Transformations of Functions

The parameters outside the function ($a$ and $c$) act on the output, so they behave the way you'd expect: $a=3$ triples every height, $c=5$ lifts everything 5 units. The parameters inside ($k$ and $d$) act on the input before $f$ sees it, and that makes them feel backwards. Take $y={(x-4)}^2$. The parent's vertex happens when the input to the squaring is $0$; now that input is $x-4$, which is $0$ when $x=4$. So the vertex moves to $x=4$ — right 4, even though the equation shows a minus sign. Likewise in $y={(2x)}^2$, the value that used to happen at $x=6$ now happens at $x=3$, because $2(3)=6$: the graph is squeezed toward the $y$-axis by a factor of $\frac{1}{2}$. In general, to reproduce a parent input $x$, the new graph needs $k(X-d)=x$, which solves to $X=\frac{x}{k}+d$.

MCR3U (Grade 11 Functions) - Transformations of Functions Overview

The cleanest way to transform a set of known points is the mapping rule: every point $(x,y)$ on the parent graph moves to a new point on the transformed graph according to

The $x$-coordinate is divided by $k$ then shifted by $d$; the $y$-coordinate is multiplied by $a$ then shifted by $c$. This works for any parent function — you never need the actual equation of the curve, just its key points. A reliable routine: list three to five key points of the parent in a table, add two columns for the images, apply the rule, plot, and join with the parent's shape. Special points stay special: the vertex of a parabola maps to the new vertex, the starting point of $\sqrt{x}$ maps to the new starting point, and asymptotes shift with $d$ and $c$.

MCR3U (1.6) - general process for mapping and graphing transformed functions

The form $y=af(k(x-d))+c$ only gives the right $d$ if $k$ has been factored out of the bracket. In $y=\sqrt{2x-6}$, it is tempting to say "shift 6 right," but first factor: $\sqrt{2x-6}=\sqrt{2(x-3)}$, so $k=2$ and $d=3$ — a horizontal compression by $\frac{1}{2}$ and a shift of only 3 right. Check with the starting point: $2x-6=0$ at $x=3$, confirming the graph starts at $x=3$. The same care applies to signs: $f(x+5)$ means $x-d=x+5$, so $d=-5$ (left 5), and $f(-x+4)=f(-(x-4))$ means $k=-1$ and $d=4$. When an equation arrives in an unusual shape, rearrange it into the standard form before reading anything: $y-2=3{(x+1)}^2$ is $y=3{(x+1)}^2+2$, and $y=5-\sqrt{x}$ is $y=-\sqrt{x}+5$, a reflection in the $x$-axis followed by a shift up 5.

• Reflection in the $x$-axis: $a$ is negative (flips up/down).
• Reflection in the $y$-axis: $k$ is negative (flips left/right).
• A horizontal stretch by factor 2 needs $k=\frac{1}{2}$, not $k=2$ (that would be a compression) — always think "$x\to x/k$."
• Points on the line of reflection do not move: under $a<0$ the $x$-intercepts stay put; under $k<0$ the $y$-intercept stays put.

Because the mapping rule moves every point, it moves the domain and range too. Horizontal parameters ($k$, $d$) change the domain; vertical parameters ($a$, $c$) change the range. For $g(x)=-2\sqrt{x-1}+4$, the parent $\sqrt{x}$ starts at $(0,0)$, which maps to $(1,4)$. The domain becomes $x\ge 1$. Since $a<0$, the graph heads down from its start, so the range becomes $y\le 4$. For $y=3{(x+2)}^2-5$ the domain stays all real numbers (horizontal moves can't create gaps in a function defined everywhere) and the range becomes $y\ge -5$, read straight from the new vertex $(-2,-5)$. Asymptotes behave the same way: for $y=\frac{1}{x-3}+2$, the parent's asymptotes $x=0$ and $y=0$ shift to $x=3$ and $y=2$, so the domain is $x\ne 3$ and the range is $y\ne 2$. Stretches and reflections never move an asymptote that passes through the origin, which is why only $d$ and $c$ matter here. On a test, finding the new "anchor" (vertex, starting point or asymptote intersection) with the mapping rule, and then asking "which way does the graph go from here?", answers most domain and range questions in two lines.