From $f(x)$ to $-2f(x-4)+1$
If $(1,3)$ lies on $y=f(x)$, then the horizontal shift sends $x=1$ to $x=5$. The output changes from $3$ to $-2(3)+1=-5$.
This point mapping is safer than relying on a memorized sketch.
Predict how changes inside and outside a function move, stretch and reflect its graph.
Why this matters. Graph transformations let you reason from a known function instead of rebuilding every graph point by point.
$f(x)+a$: up $a$
$af(x)$: vertical scale factor $|a|$
$f(x-a)$: right $a$
$f(ax)$: horizontal scale factor $1/|a|$
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.
To obtain the old input $0$, the new coordinate must be $x=-3$, so points move left.
Example 1
From $f(x)$ to $-2f(x-4)+1$
If $(1,3)$ lies on $y=f(x)$, then the horizontal shift sends $x=1$ to $x=5$. The output changes from $3$ to $-2(3)+1=-5$.
This point mapping is safer than relying on a memorized sketch.
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 $y=f(x)$.
This order matches $x\mapsto x/2+1$ and $y\mapsto3y-4$.
Example 2
A reciprocal graph
Describe $y=\dfrac{-2}{x+1}+3$ from $y=1/x$.
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.
Replacing $x$ by $-x$ reverses horizontal coordinates.
Example 3
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 $|a|=2$.
Substituting $x=-3$ confirms the vertex output is $2$.
Five questions to check the main decisions from this lesson.
Your score is not saved.
Subtracting outside the function lowers every output.
Subtracting outside the function lowers every output.
Inputs reach the old value three times sooner.
Inputs reach the old value three times sooner.
Negating outputs reflects vertically.
Negating outputs reflects vertically.
The denominator is zero at $x=4$.
The denominator is zero at $x=4$.
The input is unchanged and the output becomes $2(-1)+3=1$.
The input is unchanged and the output becomes $2(-1)+3=1$.
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