How do you sketch $y=f(x+3)$ and $y=f(3x)$ from a given graph?

Key takeaway: These are standard function transformations, but the two changes work in different directions. One is a horizontal shift, the other is a horizontal compression. That difference is where most sketching errors happen.

Let the original graph be $y=f(x)$.

(i) Sketch of $y=f(x+3)$

This is a shift 3 units left.

Why? Because the graph reaches the same $y$-value when $x$ is 3 smaller than before.

So every point $(x,y)$ on $y=f(x)$ moves to

$(x-3,\,y).$

Key points shift like this:

• $( -6,-3 ) \to ( -9,-3 )$
• $( -5,0 ) \to ( -8,0 )$
• $( 3,6 ) \to ( 0,6 )$

(ii) Sketch of $y=f(3x)$

This is a horizontal compression by factor $\frac13$.

A point $(x,y)$ on the original graph moves to

$\left(\frac{x}{3},\,y\right).$

So the $x$-coordinates are divided by 3, while the $y$-values stay the same.

That means:

• the starting point $(-6,-3)$ becomes $(-2,-3)$
• the $y$-intercept stays the same as for $f(x)$

So the new graph is squeezed toward the $y$-axis, not shifted sideways.

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Pitfalls the pros know 👇 For $f(x+3)$, students often shift right because they see the plus sign. That is the wrong direction. Inside the brackets, the sign effect is reversed.

For $f(3x)$, another common mistake is to stretch the graph horizontally instead of compressing it. Multiplying $x$ by 3 makes the graph narrower.

What if the problem changes? If the function were $y=f(x-3)$, the graph would shift 3 units right instead of left. If it were $y=f\left(\frac{x}{3}\right)$, the graph would stretch horizontally by a factor of 3 instead of compressing.

Tags: horizontal shift, horizontal compression, function transformation