Algebra 1 · Grade 9

Transformations of Functions: Shifts, Stretches and Reflections

Quick answer

A constant added outside a function moves its graph up or down. A constant added inside moves it left or right — and in the opposite direction to the sign, because the input has to work harder to reach the same value. Multiplying outside stretches vertically, multiplying inside squeezes horizontally, and a minus sign reflects.

What you'll learn

  • Predict the graph of f(x) + k, f(x + k), k·f(x) and f(kx)
  • Explain why an inside change moves the graph the opposite way
  • Read the transformation constants from a pair of graphs

One parent, many graphs

Start with a function and call it ff. Small changes to the formula move, stretch or flip its graph without changing its basic shape.

That is worth knowing because it turns a library of graphs into one graph plus a set of moves. Every parabola is y=x2y = x^2 transformed.

ChangeEffect on the graph
f(x)+kf(x) + kup kk
f(x)−kf(x) - kdown kk
f(x+k)f(x + k)left kk
f(x−k)f(x - k)right kk
k⋅f(x)k \cdot f(x), k>1k > 1vertical stretch
k⋅f(x)k \cdot f(x), 0<k<10 < k < 1vertical squeeze
f(kx)f(kx), k>1k > 1horizontal squeeze
−f(x)-f(x)reflect over the xx-axis
f(−x)f(-x)reflect over the yy-axis

Outside the parentheses acts on the output, so it behaves as written. Inside acts on the input, so it behaves backwards. That single sentence covers most of the table.

Vertical shifts read as written

g(x)=f(x)+3g(x) = f(x) + 3

Every output is 33 larger, so every point moves up 33. Nothing about the horizontal position changes.

y = x squared and the same curve shifted up 3 A grid from -4 to 4 across and -2 to 12 up. Two U-shaped curves of identical shape. The lower has its lowest point at the origin; the upper is the same curve raised so its lowest point is at (0, 3). -4-224-224681012xy
  • y = x^2
  • y = x^2 + 3
y = x squared and the same curve shifted up 3

The two curves are the same width and the same shape. Only their height differs.

Why horizontal shifts go the other way

g(x)=f(x+3)g(x) = f(x + 3)

This one surprises people, so it is worth slowing down.

gg at x=−3x = -3 computes f(−3+3)=f(0)f(-3 + 3) = f(0). So whatever ff did at 00, gg does at −3-3 — three units to the left.

xxg(x)=f(x+3)g(x) = f(x+3) evaluatesWhich ff did at
−3-3f(0)f(0)00
−2-2f(1)f(1)11
00f(3)f(3)33

Every landmark of ff appears 33 earlier in gg. Adding inside makes the input arrive sooner, which slides the picture left.

y = x squared and the same curve shifted left 3 A grid from -7 to 4 across and -2 to 12 up. Two identical U-shaped curves. One has its lowest point at the origin; the other is the same curve moved three units to the left, with its lowest point at (-3, 0). -6-4-224-224681012xy
  • y = x^2
  • y = (x + 3)^2
y = x squared and the same curve shifted left 3

This is the same fact as vertex form. In y=(x−h)2+ky = (x - h)^2 + k, the minus inside is why the vertex is at +h+h.

Stretches

g(x)=k⋅f(x)g(x) = k \cdot f(x)

Multiplying the output scales every height by kk.

kkEffect
k>1k > 1taller — a vertical stretch
0<k<10 < k < 1flatter — a vertical squeeze
k<0k < 0scaled and flipped over the xx-axis

Points already on the xx-axis do not move, because k⋅0=0k \cdot 0 = 0. That makes the xx-intercepts the fixed points of a vertical stretch — a useful check.

Multiplying the input squeezes horizontally, and again the direction is backwards:

g(x)=f(2x)g(x) = f(2x)

gg at x=3x = 3 gives f(6)f(6), so gg reaches at 33 what ff reached at 66. Everything arrives at half the xx-value, so the graph is squeezed toward the yy-axis by a factor of 22.

Reflections

FormReflects overBecause
−f(x)-f(x)the xx-axisevery output flips sign
f(−x)f(-x)the yy-axisevery input flips sign

Once more the pattern holds: the minus outside affects heights, the minus inside affects positions along the xx-axis.

For f(x)=x2f(x) = x^2 the second does nothing visible, since (−x)2=x2(-x)^2 = x^2. A parabola is already symmetric about the yy-axis, so reflecting it there lands it on itself.

Reading k off a picture

Given the original and the moved graph, compare one landmark on each.

A parabola with vertex (0,0)(0, 0) becomes one with vertex (2,−5)(2, -5).

The vertex went right 22 and down 55:

g(x)=f(x−2)−5g(x) = f(x - 2) - 5

Right 22 is a minus inside; down 55 is a minus outside.

Take the transformations in order: horizontal first, then vertical. Stated the other way, the shift amounts stay the same but any stretch has to be applied before the vertical shift, or the shift gets stretched too.

Worked examples

Common mistakes

Practice problems

  1. Describe g(x)=f(x)−2g(x) = f(x) - 2.

    Answer

    Down 22

    Full solution

    Subtracting outside lowers every output.

  2. Describe g(x)=f(x−7)g(x) = f(x - 7).

    Answer

    Right 77

    Full solution

    A minus inside shifts in the positive direction.

  3. Describe g(x)=f(x+5)g(x) = f(x + 5).

    Answer

    Left 55

    Full solution

    A plus inside shifts left.

  4. Describe g(x)=4f(x)g(x) = 4f(x).

    Answer

    Vertical stretch by 44

    Full solution

    Every output is four times as far from the xx-axis.

  5. Describe g(x)=−f(x)g(x) = -f(x).

    Answer

    Reflection over the xx-axis

    Full solution

    Every output changes sign.

  6. What is the vertex of y=(x−6)2+1y = (x - 6)^2 + 1?

    Answer

    (6,1)(6, 1)

    Full solution

    Right 66 and up 11 from the origin.

  7. What is the vertex of y=(x+2)2−3y = (x + 2)^2 - 3?

    Answer

    (−2,−3)(-2, -3)

    Full solution

    A plus inside shifts left.

  8. Write the function whose graph is y=x2y = x^2 moved left 44 and down 66.

    Hint

    Which direction needs a plus inside?

    Answer

    y=(x+4)2−6y = (x + 4)^2 - 6

    Full solution

    Left 44 is an inside change, and inside runs backwards, so it is +4+4.

    Down 66 is an outside change, and outside reads as written, so it is −6-6.

    The vertex lands at (−4,−6)(-4, -6), which is left 44 and down 66 from the origin.

  9. If f(2)=9f(2) = 9, what is g(2)g(2) for g(x)=f(x)+5g(x) = f(x) + 5? And for h(x)=f(x+5)h(x) = f(x + 5)?

    Answer

    g(2)=14g(2) = 14; h(2)=f(7)h(2) = f(7), which is unknown

    Full solution

    gg adds outside, so g(2)=f(2)+5=14g(2) = f(2) + 5 = 14.

    hh adds inside, so h(2)=f(7)h(2) = f(7). Knowing ff at 22 says nothing about ff at 77.

    That contrast is exactly the difference between changing the output and changing the input.

  10. Asked to describe g(x)=f(x+3)g(x) = f(x + 3), Kai says the graph moves right 33, because the sign is a plus. Find the error.

    Hint

    Evaluate gg at x=−3x = -3.

    Answer

    It moves left 33. An inside change runs opposite to its sign.

    Full solution

    Kai applied the outside rule to an inside change.

    Test it with a value. g(−3)=f(−3+3)=f(0)g(-3) = f(-3 + 3) = f(0).

    So whatever height ff had at x=0x = 0, gg has at x=−3x = -3 — three units to the left.

    The reason is that the +3+3 changes the input, not the output. The input now reaches any given value three units earlier, so the whole picture arrives earlier and slides left.

Frequently asked questions

What does f(x) + 3 do?

Shifts the graph up 3. Every output is 3 larger, so every point rises.

What does f(x + 3) do?

Shifts the graph left 3, not right. The input reaches any given value three units earlier.

Why does the inside change go the opposite way?

Because it changes the input, not the output. Adding 3 inside means x only has to reach -3 to produce what x = 0 used to produce.

What does -f(x) do?

Reflects the graph over the x-axis, since every output flips sign.

What does 2f(x) do?

Stretches the graph vertically by a factor of 2. Points on the x-axis stay put, because twice zero is still zero.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSF.BF.B.3Building FunctionsIdentify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.