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  3. Using the GDC

Using the GDC

What the GDC is expected to do in Papers 2 and 3, how to set it up, how to write GDC work so it earns method marks, and the errors that cost the most.

  • IB Mathematics: Analysis and Approaches
  • Exam skills
  • SL and HL

11 min readLast reviewed 14 Oct 2026

A graphic display calculator (GDC) is expected in Paper 2 and, for HL, Paper 3. The approved models draw graphs, find zeros, intersections, maxima and minima, evaluate derivatives at a point and definite integrals numerically, solve equations, analyse data, calculate probabilities from distributions and run a finance solver. They have no computer algebra system, so they cannot simplify, factorise, differentiate or integrate symbolically: every result is a number.

The GDC is a tool for the arithmetic, not a replacement for the method. Each GDC result in an answer must be supported by working that shows what was calculated and why. Menu names differ between models, so the descriptions here use generic names ("the zero function", "the normal cdf function"). Practise each one on your own GDC until it can be found without searching.

Setting up before the paper

A few settings decide whether every later answer is right.

  • Angle mode. Set radians for the paper and switch to degrees only for a triangle question given in degrees. Calculus with trigonometric functions always uses radians.

  • Number display. Use the normal floating-point display, not a fixed number of decimal places, so values are not rounded on screen before they are written down.

  • Exam mode. Many models have an exam (or press-to-test) mode that disables stored programs and files. Use it as the rules for the session require, and practise with it beforehand, because some features may behave differently.

  • Batteries. Charge or replace them before the day; a spare set is a sensible precaution.

A GDC in degree mode gives sin⁡2=0.0349sin2=0.0349 instead of sin⁡2=0.909sin2=0.909. It shows no warning, and every later result built on that value is wrong. Check the mode at the start of the paper and after any degree-mode triangle question.

Graphs and windows

The graph screen is the starting point for most GDC work: solving equations, finding turning points and areas, and checking the shape of a curve. Its main risk is the viewing window. A standard window shows −10≤x≤10−10≤x≤10 and −10≤y≤10−10≤y≤10, which hides most of the shape of many functions.

Worked example

Use your GDC to find the coordinates of the local maximum and local minimum points of the graph of f(x)=x3−30x2+200xf(x)=x3−30x2+200x.

Show the solution

In a standard window the graph appears as an almost vertical line through the origin, because the yy-values become very large very quickly. Use a table of values to choose a better window: f(10)=0f(10)=0, f(20)=0f(20)=0 and f(5)=375f(5)=375, so use −2≤x≤22−2≤x≤22 and −500≤y≤500−500≤y≤500.

−20246810121416182022−400−2000200400xy(4.23, 385)(15.8, −385)(10, 0)(20, 0)
  • y = f(x)
With a suitable window the whole shape is visible: three zeros and two turning points. A standard window shows none of the turning points.

Using the maximum and minimum functions on the graph:

local maximum (4.23,385)(4.23,385), local minimum (15.8,−385)(15.8,−385) (3 s.f.).

To choose a window, use the domain given in the question for the xx-range, then a table of values or the GDC's automatic fitting option for the yy-range. Then zoom out once to check that nothing lies outside it.

Common mistake

Reading solutions or turning points from the first window that appears. If the domain is not restricted, zoom out to make sure there are no further intersections or turning points. A cubic has up to three zeros, and a function such as sin⁡x−0.1xsinx−0.1x has many.

Zeros and intersections

Equations that cannot be solved algebraically are solved on the graph screen, in one of two ways:

  1. Graph each side as a separate function and find the points of intersection.

  2. Rearrange to the form g(x)=0g(x)=0, graph y=g(x)y=g(x) and find the zeros.

Both give the same xx-values. The first is easier to sketch in the answer; the second avoids errors when the two curves are almost tangent. For each zero or intersection, the GDC asks for a left bound and a right bound (or a starting guess), and finds the one solution inside them. Repeat for each solution.

Worked example

Solve x3−2x=cos⁡xx3−2x=cosx.

Show the solution

Graph y=x3−2xy=x3−2x and y=cos⁡xy=cosx in radian mode and find every intersection.

−2.5−2−1.5−1−0.500.511.522.5−3−2−10123xyx = −1.36x = −0.501x = 1.44
  • y = x³ − 2x
  • y = cos x
The curves cross three times. Each crossing needs its own intersection search.

Using the GDC, x=−1.36x=−1.36, x=−0.501x=−0.501 or x=1.44x=1.44 (3 s.f.).

Turning points

The maximum and minimum functions find local maxima and minima of a graph inside a chosen interval. They are useful for optimisation in context, for finding the range of a function on a restricted domain, and for checking a stationary point found by calculus.

Two cautions apply. First, the GDC finds a local maximum inside the bounds given; it does not check the end points of a domain, where the greatest value of a function may lie. Second, the results are approximate, so a value such as x=0.99999987x=0.99999987 should be recognised as x=1x=1 when an exact answer is expected.

Tip

For the greatest or least value of a function on an interval a≤x≤ba≤x≤b, compare three things: the local maxima and minima inside the interval, f(a)f(a) and f(b)f(b). A table of values gives the end points in seconds.

Numerical derivatives and integrals

The derivative at a point

The numerical derivative function gives the value of dydxdxdy​ at a specific value of xx. It answers questions such as "find the gradient of the curve at x=2x=2" or "find the rate of change when t=3t=3".

For f(x)=ln⁡(x2+1)f(x)=ln(x2+1), the GDC gives f′(2)=0.8f′(2)=0.8. By hand, f′(x)=2xx2+1f′(x)=x2+12x​ and f′(2)=45=0.8f′(2)=54​=0.8, which agrees.

It cannot give f′(x)f′(x) as a formula. When a question asks for f′(x)f′(x), or for a stationary point in terms of a parameter, differentiation must be done by hand. The GDC can still check the result at one value of xx.

Definite integrals

The numerical integral function evaluates ∫abf(x) dx∫ab​f(x)dx for numerical limits. It is the usual method on Paper 2 for areas, volumes of revolution (HL), distances travelled and probabilities from a continuous random variable.

Worked example

Find the area of the region enclosed by the curves y=sin⁡xy=sinx and y=x2y=2x​ for x≥0x≥0.

Show the solution

Find the limits: the curves intersect at x=0x=0 and, using the GDC, at x=1.8954…x=1.8954… (store this value as bb).

Between the limits, sin⁡xsinx is above x22x​, so integrate upper minus lower.

A=∫0b(sin⁡x−x2)dx=0.421 (3 s.f.)A=∫0b​(sinx−2x​)dx=0.421 (3 s.f.)

Using the stored value of bb, not 1.901.90, keeps the answer accurate.

Common mistake

Using ∫abf(x) dx∫ab​f(x)dx as an area when part of the curve lies below the xx-axis. The integral counts that part as negative. Integrate ∣f(x)∣∣f(x)∣ (the GDC accepts the absolute value function inside the integral), or split the integral at the zeros and add the sizes.

Solving equations

Besides the graph screen, most GDCs have three solving tools.

  • The numerical solver solves an equation in one unknown from a starting value. It returns one solution, the one nearest the starting value, so a graph is needed to find how many solutions there are and where to start.

  • The polynomial root finder gives all the roots of a polynomial equation up to a certain degree, including complex roots on some models.

  • The simultaneous equation solver solves systems of linear equations. For example, it solves

x+y+z=62x−y+z=3x+2y−z=2x+y+z2x−y+zx+2y−z​=6=3=2​

to give x=1x=1, y=2y=2, z=3z=3. If the system has no solution or infinitely many, the GDC reports an error or a general solution, and the algebra explains which case applies (see 1.16).

Statistics

One-variable statistics

Enter the data in a list (with a frequency list if the data are grouped or repeated) and use the one-variable statistics function. It gives the mean, the standard deviation, the quartiles, the median and the sum.

For the data 4,7,7,9,12,154,7,7,9,12,15, the GDC gives xˉ=9xˉ=9, median 88, and two standard deviations: σx=3.61σx​=3.61 and sx=3.95sx​=3.95. In IB maths, the standard deviation of a data set is σσ, so use σx=3.61σx​=3.61.

Exam tip

Use the standard deviation labelled σxσx​ (or similar) for data. Using sxsx​, which divides by n−1n−1, gives a different value and loses the accuracy mark. Quartiles can also differ slightly between GDC models and hand methods; small differences of this kind are generally accepted, but a result very different from the hand method suggests a data entry error.

Regression

The linear regression function takes two lists, xx and yy, and returns the equation of the regression line of yy on xx and Pearson's correlation coefficient rr (on some models rr must be switched on in the settings first). Check which list is which: the regression line of xx on yy needs the lists swapped (see 4.10).

Probability distributions

The distribution menu holds the functions for the binomial and normal distributions.

Function

Gives

Example

Binomial pdf

P(X=r)P(X=r) for X∼B(n,p)X∼B(n,p)

X∼B(12,0.3)X∼B(12,0.3): P(X=4)=0.231P(X=4)=0.231

Binomial cdf

P(X≤r)P(X≤r)

P(X≤4)=0.724P(X≤4)=0.724

Normal cdf

P(a<X<b)P(a<X<b) for X∼N(μ,σ2)X∼N(μ,σ2)

Lower and upper bounds, μμ and σσ

Inverse normal

The value kk with a given area to its left (or right, or centred)

X∼N(50,42)X∼N(50,42), P(X>k)=0.05P(X>k)=0.05 gives k=56.6k=56.6

The binomial cdf only gives P(X≤r)P(X≤r), so other inequalities must be rewritten first. For a discrete variable, "fewer than 4" means X≤3X≤3 and "at least 4" means 1−P(X≤3)1−P(X≤3).

P(X≥4)=1−P(X≤3)=0.507P(X≥4)=1−P(X≤3)=0.507

Common mistake

Using 1−P(X≤4)1−P(X≤4) for P(X≥4)P(X≥4). That removes the case X=4X=4, which should be included. Write the inequality in words first ("at least 4 means 4 or more"), then rewrite it as a cdf.

For the inverse normal function, check whether the GDC wants the area to the left of kk or offers a choice of tail. For P(X>k)=0.05P(X>k)=0.05, either set the tail to the right with area 0.050.05, or use area 0.950.95 to the left.

The finance solver

The finance solver (time value of money) handles compound interest and depreciation. It has fields for the number of periods (NN), the annual interest rate (I%I%), the present value (PVPV), the regular payment (PMTPMT, which is 00 in AA questions), the future value (FVFV), and the numbers of payments (P/YP/Y) and compounding periods (C/YC/Y) per year.

On most models NN is the total number of periods, not the number of years: with monthly periods (P/Y=12P/Y=12), eight years is entered as N=96N=96. Learn which convention your GDC uses.

For $5000 invested for 8 years at 3.5% per year compounded monthly, enter N=96N=96, I%=3.5I%=3.5, PV=−5000PV=−5000, PMT=0PMT=0, P/Y=12P/Y=12, C/Y=12C/Y=12 and solve for FVFV: the result is $6612.95. Solving instead for NN with FV=10 000FV=10000 gives N=238.0N=238.0 months, so the money doubles after about 19.819.8 years.

Money paid out is negative and money received is positive. The investment is paid into the account, so PV=−5000PV=−5000, and the future value comes out positive. Entering both as positive values gives an error or a meaningless answer.

Communicating GDC work

Most of the marks lost with a GDC are lost through communication, not wrong buttons. An answer that shows only a number gives the examiner nothing to award if the number is wrong, and may not earn full marks even when it is right. A complete line of GDC working states three things:

  1. What was set up: the equation, function, integral or distribution, in mathematical notation.

  2. How the GDC was used: "from the graph", "using the GDC", or a labelled sketch for a graphical solution.

  3. The result, to 3 significant figures or as the question asks, with any intermediate values to more figures.

Examples of good GDC working:

  • "P(X≥4)=1−P(X≤3)P(X≥4)=1−P(X≤3), where X∼B(12,0.3)X∼B(12,0.3); =0.507=0.507."

  • "∫05∣v(t)∣ dt=11.2∫05​∣v(t)∣dt=11.2 m."

  • "The graphs of y=exy=ex and y=4−x2y=4−x2 intersect at x=−1.96x=−1.96 and x=1.06x=1.06" with a sketch of both curves and the two points marked.

Exam tip

Never write calculator syntax as working: `normalcdf(182, 1E99, 170, 8)` or `fnInt(...)` is not mathematical notation, and an examiner may not credit it. Write P(X>182)P(X>182) with X∼N(170,82)X∼N(170,82), or the integral with its limits, instead.

Common errors

Error

Effect

How to avoid it

Degree mode for calculus or radian questions

Wrong values everywhere, no warning

Check the mode at the start and after every triangle question

Window too small

Missing solutions or turning points

Zoom out once; use a table of values

Missing brackets, e.g. `2^x+1` or `1/2x`

The GDC calculates 2x+12x+1 instead of 2x+12x+1, or 12x21​x instead of 12x2x1​

Bracket every exponent and denominator: `2^(x+1)`, `1/(2x)`

Subtraction key instead of the negative key

A syntax error or a wrong value

Use the negative sign for negative numbers

Reading `2.5E-3` as 2.52.5

An answer 1000 times too big

Convert to 2.5×10−3=0.00252.5×10−3=0.0025

Rounding before storing

Final answers wrong in the third figure

Store intermediate results and use them unrounded

sxsx​ instead of σxσx​

Wrong standard deviation

Use σxσx​ for data

Binomial pdf instead of cdf

P(X=r)P(X=r) instead of P(X≤r)P(X≤r)

Read the inequality before choosing the function

Variance entered as σσ

Wrong normal probabilities

N(μ,σ2)N(μ,σ2): enter the square root of the second parameter

Tip

After a GDC calculation, ask whether the result is sensible: a probability between 00 and 11, an area that is positive and roughly the size the graph suggests, a time inside the domain of the model. Many entry errors give answers that fail this test.

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