Graphing Calculator

A graphing calculator turns an equation into a picture. You give it a function in the form $y = f(x)$; it evaluates $f$ at hundreds of $x$ values inside the viewing window and joins the resulting points into a curve.

Two things therefore have to be settled before anything is drawn:

• The equation must be solved for $y$. An equation written as $2x + 3y = 12$ has to become $y = \frac{12 - 2x}{3}$ before it can be plotted.
• The window decides what you see. A parabola whose vertex sits at $(0, 400)$ is completely invisible in a standard $-10 \le y \le 10$ window. The graph is not wrong — the window is.

The curve is only a sample. Between the plotted points the calculator draws short straight segments, which is why a very steep function can look jagged, and why a vertical asymptote is sometimes drawn as a steep line that is not really part of the graph.

1. Solve for $y$ so the function is in the form $y = f(x)$.
2. Recognise the family. $y = mx + b$ is a straight line, $y = ax^2 + bx + c$ is a parabola, $y = \frac{1}{x - a}$ is a hyperbola with a vertical asymptote at $x = a$. Knowing the shape tells you what the picture should look like before you plot anything.
3. Find the anchor points. The $y$-intercept is $f(0)$. The $x$-intercepts come from solving $f(x) = 0$. For a parabola the vertex is at $x = -\frac{b}{2a}$.
4. Build a small table of three to five $x$ values around those anchors and compute each $y$.
5. Choose a window that contains every anchor point, then plot and join the points smoothly.

To graph a relation that is not a function — a circle such as $x^2 + y^2 = 25$, say — split it into two branches, $y = \sqrt{25 - x^2}$ and $y = -\sqrt{25 - x^2}$, and plot both.

• Blank screen, correct equation. Almost always a window problem. Zoom out, or set the window from the values in your table.
• Missing brackets. Typing 1/x+2 gives $\frac{1}{x} + 2$, not $\frac{1}{x+2}$. Wrap every denominator and every exponent in brackets.
• Degrees instead of radians. $y = \sin(x)$ looks almost flat near the origin in degree mode, because one radian is about 57 degrees.
• Treating the drawn asymptote as part of the curve. The near-vertical segment at $x = 2$ in $y = \frac{1}{x-2}$ is a drawing artefact; the function has no value there.
• Reading coordinates off the picture. Trace and pixel estimates are approximate. If you need an exact intercept or intersection, solve the equation algebraically.