point moves closer to the y-axis.

What is the difference between reflecting across the x-axis and the y-axis?

An x-axis reflection reverses output signs, as in −f(x). A y-axis reflection reverses input signs, as in f(−x). [S1]

Sources

References used for this guide.

  1. [S1] OpenStax Precalculus 2e: Transformations of Functions
    openstax.org · Checked

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