Many graphs are copies of a few basic ones, moved and stretched. You have seen this for the V of an absolute value function; it works for every function $f$. Knowing how each change moves the key features lets you sketch a graph without a table.
| New function | What happens to the graph | A point $(x, y)$ moves to |
|---|---|---|
| $f(x) + k$ | up $k$ (down if $k < 0$) | $(x, y + k)$ |
| $f(x - h)$ | right $h$ (left if $h < 0$) | $(x + h, y)$ |
| $a f(x)$ | vertical stretch by $a$ (flip if $a < 0$) | $(x, a y)$ |
| $-f(x)$ | reflect across the $x$-axis | $(x, -y)$ |
| $f(-x)$ | reflect across the $y$-axis | $(-x, y)$ |
Why it works
Changes outside $f$ act on the outputs, so they behave as they look: $+ 3$ outside raises every point $3$. Changes inside act on the inputs, and they work backward. In $g(x) = f(x - 3)$, the function $g$ gets the value that $f$ had $3$ units earlier: $g(5) = f(2)$. Whatever $f$ does at $x = 2$, $g$ does at $x = 5$, so the whole graph moves right $3$, even though there is a minus sign.
Key idea
$g(x) = a\,f(x - h) + k$ moves the graph of $f$ right $h$ and up $k$, and stretches it vertically by $a$ (flipping it if $a < 0$). The point $(x, y)$ on $f$ goes to $(x + h,\ a y + k)$.
Worked examples
Worked example: Moving a point
$(2, 5)$ is on the graph of $f$. Find a point on $g(x) = f(x + 4) - 1$.
- $x + 4 = x - (-4)$, so $h = -4$: shift left $4$. The new $x$ is $2 - 4 = -2$.
- $-1$ outside: shift down $1$. The new $y$ is $5 - 1 = 4$.
- Check: $g(-2) = f(-2 + 4) - 1 = f(2) - 1 = 4$.
Answer: $(-2, 4)$
Worked example: Writing a rule
Write a rule for the graph of $y = x^2$ reflected across the $x$-axis, then shifted left $1$ and up $3$.
- Reflect: $-x^2$.
- Shift left $1$: replace $x$ with $x + 1$: $-(x + 1)^2$.
- Shift up $3$: $y = -(x + 1)^2 + 3$, a parabola opening down with vertex $(-1, 3)$.
Answer: $y = -(x + 1)^2 + 3$
Common mistake
Shifting $f(x - 3)$ to the left because of the minus sign. Inside changes work backward: $x - 3$ moves the graph right. Check with one point: $g(x) = f(x - 3)$ gives $g(3) = f(0)$, so the point that was at $x = 0$ is now at $x = 3$.