How to use Desmos to graph a function with a slider?

Open the Desmos graphing calculator and type the expressions yourself. Nothing on this page reaches into a Desmos file. The board further down uses the same letters, on GoodMaith, so you can drag $a$, $h$, and $k$ while you read what those letters do.

Type a parabola

Click the first expression row and enter y=x^2. Desmos draws the curve as soon as the row is a graph it understands. There is no separate plot button. The row is the graph.

The curve passes through the origin, because $0^2=0$. At $x=1$ the height is $1$. At $x=2$ the height is $4$. At $x=-2$ the height is also $4$, since squaring forgets the sign. The arms are symmetric across the vertical axis, and the vertex, the lowest point, is $(0,0)$. If the curve does not appear, the usual cause is a row Desmos is treating as a label or as an unfinished expression. y=x^2 is enough. The caret is the power.

You can restrict the drawing without changing the rule. Appending a condition such as a domain in curly braces hides the parts of the curve outside that domain. For the first pass, leave the domain alone and let the arms run off the window. Zooming the window does not change the expression. It only changes how much of $y=x^2$ you see.

Add sliders

Replace that row with y=a(x-h)^2+k. Desmos does not yet have values for $a$, $h$, and $k$. It asks to add a slider for each letter. Add all three. Each slider is a number you can drag, and the curve is redrawn from the expression at every value.

The letters have fixed jobs in this form. $(h,k)$ is the vertex. $h$ moves it left and right. $k$ moves it up and down. $a$ stretches the arms away from that vertex. With $a=1$ the arms match $y=x^2$, only shifted. With $a=2$ they are twice as steep: one unit to the right of the vertex, the height is $2$ above $k$, not $1$. With $a$ between $0$ and $1$ the arms open wider. Keep $a$ positive on the first pass so the vertex stays a lowest point. A negative $a$ turns the parabola over, and the vertex becomes the highest point. Desmos will allow that if you edit the slider's lower end below zero.

Set the sliders to $a=1$, $h=1$, and $k=-1$. The vertex is $(1,-1)$. The point $(0,0)$ is still on the curve, because $$1(0-1{)}^2-1=0.$$ Drag $h$ from $1$ to $3$. The vertex moves from $(1,-1)$ to $(3,-1)$. The arms move with it. $k$ has not changed, so the vertex is no higher and no lower. Drag $k$ from $-1$ to $2$. The vertex rises to $(3,2)$ if you left $h$ at $3$, or to $(1,2)$ if you put $h$ back. Drag $a$ from $1$ to $2$ and the arms pull in toward the vertex. Put the three sliders back to $1$, $1$, and $-1$ before you add a point, so the arithmetic below still matches the picture.

The board above is not Desmos. It is the same expression on GoodMaith: $a$ starts at $1$, $h$ at $1$, $k$ at $-1$, and V is the point $(h,k)$. Drag the sliders. V follows. Drag Q along the curve. Q is a free point on $y=a(x-h{)}^2+k$, so it stays on the arms while $a$, $h$, and $k$ change. In Desmos you get those sliders by typing the letters. Here they are already on the board so the three drags above are something you can do while you read.

Add a point

In Desmos, add a new expression row and type (0,0). You can also place a point with the point tool. A point typed as a fixed pair does not know about the sliders. It sits at $(0,0)$ until you edit the row. After the sliders are at $1$, $1$, and $-1$, that fixed point lies on the parabola, by the arithmetic above. Move $h$ or $k$ and the curve leaves the point. The point was never defined from $h$ and $k$. It only happened to lie on the curve for those three numbers.

A point that should follow the vertex is typed $(h,k)$, using the same letters as the sliders. Drag $h$ or $k$ and that point moves to the new vertex. A point that should stay on the arm at a chosen input, say one unit left of the vertex, is typed $(h-1,a(1{)}^2+k)$, which is $(h-1,a+k)$. Drag $a$ and that point's height changes even though you did not drag it directly. The habit is: a coordinate written with a slider letter depends on that slider. A coordinate written as a bare number does not.

The shift itself, sending each point $(x,y)$ of a graph to $(x+a,y+b)$, is the handbook page on transformation of functions. There the sliders are named $a$ and $b$, and $a$ is the horizontal step. In the expression on this page the horizontal step is $h$ and the vertical step is $k$. The move is the same move.

When you would rather say it

Desmos draws the row you type. When you would rather describe the problem in a sentence and have the figure appear while it is explained, open this AI math solver with live graphs, or chat with a live board. You can try the tutor without an account. The graphing workbench is the board alone, with no account, for the times you already know the expression and want to drive the sliders yourself.