Function transformation explorer

A whole family of graphs, such as every parabola with a vertical axis, comes from one parent graph by shifts, stretches and reflections. Pick a parent function $f$ and set the four numbers in

The faded curve is $f$ and the bold curve is $g$. Arrows join three key points of $f$ to the points where they land on $g$, so every change has a visible cause. Drag the ring on the graph to move $h$ and $k$ together, and use the play button to watch $g$ being built from $f$ one step at a time.

The sliders for $a$ and $b$ jump from $-0.25$ straight to $0.25$. With $a = 0$ or $b = 0$ the graph would collapse onto a horizontal line, or vanish when $f(0)$ is undefined, so $g$ would no longer be a copy of $f$.

Things to try

• Choose $x^2$, set $a = 1$, $b = 1$ and $k = 0$, and move $h$ to $3$. The formula shows $(x - 3)^2$ with a minus sign, yet the vertex moves right to $(3, 0)$, because $g(3) = f(0)$.
• Keep $x^2$ and compare $b = 2$ with $a = 4$, each time with the other factor at $1$. The two graphs are identical, since $(2x)^2 = 4x^2$. The same happens with $\lvert x \rvert$ for $b = 2$ and $a = 2$, and with $\sqrt{x}$ for $b = 4$ and $a = 2$.
• Choose $\sqrt{x}$ and make $b$ negative. The graph flips across the vertical line $x = h$ and the domain turns from $[h, \infty)$ into $(-\infty, h]$. Make $a$ negative instead and the graph flips across the line $y = k$, so the range turns from $[k, \infty)$ into $(-\infty, k]$ while the domain stays where it was.
• Choose $1/x$ and drag the ring around. The dashed asymptotes $x = h$ and $y = k$ travel with it, and the domain and the range leave out exactly $h$ and $k$.
• Choose $2^x$ and set $a = -1$. The curve drops below its asymptote and the range changes from $(k, \infty)$ to $(-\infty, k)$. With $\ln x$ and $b = -1$ it is the domain that flips, to $(-\infty, h)$.
• Choose $\sin x$ and try $b = 2$ and then $b = 0.5$. The period $2\pi$ becomes $\pi$ and then $4\pi$, while the range is always $[k - \lvert a \rvert, k + \lvert a \rvert]$.
• With $x^2$, set $a = 1$, $b = 2$, $h = 3$ and $k = 0$ and play the steps. The vertex $B$ stays at the origin during the compression and then moves right to $(3, 0)$. Had the shift come first, the compression would have pulled it back to $(1.5, 0)$.
• Choose $x^2$ and switch $b$ between $1$ and $-1$. The graph does not change, but the images $A'$ and $C'$ trade places, because $x^2$ is even and $f(-x) = f(x)$. With $x^3$ the graph stays the same only when $a$ and $b$ change sign together, because $x^3$ is odd and $f(-x) = -f(x)$.

What you are seeing

Each point $(x, y)$ on the graph of $f$ lands at

on the graph of $g$. To check, put the new first coordinate into $g$.

The explorer draws $g$ exactly this way, by moving every point of $f$, and the readout under the graph spells out the rule with the current numbers. The table next to it applies the rule to the key points $A$, $B$ and $C$, with exact values.

The order of the steps

The second coordinate is multiplied by $a$ and then shifted by $k$, the same order in which $a\,f(\ldots) + k$ is worked out. The first coordinate is divided by $b$ and then shifted by $h$. So the explorer lists the steps in this order.

1. Stretch or compress horizontally by the factor $\frac{1}{\lvert b \rvert}$, and reflect across the $y$-axis when $b < 0$.
2. Shift horizontally by $h$.
3. Stretch or compress vertically by the factor $\lvert a \rvert$, and reflect across the $x$-axis when $a < 0$.
4. Shift vertically by $k$.

Horizontal and vertical steps do not affect each other, so the two pairs could come in either order. Within each pair the order matters. Compressing $y = x^2$ by the factor $\frac{1}{2}$ and then shifting it right by $3$ gives

which takes the point $(2, 4)$ to $(4, 4)$. Shifting first and then compressing gives a different parabola,

with its vertex at $(1.5, 0)$ instead of $(3, 0)$. That is why an input such as $2x - 6$ should be factored as $2(x - 3)$ before the shift is read off.

Composition gives another way to see the order. Write $u(x) = b(x - h)$ for the inside and $v(y) = ay + k$ for the outside, so that

An input passes through $u$ first, then $f$, then $v$. The lesson on function composition explains this way of building functions in general.

Why the inside acts the opposite way

The outside numbers act on the outputs of $f$ directly. Adding $k$ moves the graph up by $k$, and the factor $a$ multiplies every height by $a$. The inside numbers act on the input before $f$ sees it. To get the output that $f$ gives at some input $u$, the function $g$ needs an $x$ with $b(x - h) = u$, and solving for $x$ gives

So subtracting $h$ inside moves the graph right by $h$, and multiplying by $b$ inside scales it horizontally by the factor $\frac{1}{b}$, which is a compression when $b > 1$. Each inside change shows up on the graph undone. In $\sin(2x)$ the input runs twice as fast, so a full wave fits into half the distance.

Domain, range and asymptotes

The domain of $g$ is the set of all $x$ for which $b(x - h)$ is an allowed input of $f$. For $\sqrt{x}$ that means $b(x - h) \ge 0$, so the domain is $[h, \infty)$ when $b > 0$ and $(-\infty, h]$ when $b < 0$.

The range of $g$ is the set of all values $ay + k$ with $y$ in the range of $f$. For $x^2$, $\lvert x \rvert$ and $\sqrt{x}$ it is $[k, \infty)$ when $a > 0$ and $(-\infty, k]$ when $a < 0$. Horizontal steps never change the range, and vertical steps never change the domain.

The vertical asymptote $x = 0$ of $1/x$ and $\ln x$ moves to $x = h$, and the horizontal asymptote $y = 0$ of $1/x$ and $2^x$ moves to $y = k$. They move like any other part of the graph, by the same rule as the points.

The lessons on shifts and reflections and on stretches and compressions work through these rules with more examples, and the lesson on even and odd functions explains the symmetry behind the last thing to try. The same numbers come back in the vertex form of a parabola in Quadratic Functions and Vertex Form, in the shifted curves of Graphs of Exponential and Logarithmic Functions and as the amplitude, period and phase shift of a wave in Amplitude, Period, and Phase Shift. The rules are also collected on the precalculus reference sheet. Every lesson is listed on the Precalculus course page, and the video lessons are coming soon. For test preparation, the SAT Math lessons Shifting, Reflecting and Stretching Graphs and Quadratic Graphs and Their Forms apply the same rules.