Transforming Graphs: Shifts, Stretches & Reflections

Pick a parent graph and set the four numbers in y = a·f(bx + c) + d. The page draws the parent faint and the moved curve solid, with an arrow from each marked point to where it lands, and works out the shift, the stretches and the landing point of each one.

Worked example — Transforming Graphs: Shifts, Stretches & Reflections

Horizontal shift, where the inside is zero	3
Horizontal stretch factor	1
Vertical stretch factor	1
Vertical shift	0
Where the vertex (0, 0) lands	(3, 0)

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Transforming Graphs: Shifts, Stretches & Reflections

A transformation of a graph moves every point of it by the same rule. For y = a·f(bx + c) + d, the point (p, q) of y = f(x) goes to ((p − c)/b, a·q + d). The two numbers inside the bracket, b and c, act on x before f is applied and move the graph sideways: its horizontal shift is −c/b and its widths are multiplied by 1/b. The two outside, a and d, act on the height after f and move it up and down.

What each symbol means

a: the vertical stretch factor; a negative a also reflects the graph in the x-axis
b: the number multiplying x inside the bracket; widths are multiplied by 1/b, and a negative b also reflects the graph in the y-axis
c: the number added to bx inside the bracket
d: the vertical shift
h: the horizontal shift, −c/b, the x at which the inside of the bracket is zero

Assumes: The formula is read in its own order: bx + c is worked out first, then f, then the multiplication by a, then the addition of d. Written the other way round, a·(f(bx + c) + d), the curve is lifted by a·d rather than by d.
Breaks when: The four numbers can shift, stretch and reflect a graph in each direction, and nothing else. They cannot rotate it. A parabola turned through any angle short of a half turn is not the graph of a function of x at all, because far enough out a vertical line meets it twice.

A minus sign inside the bracket moves the graph right 🖖

Press (x − 3)². Every arrow points to the right, and the vertex lands at (3, 0). When Rina Zazkis and two colleagues asked secondary students and teachers about this very parabola in 2003, most of them found the rightward shift counterintuitive. The new curve at x takes the value that x² had at x − 3, so it reaches the parabola's lowest value only once x − 3 = 0, at x = 3: it is the old graph arriving three units late. Whatever is done to x inside the bracket happens before f sees it, and the graph shows the undoing. Subtract 3 and the graph moves 3 the other way; multiply by 2, as in sin 2x, and the graph gets there twice as early, so every width halves and the point of sine at 2π, 6.2832, lands at 3.1416. The two numbers outside, a and d, act on the height after f has done its work, and they do exactly what they say.

The 6 in √(2x − 6) is not the shift 🖖

Press 2√(2x − 6) + 1. The curve starts at x = 3. A square root starts where the thing under it is zero, and 2x − 6 is zero at 3, not at 6. Factor the inside first, 2x − 6 = 2(x − 3), and the shift is the number left in the bracket. The first row of the results table does exactly that, −c/b = −(−6)/2. There are two correct ways to read the moves off the formula: shift the graph right by 6 and then squeeze it to half its width towards the y-axis, which pulls the start from 6 in to 3, or squeeze first and shift by 3. The reading that goes wrong is the one taking the numbers in the order they are written, squeezing and then shifting by 6.

(2x)² and 4x² are the same curve 🖖

Press (2x)². The graph is the parabola squeezed to half its width, and the results table's last row writes it as 4x², the parabola stretched to four times its height. They are one curve, because (2x)² = 2²·x². For b > 0 the same identity, (bx)ⁿ = bⁿxⁿ, holds for every power, so for x², |x|, x³, √x and 1/x any horizontal stretch can be traded for a vertical one, and nothing in the graph records which move was made. The exponential trades the other move. 2ˣ⁺³ is 2ˣ shifted 3 to the left, and it is also 8·2ˣ, because 2ˣ⁺³ = 2³·2ˣ. Sine is the one parent on this page that trades neither: squeezing it, or shifting it by anything but a multiple of π, moves its zeros, and a vertical stretch leaves every zero where it was.

Problem solved in full

Moving √x to start at (3, 1) and pass through (5, 5) 4 steps
Find a, b, c and d so that y = a√(bx + c) + d starts at (3, 1) and passes through (5, 5). Then say how many answers there are. Press 2√(2x − 6) + 1 to check one of them.

Open this problem in the tool
1. A square root starts where the inside is zero, and the height there is d. The inside has to vanish at x = 3, which fixes c at −3b, and the start's height fixes d at 1.
  
  $b x + c = 0 at x = 3 \Rightarrow c = - 3 b, d = 1$
2. Now the second point. At x = 5 the inside is 5b − 3b = 2b, so the condition is a√(2b) + 1 = 5. That is one equation for two unknowns.
  
  $a 5 b - 3 b + 1 = 5 \Rightarrow a 2 b = 4$
3. Choose b = 2 and the equation gives a = 2 and c = −6, the preset. Choose b = 1 and it gives a = 4/√2 = 2.8284 and c = −3, which is the curve the results table's last row prints for the preset.
  
  $b = 2 : a = \frac{4}{2} = 2, c = - 6 b = 1 : a = \frac{4}{2} = 2.8284, c = - 3$
4. They agree because √(bx − 3b) = √b·√(x − 3), so every curve in the family is a√b·√(x − 3) + 1, and the second point fixes only the product a√b, at 2√2.
  
  $a b x - 3 b = a b x - 3, a b = 22 = 2.8284$
Answer

Infinitely many sets of four numbers, and exactly one curve. Every b > 0 works, with a = 2√2/√b, c = −3b and d = 1. Take b = 8: then a = 1 and y = √(8x − 24) + 1, which passes through (5, 5) because √16 = 4. The question "find a and b" has no single answer; the question "find the curve" has one, 2.8284√(x − 3) + 1.

$b = 8 : a = 1, y = 8 x - 24 + 1, 8 \cdot 5 - 24 + 1 = 4 + 1 = 5$

References (2)

The standard statement of the four moves, including the order the page's claim 2 turns on: in f(bx − h), shift by h first and then stretch by 1/b; in f(b(x − h)), stretch first and then shift by h: Jay Abramson, "1.5 Transformation of Functions." Precalculus 2e, OpenStax, Houston, Texas, 2021. CC BY-NC-SA 4.0.
That the rightward shift of y = (x − 3)² is counterintuitive to most of the secondary students and teachers asked about it, the observation insight 1 quotes: R. Zazkis, P. Liljedahl and K. Gadowsky, "Conceptions of function translation: obstacles, intuitions, and rerouting." The Journal of Mathematical Behavior 22(4), 2003, pp. 435-448. DOI 10.1016/j.jmathb.2003.09.003

By the 42Math team · Reviewed against published sources. Corrections welcome.