How do you sketch transformed graphs from $y=f(x)$?

Key concept: When a question asks for a sketch, the marks usually come from getting the transformation right and keeping the shape smooth. The important part is to move the right coordinates, not redraw the curve from scratch.

Step by step

You can treat the two transformations separately.

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

This shifts the graph 3 units left.

Rule:

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

So any point on the original graph moves left by 3, but keeps the same height.

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

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

Rule:

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

So the graph becomes narrower. The $y$-intercept does not change, since $x=0$ still gives the same input to $f$.

A neat way to sketch is:

1. mark the key points on $f(x)$,
2. shift or compress those points,
3. join them with the same smooth shape.

Pitfall alert

The most common slip is swapping the direction of the shift for $f(x+3)$. Another one is changing the $y$-intercept for $f(3x)$, even though $x=0$ still gives $f(0)$.

If the original graph has a steep section, keep the transformed graph smooth; do not turn it into a polygon.

Try different conditions

If the question were $y=f(x-3)$, the graph would move 3 units right. If it were $y=f\left(\frac{x}{3}\right)$, the graph would stretch horizontally by a factor of 3 instead of shrinking.

Further reading

translation, horizontal scaling, key points