Before the midterm

Why does a minus sign inside move a graph to the right?

Probe a shifted square root to see that the new graph at x copies the old one at x − 2, so every height arrives 2 later, and see why a change outside the function moves it as written.

The new graph, y: √(x)
Starts at, (x, y): (0, 0)

Step 1 of 5: Which way?

The dashed curve is $y = \sqrt{x}$ , which starts at the origin. Put $x - 2$ in place of $x$ and you have a new function, $g(x) = \sqrt{x - 2}$ .

Predict first. Where does the graph of $g$ start?

1. Which way?
The dashed curve is $y = \sqrt{x}$ , which starts at the origin. Put $x - 2$ in place of $x$ and you have a new function, $g(x) = \sqrt{x - 2}$ .
Question: Where does the graph of $g$ start?
At $x = 2$ . The root needs an input of at least $0$ , and $g$ feeds it $x - 2$ , which is $0$ at $x = 2$ and negative before that. So the whole graph sits $2$ to the right, although the formula has a minus.
The figure: The dashed graph of y = √x, and the graph of y = √(x − 2), which starts at (2, 0).
2. The probe
The probe at $x$ on the new graph shows the input it feeds the root, $x - 2$ , and the hollow point of $y = \sqrt{x}$ whose height it copies. So $g$ at $x$ shows what $\sqrt{x}$ showed at $x - 2$ .
Drag the probe to where the new graph is at height $2$ .
At $x = 6$ : it feeds the root $6 - 2 = 4$ , and $\sqrt{4} = 2$ . The root reaches $2$ at $4$ , so $g$ reaches it $2$ later, at $6$ . Every height of $\sqrt{x}$ turns up $2$ further right, because $x$ has to be $2$ bigger to feed the root the same input.
$g(x) = f(x - 2) \text{ is the graph of } f \text{ moved } 2 \text{ to the right}$
The figure: The dashed graph of y = √x, and the graph of y = √(x − 2), which starts at (2, 0). The probe at x = 6.00 feeds the root 4.00; the height 2.00 is copied across from the point (4.00, 2.00) of y = √x, to 2.00.
3. The wrong way
A common slip reads the minus as a move left, and draws the amber graph, $y = \sqrt{x + 2}$ . The probe is now at $x = 1$ .
Question: What is $g(1) = \sqrt{1 - 2}$ ?
Not defined: $1 - 2 = -1$ , and a negative number has no real square root. The amber graph gives $\sqrt{1 + 2} \approx 1.73$ because it is $\sqrt{x + 2}$ , the shift the other way. Inside the function the sign reads backwards: $x - 2$ moves the graph right and $x + 2$ moves it left.
The figure: The dashed graph of y = √x, and the graph of y = √(x − 2), which starts at (2, 0). The slip's graph, y = √(x + 2), starts at (−2, 0). The probe at x = 1.00 feeds the root −1.00, which has no square root: the new graph has no point there. The slip's graph is at 1.73 there.
4. Set the shift
Now choose the shift yourself. Drag the start of the new graph along the axis, or use the slider, and watch its formula change under the figure.
Move the graph so that it starts at $x = -3$ .
To start at $-3$ , the root must be fed $0$ when $x = -3$ , so its input is $x + 3$ and the graph is $y = \sqrt{x + 3}$ . A plus sign inside moves the graph left.
The figure: The dashed graph of y = √x, and the graph of y = √(x + 3), which starts at (−3, 0).
5. Outside the root
Every change so far was inside the root, made to the input. A change outside, made to the output after the root, acts on heights instead.
Question: What does $y = \sqrt{x - 2} - 1$ do to the graph of $y = \sqrt{x}$ ?
Right $2$ and down $1$ . Follow the probe at $x = 6$ : the root is fed $6 - 2 = 4$ and gives $2$ , then $1$ is taken off, leaving $1$ . The $-1$ acts after the root, so it moves the graph as written; the $-2$ acts before, and is undone.
$y = f(x - c) + d \text{ is } y = f(x) \text{ moved } c \text{ right and } d \text{ up}$
The figure: The dashed graph of y = √x, and the graph of y = √(x − 2) − 1, which starts at (2, −1). The probe at x = 6.00 feeds the root 4.00; the height 2.00 is copied across from the point (4.00, 2.00) of y = √x and then moved by −1.00, to 1.00.

Step 1 of 5: Which way?

The lesson as text

1. Which way?

2. The probe

3. The wrong way

4. Set the shift

5. Outside the root