Moving, stretching and flipping a graph

Start from one graph

Once you know the graph of $y = f(x)$, you can draw many other graphs without a table of values. A small change in the formula moves, stretches or flips the same picture. This saves a great deal of work.

In the picture, the dashed grey curve is $f(x) = (x - 1)^2$. Each blue curve is a change of it.

Changes outside the function: up and down

These change the height of each point, so they do exactly what you expect.

Changes inside the function: left and right

These change the input before $f$ sees it. They work sideways, and the effect is the opposite of what the sign suggests.

Why right? The graph of $f(x - 2)$ gives at $x = 5$ the value that $f$ gave at $x = 3$. So every point has moved $2$ units to the right. In the same way, $f(2x)$ at $x = 1$ gives the value that $f$ gave at $2$, so the picture has been squeezed to half the width.

Set $a = -1$. The parabola turns upside down about its vertex, which stays at $(h, k)$.

Putting several changes together

For $y = a\,f\big(b(x - h)\big) + k$, work in this order. First the changes inside: scale by $b$, then move by $h$. Then the changes outside: scale by $a$, then move by $k$.

A point moves too. A point $(x, y)$ on the graph of $f$ becomes the point $\left(\dfrac{x}{b} + h,\ a y + k\right)$ on the new graph. For $y = 3f(x - 1) + 2$, the point $(2, 5)$ goes to $(3, 17)$.

Three worked examples

Example 1. Draw $y = (x + 3)^2 - 2$. Start from $f(x) = x^2$. Replace $x$ by $x + 3$: the graph moves $3$ units to the left. Subtract $2$: it moves $2$ units down. The vertex goes from $(0, 0)$ to $(-3, -2)$. The domain is $\mathbb{R}$ and the range is $[-2, \infty)$.

Example 2. Draw $y = 2\sqrt{x - 1} + 3$. Start from $\sqrt{x}$. Replace $x$ by $x - 1$: move $1$ to the right. Multiply by $2$: stretch upward. Add $3$: move up. The starting point $(0, 0)$ goes to $(1, 3)$. The domain is $[1, \infty)$ and the range is $[3, \infty)$.

Example 3. The graph of $y = |x - 2|$. Start from $|x|$ and move it $2$ units to the right. The point of the "V" is now at $(2, 0)$. The right arm of the "V" starts there, and the left arm comes from the $y$ axis at $y = 2$.

Two more changes with the modulus

• $y = |f(x)|$: any part of the graph below the $x$ axis is flipped up above it.
• $y = f(|x|)$: the part for $x \ge 0$ is kept and copied as a mirror image to the left side. The graph is then symmetric about the $y$ axis.

Practise with new numbers

Try it in Python

This program draws $f(x) = x^2$ and three changes of it on one picture.

Remember these

• Outside the function: up and down. $f(x) + k$ moves up, $a f(x)$ stretches up, $-f(x)$ flips over the $x$ axis.
• Inside the function: sideways, and opposite to what you may expect. $f(x - h)$ moves right, $f(bx)$ squeezes, $f(-x)$ flips over the $y$ axis.
• For several changes, work from the inside out.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How is the graph of $y = x^2 + 5$ different from the graph of $y = x^2$?

2. Where is the vertex of $y = (x - 4)^2 + 1$?

3. The point $(2, 6)$ is on the graph of $f$. What point is on the graph of $y = -f(x)$? Of $y = f(-x)$?

4. Where is the corner of the graph of $y = |x + 3| - 2$?