Quick Summary

• You can type any function into your GDC as f(x) = and it will draw the graph for you — no table of values needed in the exam.
• The "analyze graph" / G-Solv menu finds roots (x-intercepts), maximum/minimum points, and intersections automatically.
• Solving f₁(x) = f₂(x) graphically just means: draw both graphs, find where they cross, and read off the x-coordinates.
• Solving f(x) = 0 is a special case — it's just finding where the graph crosses the x-axis (y = 0).
• Solving f(x) = k means finding where the graph crosses the horizontal line y = k.
• To solve an unfamiliar equation using a graph you've already drawn, rearrange it so one side matches your original function.
• For inequalities, solve the "=" version first to find the crossover points, then check which curve is on top in each region.
• When sketching from your GDC onto exam paper: freehand smooth curves, no ruler, mark all key coordinates.

1. Using Calculators to Sketch Graphs

Getting a graph onto your screen

Every GDC has an "Add Graphs" (or "Graph Mode") screen with a line that says f(x) = — sometimes labelled f1(x) = or Y1:. Whatever the label, the idea is identical: you type in the for your function, and the calculator plots every point that satisfies it.

Think of it like this — the calculator is testing thousands of x-values in a split second, working out the matching y-value for each one, and joining all the dots into a smooth curve. That's exactly what you'd do by hand with a table of values, just millions of times faster.

Don't skip the zoom step! A graph can look like a flat line or be totally off-screen if the window (the visible x and y range) isn't set sensibly. The "zoom-fit" button is your best friend here — it automatically finds a window that shows off all the key features (turning points, intercepts) without you guessing numbers.

Finding key features on the graph

Once your curve is on screen, go into the "analyze graph" menu (sometimes called G-Solv). This is where the real power is — instead of calculating things algebraically, you're asking the calculator "show me the interesting points."

Maximum and minimum points together are called turning points (because the graph "turns" direction there) — and if you're specifically talking about a quadratic, that single turning point also has a special name: the vertex.

Sketching the graph onto exam paper

When a question says "draw this graph on your calculator" and then asks you to sketch it on given axes, examiners are not expecting a ruler-perfect, pixel-accurate copy. They want:

• One smooth, continuous freehand curve (no ruler for curved parts!)
• The correct overall shape and symmetry copied from your GDC screen
• All axis intercepts and turning points clearly labelled with coordinates

You do not need to plot lots of individual points or build a table of values by hand — that's the entire point of using the calculator. Your job is to transfer what you see on screen onto paper accurately enough that someone could tell exactly which function it represents.

2. Using Calculators to Solve Equations

The core idea: intersections = solutions

This is the single most important idea in this whole chapter, so let's slow down and really unpack it. Suppose you want to solve an equation like x² + 3x + 1 = 2x + 1. Instead of rearranging and factorising by hand, you can treat each side of the equation as its own separate graph:

• Left-hand side becomes y = f₁(x) = x² + 3x + 1
• Right-hand side becomes y = f₂(x) = 2x + 1

Now here's the magic: wherever these two graphs cross, the y-values are equal — which means the original left-hand side equals the original right-hand side at that x-value. In other words, the x-coordinates of the intersection points ARE the solutions to the equation.

For the example above, plotting both graphs and using "Analyze graph → Intersection" (or G-Solv → INTSECT) gives intersection points (−1, −1) and (0, 1). So the solutions are x = −1 and x = 0.

Two special (very common) cases

You'll almost always meet equation-solving in one of these two disguises:

Case A — Solving f(x) = 0

This just means finding where your curve crosses the line y = 0 — and y = 0 the x-axis. So solving f(x) = 0 is exactly the same as finding the x-intercepts (roots) of y = f(x). No second graph needed — just use "Zeros" on your calculator.

Case B — Solving f(x) = k (a constant)

Here you're finding where your curve crosses the horizontal line y = k. For example, to solve x² + 3x + 1 = 1, draw y = x² + 3x + 1 together with the horizontal line y = 1, then find where they intersect.

Using an existing graph to solve a different equation

Sometimes an exam question gives you a graph you've already drawn (say y = x² + 3x + 1) and then asks you to solve a equation, like x² + 3x − 4 = 0. The trick is to manipulate the new equation with the exact same add/subtract-to-both-sides moves until the left-hand side matches your existing graph exactly.

Now the left-hand side matches your existing graph y = x² + 3x + 1 perfectly, and you just need to find where it crosses the horizontal line y = 5.

Graphical method vs. equation solver (algebraic method)

Your GDC actually offers two completely different routes to a solution, and it's crucial you know the difference — because exam questions often specify which one they want.

3. Using Calculators to Solve Inequalities

Building on what you already know

Here's the good news: if you understood section 2, inequalities are barely a new topic — they're just one extra step tacked onto equation-solving. Instead of asking "where are the two graphs ?", you're asking "over which x-values is one graph than the other?"

Reading "top curve" vs "bottom curve"

This is the one genuinely new skill here, so let's make sure it clicks. If you're solving f1(x) > f2(x), you want the x-values where the f1 curve sits the f2 curve on the graph — because "above" on a graph literally means "bigger y-value," and that's exactly what the inequality is asking for.

Flip it round for f1(x) < f2(x): now you're looking for the regions where f1 sits f2 instead.

Worked walk-through

Let's solve x² + 3x + 1 > 2x + 1 using a graphical method, step by step:

1. Solve the "=" version first: x² + 3x + 1 = 2x + 1 gives x = −1 and x = 0 (the two crossover points).
2. Plot both graphs on your GDC — the quadratic y = x² + 3x + 1 and the line y = 2x + 1.
3. Check each region: to the left of x = −1, the quadratic is above the line. Between x = −1 and x = 0, the line is above the quadratic. To the right of x = 0, the quadratic is above the line again.
4. Pick the regions where the quadratic (f1) is on top, since we want f1(x) > f2(x).

What to Memorise

Key Terms

Core Rules

Calculator Button Shortcuts

Concepts Checklist

Tick off each concept once you're confident you could explain it to someone else without looking at your notes.

Exam Tips & Common Mistakes