Lesson 1.3 · Unit 1 · F-BF.3

Four moves, and why the horizontal ones run backward

A transformation changes a graph in a predictable way, and there are only four kinds. The one thing that reliably confuses students is that changes inside the function act opposite to their sign, so this lesson explains why rather than asking you to memorize it.

The method
  1. The general form is \( g(x) = a \cdot f\big(b(x - h)\big) + k \), and each letter does one job.
  2. \( k \) shifts vertically, up for positive, and it acts as its sign suggests because it is applied after the function.
  3. \( h \) shifts horizontally, right for positive, because the form subtracts it.
  4. \( a \) stretches vertically by a factor of \( \left| a \right| \), and reflects in the \( x \)-axis if it is negative.
  5. \( b \) compresses horizontally by a factor of \( \dfrac{1}{\left| b \right|} \), and reflects in the \( y \)-axis if it is negative.
  6. Anything inside the function acts on the input, so it runs opposite to intuition; anything outside acts on the output and runs as expected.
  7. Factor out \( b \) before reading \( h \). In \( f(2x - 6) \) the shift is 3, not 6, because \( 2x - 6 = 2(x - 3) \).
  8. Order: horizontal stretch, then horizontal shift, then vertical stretch, then vertical shift. Inside-out.

Where students lose marks: reading the shift straight off \( f(2x - 6) \) as 6 units. The horizontal shift is whatever makes the inside zero, here \( x = 3 \). Factoring first makes it visible and removes the guesswork.

Worked example

The problem. (a) Describe the transformations taking \( f(x) = x^2 \) to \( g(x) = -2(x + 3)^2 + 5 \), and give the vertex. (b) Describe \( y = \sqrt{2x - 6} \) as a transformation of \( \sqrt{x} \). (c) Explain why the horizontal transformations act opposite to their sign. (d) Write the function whose graph is \( \left| x \right| \) reflected in the \( x \)-axis, stretched vertically by 3, and moved 2 left and 1 down.

Step one: match (a) to the general form. \( g(x) = -2(x + 3)^2 + 5 \) has \( a = -2 \), \( b = 1 \), \( h = -3 \) and \( k = 5 \), reading \( (x + 3) \) as \( (x - (-3)) \).

Step two: list the transformations for (a). \( a = -2 \): reflect in the \( x \)-axis, then stretch vertically by 2. \( h = -3 \): shift 3 units left. \( k = 5 \): shift 5 units up.

Step three: find the vertex. The parent's vertex is at the origin. Moving 3 left and 5 up puts it at \( (-3, 5) \), and the negative \( a \) means the parabola opens downward, so that vertex is a maximum. Check by substituting: \( g(-3) = -2(0) + 5 = 5 \). Correct. A second check: \( g(-2) = -2(1) + 5 = 3 \), which is below 5. Consistent with a maximum.

Step four: factor before reading (b). \( \sqrt{2x - 6} = \sqrt{2(x - 3)} \). Now \( b = 2 \) and \( h = 3 \) are visible.

Step five: describe (b). A horizontal compression by a factor of \( \dfrac{1}{2} \), then a shift 3 units right. Check the starting point: the parent starts at \( (0, 0) \), and this graph starts where the radicand is zero, at \( x = 3 \). Correct. Had the shift been read as 6, the graph would have been drawn starting at \( x = 6 \), where the actual value is \( \sqrt{6} \approx 2.449 \), not 0.

Step six: begin (c). Ask what input the new function needs in order to do what the old one did at a particular value. Take \( g(x) = f(x - 3) \). For \( g \) to produce what \( f \) produced at 0, the inside must equal 0, so \( x - 3 = 0 \) and \( x = 3 \).

Step seven: draw the conclusion for (c). The behavior that used to happen at \( x = 0 \) now happens at \( x = 3 \), so the graph has moved 3 units right, even though the formula says minus 3. The same reasoning covers the stretch. In \( f(2x) \), the inside reaches a given value at half the \( x \), so everything happens twice as early and the graph is squeezed toward the \( y \)-axis by a factor of \( \dfrac{1}{2} \), despite the 2. Vertical changes need no such reasoning because they are applied to the answer after the function has done its work, so adding 5 genuinely raises the output by 5. The one-sentence version: horizontal changes are demands on the input, and to meet a larger demand you supply a smaller input.

Step eight: build the function for (d). Reflection and vertical stretch by 3 give \( a = -3 \). Two left gives \( h = -2 \). One down gives \( k = -1 \). \( y = -3\left| x + 2 \right| - 1 \). Check the corner: the parent's corner is at the origin, so it moves to \( (-2, -1) \). Substituting: \( -3\left| 0 \right| - 1 = -1 \). Correct. Check the shape: at \( x = -1 \), \( y = -3(1) - 1 = -4 \), which is below the corner, so the V opens downward as the reflection requires. Correct.

Practice · 10 problems

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

  1. Describe the transformation from \( x^2 \) to \( x^2 + 4 \).
    Show the full solution

    Up 4

  2. Describe the transformation from \( x^2 \) to \( (x - 5)^2 \).
    Show the full solution

    Right 5

  3. Describe the transformation from \( \left| x \right| \) to \( -\left| x \right| \).
    Show the full solution

    Reflection in the \( x \)-axis

  4. Describe the transformation from \( \sqrt{x} \) to \( 4\sqrt{x} \).
    Show the full solution

    Vertical stretch by 4

  5. Give the vertex of \( y = (x + 1)^2 - 7 \).
    Show the full solution

    \( (-1, -7) \)

  6. Describe \( y = \dfrac{1}{x - 2} + 3 \) as a transformation, and give its asymptotes.
    Show the full solution

    From the reciprocal parent: right 2, up 3. The parent's asymptotes are \( x = 0 \) and \( y = 0 \). Moving right 2 carries the vertical asymptote to \( x = 2 \); moving up 3 carries the horizontal one to \( y = 3 \). Check: at \( x = 3 \), \( y = 1 + 3 = 4 \). At \( x = 102 \), \( y = 0.01 + 3 \), close to 3. Consistent with a horizontal asymptote at 3. Right 2 and up 3; asymptotes \( x = 2 \) and \( y = 3 \)

  7. Describe \( y = \sqrt{3x + 12} \) as a transformation of \( \sqrt{x} \).
    Show the full solution

    Factor the inside first: \( \sqrt{3x + 12} = \sqrt{3(x + 4)} \). So \( b = 3 \) and \( h = -4 \): a horizontal compression by \( \dfrac{1}{3} \), then a shift 4 units left. Check the starting point: the radicand is zero at \( x = -4 \), and the graph starts there. Reading the shift as 12 would have put the start at \( x = -12 \), where the radicand is \( -24 \) and the function is undefined. Compression by \( \frac{1}{3} \), then left 4

  8. Write the equation of \( y = x^3 \) reflected in the \( y \)-axis, then moved 1 right and 2 up. Simplify if possible.
    Show the full solution

    Reflection in the \( y \)-axis replaces \( x \) with \( -x \): \( y = (-x)^3 = -x^3 \). Right 1: replace \( x \) with \( x - 1 \), giving \( y = -(x - 1)^3 \). Up 2: \( y = -(x - 1)^3 + 2 \). A simplification worth noticing. Since the cubic is odd, \( -(x-1)^3 = (1 - x)^3 \), so the answer can also be written \( y = (1 - x)^3 + 2 \). Both are correct; the first shows the transformations more clearly. Check at \( x = 1 \): \( y = 0 + 2 = 2 \). The parent's inflection point at the origin has moved to \( (1, 2) \), as the shifts require. Check at \( x = 2 \): \( y = -(1) + 2 = 1 \), below the inflection point, which is right for a reflected cubic. \( y = -(x - 1)^3 + 2 \)

  9. Explain why \( f(x) + 3 \) and \( f(x + 3) \) are different transformations.
    Show the full solution

    They act at different stages, and that is the whole difference. \( f(x) + 3 \) feeds \( x \) into the function, gets an answer, then adds 3 to that answer. Every output rises by 3, so the graph moves up 3. \( f(x + 3) \) adds 3 to the input before the function sees it. The function then behaves at \( x = -3 \) the way it used to behave at \( 0 \), so the graph moves left 3. A concrete comparison. Take \( f(x) = x^2 \). \( f(x) + 3 = x^2 + 3 \), with vertex \( (0, 3) \). \( f(x + 3) = (x + 3)^2 \), with vertex \( (-3, 0) \). Different graphs entirely: one moved vertically, one horizontally, in opposite apparent directions from their signs. The general rule this illustrates. Outside the function means the output, and acts as written. Inside the function means the input, and acts opposite. Everything in this lesson is that one distinction applied four times. One adds to the output and moves the graph up; the other adds to the input and moves it left

  10. A graph of \( y = f(x) \) passes through \( (2, 5) \) and \( (-1, 0) \). Find the corresponding points on \( y = -2f(x - 1) + 4 \).
    Show the full solution

    Work out what happens to a general point. Suppose \( (p, q) \) is on \( y = f(x) \), so \( f(p) = q \). For the new function to use that value, its input must satisfy \( x - 1 = p \), so \( x = p + 1 \). The \( x \)-coordinate increases by 1. The output is then \( -2q + 4 \). So \( (p, q) \) maps to \( (p + 1,\; -2q + 4) \). Apply it to \( (2, 5) \). New \( x \): \( 2 + 1 = 3 \). New \( y \): \( -2(5) + 4 = -6 \). The point is \( (3, -6) \). Apply it to \( (-1, 0) \). New \( x \): \( -1 + 1 = 0 \). New \( y \): \( -2(0) + 4 = 4 \). The point is \( (0, 4) \). Check the order of operations on the output. The stretch and reflection apply before the shift, because the form is \( -2f(\cdot) + 4 \): multiply first, then add. Adding 4 first and then multiplying would give \( -2(5 + 4) = -18 \), which is wrong. Sanity check on the \( x \)-coordinate. The transformation is a shift right by 1, so both \( x \)-coordinates should increase by exactly 1, and they do. A note on what does not change. The horizontal shift does not touch the \( y \)-coordinate and the vertical work does not touch the \( x \)-coordinate. Keeping the two columns separate is the reliable way to do these. \( (3, -6) \) and \( (0, 4) \)

← 1.2 The parent functions of the year1.4 Even and odd functions →

Keep going in the full Algebra 2 course

All of U1 (Unit 1: Functions, Transformations and Inverses) 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