Transforming Graphs

Start from $y=x^2$y = x²

Take $f\left(x\right)=x^2$f(x) = x², whose graph is the parabola $y=x^2$y = x². Its lowest point, the vertex, is at the origin, (0, 0), and it passes through (1, 1), (2, 4) and (3, 9), and through (−1, 1), (−2, 4) and (−3, 9).

Changing the equation of a graph in a simple way moves the graph, or changes its shape. To see what a change does, follow the points: where does each point of $y=x^2$y = x² end up?

A change outside the bracket

Add 3 to the function: y = f(x) + 3, which is $y=x^2+3$y = x² + 3. The 3 is added after the squaring is done, so it changes the output and not the input. At every x, the new y is 3 more than before: (1, 1) moves to (1, 4), (2, 4) moves to (2, 7), and the vertex (0, 0) moves to (0, 3).

Every point moves 3 straight up, so the whole graph moves up 3 without changing its shape. A move like this, where every point slides the same distance in the same direction, is called a translation. In general, y = f(x) + a is the graph of y = f(x) moved up by a. When a is negative the move is down: $y=x^2-4$y = x² − 4 has its vertex at (0, −4).

The dashed curve is $y=x^2$y = x² and the gold curve is $y=x^2+3$y = x² + 3. Each arrow moves a point 3 straight up.

A change inside the bracket

Now put the 3 inside the bracket: y = f(x − 3), which is $y={\left(x-3\right)}^2$y = (x − 3)². The 3 is subtracted before the squaring, so it changes the input. Work out some values. At x = 3, $y={\left(3-3\right)}^2=0$y = (3 − 3)² = 0. At x = 4, $y={\left(4-3\right)}^2=1$y = (4 − 3)² = 1, and at x = 5, $y={\left(5-3\right)}^2=4$y = (5 − 3)² = 4. On the other side, x = 2 gives ${\left(-1\right)}^2=1$(−1)² = 1 and x = 1 gives ${\left(-2\right)}^2=4$(−2)² = 4.

Compare these with $y=x^2$y = x². The old graph reached the height 0 at x = 0; the new one reaches it at x = 3. The old graph reached the height 4 at x = 2 and at x = −2; the new one reaches it at x = 5 and at x = 1. Every height is reached 3 later, at an x that is 3 bigger. So the whole graph has moved 3 to the right, even though the bracket holds a minus sign. The vertex is at (3, 0).

The reason is that the square receives x − 3, which is always 3 less than x. For the square to receive any value it received before, x has to be 3 bigger than it was.

The dashed curve is $y=x^2$y = x² and the gold curve is $y={\left(x-3\right)}^2$y = (x − 3)². Each arrow moves a point 3 to the right.

A plus sign inside moves the graph left

Now take y = f(x + 2), which is $y={\left(x+2\right)}^2$y = (x + 2)². Here the square receives x + 2, which is always 2 more than x. At x = −2 the square receives 0, and y = 0. $y=x^2$y = x² had to wait until x = 0 for that. At x = 0 the square receives 2, and y = 4. $y=x^2$y = x² reached 4 only at x = 2.

So the new graph reaches every height 2 earlier, at an x that is 2 smaller, and the whole graph lies 2 to the left. Its vertex is at (−2, 0).

The same argument works for any function f and any number a. Suppose (p, q) is on the graph of y = f(x), so f(p) = q. On y = f(x + a), put in x = p − a: the bracket holds (p − a) + a = p, so y = f(p) = q. So the point (p − a, q) is on the new graph: every point has moved a to the left. A number added inside the bracket moves the graph left by that amount, and a number subtracted inside the bracket moves it right.

The dashed curve is $y=x^2$y = x² and the gold curve is $y={\left(x+2\right)}^2$y = (x + 2)². Each arrow moves a point 2 to the left, and the vertex is at (−2, 0).

The faint curve is a function y = f(x) with its peak at x = 0, and the gold curve is y = f(x − c), here with c = 1. The peak of the gold curve is where the input x − c is 0, which is at x = c. Drag c to 3 and the peak moves to x = 3, to the right. Drag c below 0, to −2, and the peak moves to x = −2, to the left: then x − c is x − (−2) = x + 2.

Two changes together

Take $y={\left(x-3\right)}^2+2$y = (x − 3)² + 2. The change inside the bracket moves $y=x^{2}3$y = x² 3 to the right, and the change outside moves it 2 up. So the vertex moves from (0, 0) to (3, 2).

Check with one more point. (1, 1) is on $y=x^2$y = x². Moved 3 right and 2 up, it goes to (4, 3). On the new graph, x = 4 gives ${\left(4-3\right)}^2+2=1+2=3$(4 − 3)² + 2 = 1 + 2 = 3, so (4, 3) is on it.

This is the vertex form of a quadratic. Completing the square writes $x^2-6x+11$x² − 6x + 11 as ${\left(x-3\right)}^2+2$(x − 3)² + 2, since ${\left(x-3\right)}^2+2=x^2-6x+9+2$(x − 3)² + 2 = x² − 6x + 9 + 2. So the graph of $y=x^2-6x+11$y = x² − 6x + 11 is the graph of $y=x^2$y = x² moved 3 right and 2 up.

The rules hold for every function, not only $x^2$x². The graph of y = |x + 1| is the V of y = |x| moved 1 to the left, with its corner at (−1, 0).

The white curve is $y=x^2$y = x², with its vertex at (0, 0) and the point (1, 1). The gold curve is $y={\left(x-3\right)}^2+2$y = (x − 3)² + 2: its vertex is at (3, 2), and (1, 1) has moved to (4, 3).

The usual mistakes

Reading the sign inside the bracket at face value. $y={\left(x-3\right)}^2$y = (x − 3)² does not move 3 to the left. Its vertex is where x − 3 = 0, at x = 3, so the graph moves 3 to the right.

Putting the vertex of $y={\left(x+2\right)}^2$y = (x + 2)² at (2, 0). There the bracket holds 4, and y = 16. The vertex is where x + 2 = 0, at (−2, 0).

Mixing up the two directions. A number outside the function, as in $x^2+3$x² + 3, changes every height and moves the graph up or down. A number inside the bracket changes the input and moves the graph left or right.

Multiplying the whole function

A third kind of change multiplies the whole function. In y = kf(x), every output of f is multiplied by k. At each x the height is k times what it was, so a point (p, q) moves to (p, kq): it stays above the same x, and only its height changes. For y = 0.6f(x), every height becomes 0.6 of what it was, so the point (2, 10) moves to (2, 0.6 × 10) = (2, 6), and, since 0.6 is positive, the highest point is still the highest point, above the same x. This is called a vertical stretch.