Finding the rule of a transformed graph or table
You can see a curve, or a table of its values. It looks like a parabola, or a square-root curve, that has been moved. What rule draws exactly that curve?
Where you are 7 in Transformations of functions
When you reach for it
Combining transformations reads the moves off a rule, and graphing by transformations turns the rule into a curve. This lesson runs the other way: you have the curve, or a table of values, and you want the rule. Reach for it whenever a graph or a set of measurements has the shape of a parent you know, moved, stretched or flipped, and you need a formula that reproduces it.
The procedure
Name the parent from the shape
Each parent has a shape you can recognize (The squaring and cubing functions, The square root and reciprocal functions):
| Shape | Parent | Its anchor point |
|---|---|---|
| a U with one turning point | y = x^2 | the vertex (0, 0) |
| an S that levels off for a moment | y = x^3 | the center (0, 0) |
| half a curve, leaving a start point | \displaystyle y = \sqrt{x} | the start (0, 0) |
| two branches, each hugging two lines | \displaystyle y = \frac{1}{x} | where the asymptotes x = 0 and y = 0 cross |
The anchor gives h and k
Write the rule as y = a\,f(x - h) + k. Putting x = h gives y = a\,f(0) + k, and every parent above has f(0) = 0, so the anchor (0, 0) lands at (h, k). For \displaystyle \frac{1}{x}, which has no value at 0, the asymptote x = 0 moves to x = h and the asymptote y = 0 moves to y = k. So read the anchor’s coordinates and they are h and k. Mind the sign: an anchor at x = -4 means h = -4, and the rule shows x + 4.
One more point gives a
Any other point (x, y) on the curve must satisfy y = a\,f(x - h) + k. With h and k known, that leaves one unknown:
for any point where f(x - h) \ne 0. All four parents have f(1) = 1, so the easiest point is one step to the right of the anchor: its height above the anchor’s level is a. If it sits below, a is negative and the graph has been flipped across a horizontal line.
Why the rule needs no b, with one exception
For these parents a horizontal stretch can always be traded for a vertical one: (bx)^2 = b^2x^2, (bx)^3 = b^3x^3, \displaystyle \frac{1}{bx} = \frac{1}{b} \cdot \frac{1}{x}, and \displaystyle \sqrt{bx} = \sqrt{b}\,\sqrt{x} when b > 0. A picture cannot tell the two apart, so write the rule with b = 1. The exception is a square-root curve that runs left from its start point. No vertical move turns a half-curve round, so it needs b = -1: \displaystyle y = a\sqrt{-(x - h)} + k.
Check with a point you did not use
Two points fixed a, h and k, so any third point is a real test: of your reading, and of the parent you chose.
Reading a table
If g(x) = a\,f(x - h) + k, then each entry (p, q) of f appears in g’s table as (p + h,\ aq + k). With a = 1 (a translation) the outputs of g repeat the pattern of f’s outputs, slid along by h and up by k. Match one distinctive entry, such as the smallest output, to read off h and k, then check the rest.
Worked examples
A parabola moved by two shifts
Write a rule for the solid parabola in the figure. It has vertex (2, -2) and passes through (3, -1) and (4, 2).
The U shape names the parent y = x^2, so the rule is y = a(x - h)^2 + k. The vertex is the anchor, so h = 2 and k = -2. One step right of the vertex, at (3, -1), the curve is -1 - (-2) = 1 above the vertex, so a = 1. The rule is
Check with (4, 2), which was not used: (4 - 2)^2 - 2 = 4 - 2 = 2. The curve is the parent moved 2 right and 2 down, with no stretch, which is the vertex form of a parabola.
A cubic that falls
Write a rule for the cubic in the figure: center (-4, 8), passing through (-5, 11) and (-3, 5).
An S that pauses at one point names y = x^3, and the pause point is the center, so h = -4 and k = 8. The rule so far is y = a(x + 4)^3 + 8, the +4 coming from x - (-4).
One step right of the center the curve is at 5, which is 3 below the center. So a = -3: a vertical stretch by 3 and a flip, which is why the curve falls where x^3 rises.
Check with (-5, 11): -3(-1)^3 + 8 = 3 + 8 = 11.
A square root, using a point further out
A square-root curve starts at (-4, -2), runs to the right, and passes through (0, 1) and (5, 2.5). Write its rule.
The start point is the anchor, so h = -4, k = -2, and the curve runs right, so b = 1: \displaystyle y = a\sqrt{x + 4} - 2. The point one step right of the start, x = -3, falls between grid lines, so use (0, 1), which sits on one:
Check with (5, 2.5): \displaystyle \frac{3}{2}\sqrt{9} - 2 = 4.5 - 2 = 2.5. Any point except the start would have done; the formula for a divides by \displaystyle \sqrt{x + 4}, which is 0 only at the start.
Asymptotes as the anchor
The reciprocal curve in the figure has asymptotes x = 2 and y = 8 and passes through (3, 13), (-3, 7) and (1, 3). Write its rule.
Two branches hugging two lines name \displaystyle y = \frac{1}{x}. Its asymptotes x = 0 and y = 0 have moved to x = 2 and y = 8, so h = 2 and k = 8, and the rule is \displaystyle y = \frac{a}{x - 2} + 8. At (3, 13), one step right of x = 2, the curve is 13 - 8 = 5 above the horizontal asymptote, so a = 5:
Check with (-3, 7): \displaystyle \frac{5}{-5} + 8 = 7; and with (1, 3): \displaystyle \frac{5}{-1} + 8 = 3.
The same curve is \displaystyle y = \frac{1}{(x - 2)/5} + 8, a horizontal stretch by 5 instead of a vertical one. Both rules are correct; writing b = 1 just picks the simpler.
A translation hiding in a table
Here are tables for a function f and a function g. Find h and k with g(x) = f(x - h) + k.
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| f(x) | 3 | 0 | -1 | 4 | 7 |
| x | -6 | -5 | -4 | -3 | -2 |
|---|---|---|---|---|---|
| g(x) | 1 | -2 | -3 | 2 | 5 |
The outputs of g go down 3, down 1, up 5, up 3, the same steps as f, so g is a translation of f. Match the smallest outputs: f has -1 at x = 0, and g has -3 at x = -4. The point (0, -1) moved to (-4, -3): left 4 and down 2. So h = -4, k = -2, and
Check another entry: g(-2) = f(2) - 2 = 7 - 2 = 5. The sign inside is worth a second look. The graph of g sits 4 to the left, so at any x, g looks up f at x + 4, four steps further right.
Hilka’s container dip
Hilka works a harbor crane that lowers a container into a ship’s hold and lifts it out in one smooth dip. She logs the height H(t) of the container’s base, in meters relative to the quay edge (negative is below it), every 2 seconds:
| t | H(t) |
|---|---|
| 0 | 6.5 |
| 2 | -1.5 |
| 4 | -5.5 |
| 6 | -5.5 |
| 8 | -1.5 |
| 10 | 6.5 |
The path is a transformed parabola. Find its rule and the lowest point of the dip.
SolutionThe table does not show the vertex, but it shows the symmetry: the heights repeat in reverse about the midpoint of t = 4 and t = 6. A parabola’s vertex lies on that line of symmetry, so h = 5, and H(t) = a(t - 5)^2 + k.
Two unknowns need two entries. At t = 4, (t - 5)^2 = 1, and at t = 2, (t - 5)^2 = 9:
Subtracting, 8a = 4, so \displaystyle a = \frac{1}{2} and k = -6:
Check with t = 0: \displaystyle \frac{1}{2}(25) - 6 = 6.5. The lowest point is the vertex: at t = 5 seconds the container’s base is 6 meters below the quay edge.
Tables with a stretch and a flip
A function f is known only from its table, and g(x) = a\,f(x - h) + k. Find a, h and k.
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| f(x) | 0 | 2 | 2 | -1 |
| x | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| g(x) | 8 | 2 | 2 | 11 |
The repeated pair of equal outputs sits at x = 1, 2 for f and x = 3, 4 for g, so the inputs moved right 2: h = 2, and g(x) uses f(x - 2). Now compare heights entry by entry. At x = 2, g(2) = a\,f(0) + k = k, so k = 8. At x = 3, 2 = a \cdot 2 + 8, so a = -3:
Check the last entry: g(5) = -3\,f(3) + 8 = 3 + 8 = 11. The outputs did not repeat f’s pattern here: f’s steps up 2, then 0, then down 3 became down 6, 0, up 9, each multiplied by -3. A stretch shows in a table as every step scaled by the same number.
Boundaries
- The method needs the parent to be known, or recognized. A handful of points fits many curves: the third point checks the guess but cannot prove it.
- Values read from a picture are only as exact as its grid. The rule is right only if every marked point satisfies it exactly.
- The anchor must be visible. A piece of parabola that never turns, or a table with no symmetry or repeated pattern, leaves h unfixed by this reading, and then it must be solved for with more points.
- One graph has more than one correct rule: a and b trade off as above, and since \big(b(x - h)\big)^2 = b^2(x - h)^2, the sign of b in a parabola’s rule cannot be seen at all. Writing b = 1 is a choice, not a fact about the graph.
- A table describes a function only at its listed inputs. A rule matching every entry agrees with the function there and says nothing certain between them.
Where this goes next
Transforming the absolute value graph applies the same reading to V-shaped graphs, whose anchor is the corner. Asking whether a graph matches itself after a flip leads to Symmetry about the vertical axis and the origin. And when measured data only roughly follow a curve, the anchor and one point give way to choosing a, h and k to fit all the points at once, the job of Fitting linear and exponential models with technology.
Exercises
Rules from a description of the graph
- A parabola has vertex (-1, 3) and passes through (1, 7).
- A cubic has center (3, -1) and passes through (4, -3).
- A square-root curve starts at (1, 4), runs right, and passes through (10, 6).
- A reciprocal curve has asymptotes x = -2 and y = 3 and passes through (-1, 1).
- A square-root curve starts at (2, 1), runs left, and passes through (-7, -5).
- A V-shaped graph of the absolute value family has its corner at (4, -1) and passes through (6, 5).
Translations in tables
Each table is given as its list of (x, y) pairs.
- f: (-1, 4), (0, 1), (1, 0), (2, 3), (3, 6). g: (2, 7), (3, 4), (4, 3), (5, 6), (6, 9). Find h and k with g(x) = f(x - h) + k.
- A translation of y = x^2: (1, 3), (2, 0), (3, -1), (4, 0), (5, 3). Write its rule.
- A translation of y = x^3: (-3, -14), (-2, -7), (-1, -6), (0, -5), (1, 2). Write its rule.
- A translation of \displaystyle y = \sqrt{x} that starts at its first pair: (-7, -3), (-6, -2), (-3, -1), (2, 0). Write its rule.
Tables with a stretch
- f: (0, 1), (1, 3), (2, 0), (3, -2). g: (-5, 6), (-4, 10), (-3, 4), (-2, 0). Find a, h and k with g(x) = a\,f(x - h) + k.
- A stretched, translated parabola: (1, 8), (2, -1), (4, -1), (5, 8). Write its rule.
Problems
- Padma times a load lifted by a harbor crane. It rises quickly, then more slowly, and its height in meters follows a square-root curve starting at the first pair (t, H): (3, 1), (4, 3), (7, 5), (12, 7). Write the rule and find the height at t = 28.
- Along the quay, Dalibor strings a cable between two lamp posts 20 meters apart. It hangs in a parabola, held 8 meters up at each post, and its lowest point is 3 meters up, halfway between. Measuring x in meters from the left post, write the rule for the cable’s height.
Reasoning
- Explain why the anchor point of each parent, the vertex, center, start or asymptote crossing, lands at (h, k) on the graph of y = a\,f(x - h) + k.
- Filipa says the graph of \displaystyle y = \sqrt{16x} is \displaystyle y = \sqrt{x} compressed horizontally by a factor of \displaystyle \frac{1}{16}. Kwesi says it is \displaystyle y = \sqrt{x} stretched vertically by 4. Who is right?
- A parabola passes through (1, 5) and (3, 5), and nothing else is known. Explain why that is not enough to write its rule, and what one more fact would settle it.
- A table lists x = 1, 2, 3 with outputs 4, 5, 6. Show that both y = (x - 2)^3 + 5 and y = x + 3 fit it, and give an entry that would tell them apart.
Answers
- y = (x + 1)^2 + 3. h = -1, k = 3; \displaystyle a = \frac{7 - 3}{(1 + 1)^2} = 1.
- y = -2(x - 3)^3 - 1. One step right of the center the curve is 2 lower, so a = -2.
- \displaystyle y = \frac{2}{3}\sqrt{x - 1} + 4. \displaystyle a = \frac{6 - 4}{\sqrt{10 - 1}} = \frac{2}{3}.
- \displaystyle y = -\frac{2}{x + 2} + 3. At x = -1, one step right of x = -2, the curve is 2 below y = 3, so a = -2.
- \displaystyle y = -2\sqrt{2 - x} + 1. Running left needs \displaystyle \sqrt{-(x - 2)} = \sqrt{2 - x}; h = 2, k = 1, and \displaystyle a = \frac{-5 - 1}{\sqrt{2 - (-7)}} = \frac{-6}{3} = -2.
- y = 3\lvert x - 4 \rvert - 1. \displaystyle a = \frac{5 - (-1)}{\lvert 6 - 4 \rvert} = 3.
- h = 3, k = 3. The smallest output 0 at x = 1 matches 3 at x = 4. Check: g(6) = f(3) + 3 = 9.
- y = (x - 3)^2 - 1. The vertex is the lowest entry, (3, -1), and the outputs 1 step either side are 1 higher.
- y = (x + 1)^3 - 6. The center is (-1, -6), where the steps are smallest: up 1 on each side. Check: (1 + 1)^3 - 6 = 2.
- \displaystyle y = \sqrt{x + 7} - 3. Start (-7, -3); the next entries are 1, 4 and 9 steps along, as on \displaystyle \sqrt{x}. Check: \displaystyle \sqrt{2 + 7} - 3 = 0.
- a = 2, h = -5, k = 4. The tables line up with g’s inputs 5 less, so h = -5. Then 6 = a + k and 10 = 3a + k give a = 2, k = 4. Check: g(-2) = 2(-2) + 4 = 0.
- y = 3(x - 3)^2 - 4. Symmetry about x = 3 gives h = 3; then a + k = -1 and 4a + k = 8 give a = 3, k = -4.
- \displaystyle H(t) = 2\sqrt{t - 3} + 1, and H(28) = 2 \cdot 5 + 1 = 11 meters. One second after the start the load is 2 higher, so a = 2.
- \displaystyle y = \frac{1}{20}(x - 10)^2 + 3. The vertex is (10, 3), and the post at (0, 8) gives 8 = 100a + 3, so \displaystyle a = \frac{1}{20}.
- At x = h the rule gives a\,f(0) + k = k, since each parent with an anchor on its graph has f(0) = 0. For \displaystyle \frac{1}{x}, the rule is undefined exactly where x - h = 0, so the vertical asymptote is x = h; and \displaystyle \frac{a}{x - h} gets close to 0 far out, so the output gets close to k.
- Both. For x \ge 0, \displaystyle \sqrt{16x} = \sqrt{16}\,\sqrt{x} = 4\sqrt{x}, and for x < 0 neither is defined, so the two descriptions give the same graph.
- Equal heights at x = 1 and x = 3 put the vertex on x = 2, but every rule y = a(x - 2)^2 + 5 - a with a \ne 0 passes through both points. One more point, or the vertex itself, fixes a and then k = 5 - a.
- The cubic gives -1 + 5 = 4, 0 + 5 = 5, 1 + 5 = 6, and the line gives 4, 5, 6. At x = 4 the cubic gives 13 and the line 7.