Predict how changes inside and outside a function move, stretch and reflect its graph.
AA HL
Graph transformations
Papers 1 and 2
GDC useful for checking
About 20 min
By the end of this lesson
distinguish horizontal from vertical transformations
apply shifts, stretches and reflections
track the order of combined transformations
Inside changes inputs; outside changes outputs
Compare a point before and after changing the function. Adding outside $f$ changes its output, so the point moves vertically. Changing the input inside $f$ changes where the old output is reached.
Check the logic
For g(x)=f(x+3), where does an original point with x=2 move?
Explore the result
The new input must satisfy x+3=2, so x=−1.
Compare h = −3 and h = 3. Follow point 1 to point 2, then change k: which coordinate changes?
Why does f(x + 3) move left? Compare corresponding points, then change h.
Example f(x) = x²: (0,0) moves to (−3,0), and (2,4) moves to (−1,4). The new input x satisfies x + 3 = the old input. This input rule applies to any suitable function, not only this parabola.
For $f(x+3)$ to receive an old input $a$, we need $x+3=a$, so $x=a-3$. Every point moves three units left. Try $h=-3$ and $h=3$ in $f(x-h)$; then change $k$ to compare a vertical move.
Why does the horizontal sign seem reversed?
The expression describes the input being fed into $f$, not an instruction to move a point. Solve for the new horizontal coordinate: $x-h=a$ gives $x=a+h$. Thus positive $h$ moves right in $f(x-h)$.
Outside: outputs
$f(x)+a$: vertical translation by $a$.
$af(x)$: vertical scale factor $|a|$; reflect in the $x$-axis if $a<0$.
Inside: inputs
$f(x-a)$: horizontal translation by $a$.
$f(ax)$: horizontal scale factor $1/|a|$ for $a\ne0$; reflect in the $y$-axis if $a<0$.
Checkpoint
Correct. To obtain the old input $0$, the new coordinate must be $x=-3$, so points move left.
Not quite. Horizontal shifts act opposite to the visible sign inside the function.
Show explanation
To obtain the old input $0$, the new coordinate must be $x=-3$, so points move left.
Example 1
Track a key point
From $f(x)$ to $-2f(x-4)+1$
Check the logic
If (1,3) lies on f, what is its new horizontal coordinate on −2f(x−4)+1?
Show the working
The input x−4 must equal 1, so x=5. The outside operations affect the output separately.
Now follow the output operations and compare the full mapped point.
If $(1,3)$ lies on $y=f(x)$, then the horizontal shift sends $x=1$ to $x=5$. The output changes from $3$ to $-2(3)+1=-5$.
$$(1,3)\longmapsto(5,-5).$$
This point mapping is safer than relying on a memorized sketch.
Combine transformations carefully
For $y=a f(b(x-h))+k$, horizontal coordinates change by $x\mapsto x/b+h$, while outputs change by $y\mapsto ay+k$.
Order the method
For $y=3f(2(x-1))-4$, order the operations applied to points of $y=f(x)$.
Compress horizontally by factor $1/2$
Shift right $1$
Stretch vertically by factor $3$
Shift down $4$
Show a valid order
Compress horizontally by factor $1/2$
Shift right $1$
Stretch vertically by factor $3$
Shift down $4$
This order matches $x\mapsto x/2+1$ and $y\mapsto3y-4$.
Example 2
Transform an asymptote
A reciprocal graph
Describe $y=\dfrac{-2}{x+1}+3$ from $y=1/x$.
$x+1$ shifts the graph left $1$.
The factor $-2$ reflects it in the $x$-axis and stretches vertically by $2$.
$+3$ shifts it up $3$.
The asymptotes become $x=-1$ and $y=3$.
Checkpoint
Correct. Replacing $x$ by $-x$ reverses horizontal coordinates.
Not quite. $-f(x)$ reverses outputs and therefore reflects in the $x$-axis.
Show explanation
Replacing $x$ by $-x$ reverses horizontal coordinates.
Example 3
Recover an equation
Match the vertex and width
A transformed parabola has vertex $(-3,2)$, opens downward, and is twice as steep as $y=x^2$.
Vertex form gives $y=a(x+3)^2+2$. Opening downward makes $a<0$, and the vertical stretch gives $|a|=2$.
$$y=-2(x+3)^2+2.$$
Substituting $x=-3$ confirms the vertex output is $2$.
In the exam
Quick check
Your score is not saved.
Subtracting outside the function lowers every output.
Show explanation
Subtracting outside the function lowers every output.
Inputs reach the old value three times sooner.
Show explanation
Inputs reach the old value three times sooner.
Negating outputs reflects vertically.
Show explanation
Negating outputs reflects vertically.
The denominator is zero at $x=4$.
Show explanation
The denominator is zero at $x=4$.
The input is unchanged and the output becomes $2(-1)+3=1$.
Show explanation
The input is unchanged and the output becomes $2(-1)+3=1$.
Want to leave a comment?
Log in