Lesson 1.3 · Unit 1 · F-BF.3

Shifting, stretching and reflecting a graph by changing its rule

Every function in this course is a parent with four kinds of change applied to it. The skill is to read a rule such as \( -2\sqrt{x + 3} + 1 \) and see a square root that has been moved and flipped, then track a few key points through those changes instead of plotting from scratch.

The method
  1. The general form is \( y = a\,f\big(b(x - h)\big) + k \).
  2. \( k \) shifts the graph up (positive) or down (negative). It acts on the output.
  3. \( h \) shifts the graph right (positive \( h \)) or left. The shift is opposite to the sign as written: \( x + 3 \) moves the graph 3 to the left.
  4. \( a \) stretches vertically by \( |a| \), and reflects in the \( x \)-axis if \( a \lt 0 \).
  5. \( b \) compresses horizontally by a factor of \( |b| \), and reflects in the \( y \)-axis if \( b \lt 0 \).
  6. Factor the inside so that \( x \) stands alone before reading the shift: \( f(2x - 6) = f\big(2(x - 3)\big) \) shifts by 3, not by 6.
  7. To move a key point \( (x, y) \), apply the inside changes to \( x \) and the outside changes to \( y \): the new point is \( \left(\dfrac{x}{b} + h,\ ay + k\right) \).
  8. The domain and range move with the graph: horizontal changes act on the domain, vertical changes act on the range.

Where students lose marks: reading the shift off an unfactored inside. For \( f(2x - 6) \) the graph moves right by 3, because \( 2x - 6 = 2(x - 3) \). Reading the 6 directly gives a shift twice as large.

Worked example

The problem. (a) Describe how \( g(x) = -2\sqrt{x + 3} + 1 \) comes from \( f(x) = \sqrt{x} \), then track the key points \( (0, 0), (1, 1), (4, 2), (9, 3) \). (b) State the domain and range of \( g \). (c) Find the point of \( y = \sqrt{2x} \) that comes from \( (4, 2) \) on \( f \). (d) Describe how \( y = |2x - 6| + 1 \) comes from \( y = |x| \).

Step one: read the pieces for (a). Inside: \( x + 3 \), a shift of 3 to the left. Outside: the factor \( -2 \), a reflection in the \( x \)-axis with a vertical stretch of 2; and the \( +1 \), a shift up by 1.

Step two: write the point rule. Horizontal change first: each \( x \) becomes \( x - 3 \). Then vertical: each \( y \) becomes \( -2y + 1 \).

Step three: apply it to each key point. \( (0, 0) \to (-3, 1) \). \( (1, 1) \to (-2, -1) \). \( (4, 2) \to (1, -3) \). \( (9, 3) \to (6, -5) \). Check the third with the rule directly: \( g(1) = -2\sqrt{4} + 1 = -4 + 1 = -3 \). Correct.

Step four: read the domain and range for (b). The starting point of the graph moved to \( (-3, 1) \), and the graph runs to the right and downward from there. Domain \( [-3, \infty) \), range \( (-\infty, 1] \). Check the domain from the rule: \( x + 3 \ge 0 \) gives \( x \ge -3 \). Correct.

Step five: handle the compression in (c). For \( y = \sqrt{2x} \) the inside is \( 2x \), so \( b = 2 \) and each \( x \) is divided by 2. The point \( (4, 2) \) becomes \( (2, 2) \). Check: \( \sqrt{2 \cdot 2} = \sqrt{4} = 2 \). Correct.

Step six: a caution about what compression means. A horizontal compression by 2 pulls every point toward the \( y \)-axis, so the graph rises faster. Points move to half their original \( x \)-coordinate, not double. The opposite-looking direction is the usual source of error: a factor inside that is larger than 1 makes the graph narrower.

Step seven: factor the inside for (d). \( |2x - 6| + 1 = |2(x - 3)| + 1 \). Now the inside reads clearly: a horizontal compression by 2 and a shift right by 3.

Step eight: locate the vertex and verify. The vertex of \( |x| \) at \( (0, 0) \) becomes \( (0/2 + 3, 0 + 1) = (3, 1) \). Check in the rule: \( |2(3) - 6| + 1 = 0 + 1 = 1 \), and the minimum of an absolute value is at the zero of its inside, \( 2x - 6 = 0 \), which is \( x = 3 \). Correct. The result is that the order of the change matters only when a horizontal stretch and a horizontal shift occur together, and factoring the inside removes the ambiguity.

Practice · 10 problems

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

  1. Describe the graph of \( y = f(x - 4) + 2 \) relative to \( y = f(x) \).
    Show the full solution

    The \( x - 4 \) moves the graph to the right by 4, and the \( +2 \) moves it up by 2. Right 4, up 2

  2. What does \( y = -f(x) \) do to the graph of \( f \)?
    Show the full solution

    Every output changes sign. Reflection in the \( x \)-axis

  3. What does \( y = f(-x) \) do to the graph of \( f \)?
    Show the full solution

    Every input changes sign before the function sees it. Reflection in the \( y \)-axis

  4. The point \( (3, 5) \) is on \( y = f(x) \). Find the corresponding point on \( y = f(x + 2) - 1 \).
    Show the full solution

    The \( x + 2 \) shifts left by 2: \( x \) becomes 1. The \( -1 \) lowers \( y \) to 4. \( (1, 4) \)

  5. The point \( (3, 5) \) is on \( y = f(x) \). Find the corresponding point on \( y = 3f(x) \).
    Show the full solution

    The factor 3 multiplies the output only. \( (3, 15) \)

  6. The point \( (4, 8) \) is on \( y = f(x) \). Find the corresponding point on \( y = f(2x) \).
    Show the full solution

    We need \( 2x = 4 \), so \( x = 2 \); the output stays 8. \( (2, 8) \)

  7. Write the equation of the parabola obtained from \( y = x^2 \) by shifting 3 left, reflecting in the \( x \)-axis and shifting up 5.
    Show the full solution

    Left 3: \( (x + 3)^2 \). Reflect: \( -(x + 3)^2 \). Up 5: add 5. Vertex \( (-3, 5) \), opening downward. Check at \( x = -3 \): the value is 5. Correct. \( y = -(x + 3)^2 + 5 \)

  8. The point \( (4, 1) \) is on \( y = f(x) \). Find the corresponding point on \( y = f(2x - 6) \).
    Show the full solution

    We need the inside to equal 4: \( 2x - 6 = 4 \), so \( x = 5 \). The output stays 1. Confirm with the factored form \( f\big(2(x - 3)\big) \): halve the original \( x \), \( 4/2 = 2 \), then shift right by 3 to get 5. Both routes agree. \( (5, 1) \)

  9. Find the domain and range of \( y = -3\sqrt{x - 2} + 4 \).
    Show the full solution

    Domain: \( x - 2 \ge 0 \), so \( x \ge 2 \). The root is at least 0, so \( -3\sqrt{x - 2} \le 0 \) and the whole expression is at most 4. The largest value, 4, occurs at \( x = 2 \). Domain \( [2, \infty) \), range \( (-\infty, 4] \)

  10. Does it matter whether a vertical stretch or a vertical shift is applied first? Test with \( f(x) = x \) at \( x = 1 \), comparing \( 2f(x) + 3 \) with \( 2\big(f(x) + 3\big) \).
    Show the full solution

    \( 2f(1) + 3 = 2 + 3 = 5 \), but \( 2\big(f(1) + 3\big) = 2(4) = 8 \). They differ, so the order matters, and the notation records it. In \( a\,f(x) + k \) the stretch is applied to the function and the shift afterward. The second expression is really \( 2f(x) + 6 \), a shift of 6. The same caution applies to the inside: the order of a horizontal stretch and a shift is settled by factoring so that \( x \) stands alone. Order matters; the written form fixes it

← 1.2 Parent functions and their graphs1.4 Symmetry: even and odd functions →

Keep going in the full Precalculus course

All of U1 (Unit 1: Functions and Their Transformations) is free to everyone, with no account. The course has 75 lessons, each with worked examples and practice problems that come with full solutions. A free account can also start a free month of the whole course, with no card.

Open this lesson in the courseSee the whole course