Graphs — AQA GCSE Mathematics
Test yourself on Graphs with AQA GCSE practice questions.
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Graphs explained
The two axes cut the plane into four regions.
Read the full explanation
A point is written as an ordered pair, across first and then up, with right and up counted as positive and left and down as negative. So the top right region holds points like (3, 4), the top left holds (−3, 4), the bottom left holds (−3, −4) and the bottom right holds (3, −4). To read a point off a grid, start at the origin, count along the horizontal axis, then count parallel to the vertical axis, keeping the sign of each direction. You also need the midpoint of a line segment, which is the average of the two horizontal values paired with the average of the two vertical values: the midpoint of (−2, 6) and (4, −2) is (1, 2). Always check the scale, because one square is often worth more than one unit.
plot graphs of equations that correspond to straight-line graphs in the coordinate plane
A straight-line graph is the set of points whose coordinates fit a linear equation. The usual method is a table of values: take each horizontal value the question gives, substitute it into the equation and record the result. For y = 3x − 2, an input of −1 gives −5, an input of 0 gives −2 and an input of 2 gives 4. Plot those pairs, then join them with one ruled straight line that runs across the whole grid rather than stopping at the outer points. Some equations arrive arranged differently, such as 2y + x = 6. You can rearrange to leave y on its own, or find the two axis crossings by putting x = 0 and then y = 0, and rule the line through them.
use the formy = mx + c to identify parallel lines find the equation of the line through two given points, or through one point with a given gradient
The equation of a straight line is often written as y = mx + c, where 'm' is the gradient (steepness) and 'c' is the y-intercept (where the line crosses the y-axis). Two distinct lines are parallel if and only if they have the same gradient. For example, y = 2x + 3 and y = 2x − 5 are parallel. To find the equation of a line through two points, (x₁, y₁) and (x₂, y₂), first calculate the gradient using m = (y₂ − y₁) / (x₂ − x₁). For points (1, 5) and (3, 11), the gradient m = (11 − 5) / (3 − 1) = 6/2 = 3. Then, substitute this gradient and one of the points into y = mx + c to find c. Using (1, 5): 5 = 3(1) + c, so c = 2. The final equation is y = 3x + 2. If given one point and the gradient, the process is simpler: just substitute m and the point's coordinates to solve for c.
use the form y = mx + c to identify perpendicular lines (Higher tier only)
This is Higher tier only. Two lines are perpendicular when the product of their gradients is −1, which means each gradient is the negative reciprocal of the other: flip the fraction and change the sign. A gradient of 2 pairs with −½, and a gradient of −3/4 pairs with 4/3. A typical question gives a line such as y = 2x + 5 and a point such as (4, 3), and asks for the perpendicular line through that point. Take the new gradient as −½, substitute the point to get 3 = −½ × 4 + c, so c = 5 and the answer is y = −½x + 5. Lines on a grid can look perpendicular without being so, especially when the two axes are scaled differently, so test the gradients rather than trusting the picture.
identify and interpret gradients and intercepts of linear functions graphically and algebraically
Reading a line from a picture means choosing two points that sit exactly on gridline crossings, then dividing the rise by the run. A line going up from left to right has a positive gradient, one going down has a negative gradient, and a horizontal line has a gradient of zero. The intercept is the value where the line meets the vertical axis. Working from the equation instead, rearrange until y stands alone: 3y − 6x = 12 becomes y = 2x + 4, so the gradient is 2 and the intercept is 4. In a real context the gradient is a rate, such as pounds per hour of hire, and the intercept is the starting value, such as a fixed call-out charge. An interpretation question wants that meaning in words with units, not only the number.
identify and interpret roots, intercepts and turning points of quadratic functions graphically deduce roots algebraically
A quadratic graph is a parabola: it is U-shaped when the squared term is positive and n-shaped when it is negative. The roots are the horizontal values where the curve meets the horizontal axis, and they are the solutions you get by setting the expression to zero. The curve meets the vertical axis at the constant term, because putting zero in for x leaves it alone. The turning point is the lowest point of a U-shaped curve or the highest point of an n-shaped one, and it sits on the line of symmetry, which is halfway between the roots when the curve has two of them. For y = x² − 5x + 6, factorising gives (x − 2)(x − 3), so the roots are x = 2 and x = 3, the curve meets the vertical axis at 6, and the line of symmetry is x = 2.5.
deduce turning points by completing the square (Higher tier only)
This is Higher tier only. Writing a quadratic as a completed square hands you the turning point with no plotting at all. Halve the coefficient of x to get the number inside the bracket, then subtract the square of that number to undo what the bracket added. For y = x² − 6x + 5, half of −6 is −3, and (x − 3)² expands to x² − 6x + 9, which is 4 too much, so y = (x − 3)² − 4. Whenever a quadratic is written as a squared bracket plus a constant, the turning point sits where the bracket is zero, with the constant as its height: here the minimum is at (3, −4) and the line of symmetry is x = 3. A squared bracket is never negative, so where the squared term is positive that constant is the least value y can take, and where it is negative the same point is a maximum instead. If the squared term has a coefficient, factor it out first.
recognise, sketch and interpret graphs of linear functions and quadratic functions
A sketch is not a plotted graph. You need no table of values and no accurate squares; you need the right shape in the right place with the important features labelled. For a straight line, mark where it crosses each axis and rule a line through them, so y = 2x − 6 meets the vertical axis at −6 and the horizontal axis at x = 3. For a quadratic, decide first whether the curve opens upwards or downwards from the sign of the squared term, then mark the roots, the vertical axis crossing and the turning point, and draw a smooth symmetrical curve through them. Recognition questions run the other way: you are shown a picture and asked which equation fits, so check the shape first, then the crossings, then the steepness.
including simple cubic functions and the reciprocal function y = 1/x with x ≠ 0
A cubic has a highest power of three, and its graph is a smooth curve that either turns twice, giving a hump and a dip, or never turns at all, and it heads off in opposite directions at its two ends. The graph of y = x³ climbs through the origin, flattening for a moment there before rising again, while y = −x³ is that curve reflected, falling from left to right. A reciprocal graph such as y = 1/x comes in two separate branches, one in the top right region and one in the bottom left, and it crosses neither axis. As x grows the curve creeps towards the horizontal axis without touching it, and as x nears zero the curve shoots away. Zero is excluded because dividing by zero has no meaning. Both types need more plotted points than a straight line does, joined smoothly.
including exponential functions y = kˣ for positive values of k, and the trigonometric functions (with arguments in degrees) y = sin x, y = cos x and y = tan x for angles of any size (Higher tier only)
This is Higher tier only. An exponential curve of the form y = kˣ, with k positive, passes through (0, 1), because any positive number raised to the power zero is one. When k is bigger than one the curve grows ever more steeply to the right and flattens towards the horizontal axis on the left; when k lies between zero and one it decays instead. It never dips below the axis. The trigonometric curves repeat: y = sin x and y = cos x both wave between −1 and 1 with a period of 360°, the sine curve starting at zero and the cosine curve at one. The tangent curve repeats every 180° and races away either side of 90° and 270°, where it has no value. Knowing these shapes lets you read extra solutions off them.
sketch translations and reflections of a given function (Higher tier only)
This is Higher tier only. Start from a curve you already know and move it as a whole. A number added outside the function, y = f(x) + 3, lifts every point up by three. A number added inside the bracket, y = f(x + 3), shifts the curve three to the left, which feels backwards until you notice that the input must be smaller to give the same output. A minus sign outside, y = −f(x), flips the curve in the horizontal axis, and a minus sign inside, y = f(−x), flips it in the vertical axis. The quickest method is to track one labelled point: if y = f(x) has a maximum at (2, 5), then on y = f(x) + 3 the maximum sits at (2, 8), and on y = f(x − 1) it sits at (3, 5). The shape never changes, only where it sits.
plot and interpret graphs, and graphs of non-standard functions in real contexts, to find approximate solutions to problems such as simple kinematic problems involving distance, speed and acceleration
This topic involves both plotting and interpreting graphs that model real-world situations. To plot a graph, you must choose and label appropriate linear scales for the axes, plot points accurately from data, and draw a suitable line or curve. Interpretation focuses on what the graph's features mean. For kinematic problems, the gradient of a distance-time graph represents speed, while the gradient of a speed-time graph represents acceleration. The area under a speed-time graph represents the distance travelled, often found by splitting the area into rectangles and trapeziums. For example, for an object accelerating from 10 m/s to 20 m/s in 5s, the distance is the area of a trapezium: ½(10 + 20) × 5 = 75m. For non-standard graphs, like a container filling with water, the gradient shows the rate of change; a steepening curve might mean the container is getting narrower.
including reciprocal graphs
A reciprocal relationship is one where multiplying one quantity by a number divides the other by the same number. For a fixed journey of 60 km, the time taken is 60 divided by the speed, so doubling the speed halves the time. Drawn as a graph, this gives a curve that drops steeply at first and then flattens as it moves right, approaching the horizontal axis without ever reaching it. In context that means large gains at the start and much smaller gains later: raising a speed from 20 km/h to 40 km/h saves far more time than raising it from 80 km/h to 100 km/h. Take readings off the curve as you would from any graph, and expect an estimate rather than an exact figure, because the left-hand end is so steep.
including exponential graphs (Higher tier only)
This is Higher tier only. An exponential context is one where a quantity is multiplied by the same factor in every equal time step, rather than having a fixed amount added. Savings of £2000 growing at 5% a year reach 2000 × 1.05ⁿ pounds after n years, and a car losing 15% of its value each year is multiplied by 0.85 each time. Drawn against time, growth gives a curve that starts gently and steepens, while decay gives a curve that falls fast and then flattens towards the axis without reaching it. To take a doubling time or a half-life off the graph, find the height on the vertical axis, go across to the curve and drop down to the time axis. The steepness changes constantly, so any rate you take from such a curve belongs to that moment only.
calculate or estimate gradients of graphs and areas under graphs (including quadratic and other non- linear graphs), and interpret results in cases such as distance-time graphs, velocity-time graphs and graphs in financial contexts (Higher tier only)
This is Higher tier only. On a curve the steepness changes from point to point, so to find the rate at one instant you rule a tangent that touches the curve at that point without crossing it, build a large right-angled triangle on it and divide the rise by the run. A straight line joining two points on the curve gives an average rate over that interval instead, which is a different answer to a different question. Areas under graphs are found by splitting the region into rectangles, triangles and trapeziums and adding the parts, or by counting squares once you know what one square is worth. On a distance-time graph the gradient is speed; on a velocity-time graph the gradient is acceleration and the area underneath is the distance travelled. In a financial graph the gradient might be pounds per month. Units come from the axes every time.
recognise and use the equation of a circle with centre at the origin find the equation of a tangent to a circle at a given point (Higher tier only)
This is Higher tier only. A circle centred on the origin with radius r has equation x² + y² = r², so x² + y² = 25 is the circle of radius 5. To test whether a point lies on it, substitute: (3, 4) gives 9 + 16 = 25, so it does. A tangent touches the circle at exactly one point and is perpendicular to the radius drawn to that point, which gives you the method. Find the gradient of the radius from the origin to the point of contact, take its negative reciprocal for the tangent, then substitute the point to find the intercept. Where the point of contact sits on an axis, the radius is horizontal or vertical and the tangent is the perpendicular line through that point. At (3, 4) the radius has gradient 4/3, so the tangent has gradient −3/4; substituting gives an intercept of 25/4, and clearing the fraction leaves the tangent as 3x + 4y = 25.
Your focus
- Write down a point as an ordered pair with the horizontal value first, keeping the sign of each direction.
- Find the midpoint of a segment by averaging each pair of values, so (−2, 6) and (4, −2) give (1, 2).
- Explain why the scale on each axis must be read before plotting, since one square is often worth more than one unit.
Show all 50 objectives
- Complete a table of values for an equation such as y = 3x − 2 by substituting each input the question gives.
- Draw one ruled straight line through the plotted points, running across the whole grid rather than stopping at the outer ones.
- Explain how to draw 2y + x = 6, either by rearranging for y or by finding where the line crosses each axis.
- Write down the gradient and the height of the vertical intercept from an equation given in the form y = mx + c.
- Find the equation through (1, 5) and (3, 11) by working out the gradient as 6/2, then substituting to get c = 2.
- Work out the equation of a line through one given point with a stated gradient by substituting to find the intercept.
- Explain why y = 4x − 1 and y = 4x + 7 never meet, using the fact that parallel lines have equal gradients.
- Write down the rule that perpendicular gradients multiply to −1, making each one the negative reciprocal of the other.
- Find the line perpendicular to y = 2x + 5 through (4, 3), taking the gradient as −½ and reaching y = −½x + 5.
- Explain why lines that look perpendicular on a grid may not be, and test the two gradients rather than the picture.
- Describe what a positive, a negative and a zero gradient look like on a drawn line and where the intercept is read.
- Work out the gradient from the rise and run of two grid points, and from an equation rearranged into y = mx + c.
- Interpret the gradient as a rate and the intercept as a starting value, stating both in the units of the context.
- Describe the shape of a parabola from the sign of the squared term and state where it meets the vertical axis.
- Find the roots of a quadratic by setting the expression to zero and factorising, giving both values of x.
- Explain why the turning point lies midway between two roots and use that to write it as a coordinate pair.
- Describe how a completed square form hands you the turning point, the bracket at zero and the constant as its height.
- Work out the completed square form by halving the coefficient of x and correcting with the square of that half.
- Explain why a squared bracket is never negative and justify the least value the expression can take.
- Describe what a sketch must show that a plotted graph need not, naming the features that have to be labelled.
- Sketch a straight line from its two axis crossings and a quadratic from its shape, roots and turning point.
- Explain which equation fits a graph you are shown by checking the shape first, then the crossings, then the steepness.
- Describe the shape of a cubic curve and of the two branches of y = 1/x, including why x cannot take the value zero.
- Complete a table of values for a cubic, cubing negative inputs correctly, and join the plotted points as one smooth curve.
- Justify which of two given sketches is the cubic and which is the reciprocal by naming the feature that separates them.
- Describe the shapes of y = sin x, y = cos x and y = tan x, stating the period of each and where tan x has no value.
- Sketch y = kˣ through (0, 1), showing growth where k is above one and decay where it lies between zero and one.
- Use the symmetry of a sine or cosine curve to write down a further solution once one angle is known.
- Describe the effect of f(x) + a, f(x + a), −f(x) and f(−x) on where a curve sits.
- Sketch the transformed curve and work out where a labelled point such as a maximum moves to.
- Explain why a number added inside the bracket shifts the curve to the left rather than to the right.
- Describe what the gradient, a horizontal section and a steeper section each mean on a distance-time graph.
- Estimate a speed from a distance-time graph by reading off a distance and a time and dividing, with units attached.
- Interpret a flat or steep stretch of a container filling graph in the words of the context rather than in general terms.
- Describe a reciprocal relationship as one where multiplying one quantity divides the other by the same number.
- Work out a missing quantity from a reciprocal relation, such as dividing a fixed distance by a speed to get a time.
- Explain why raising a low speed saves far more time than raising an already high one, using the flattening of the curve.
- Describe what makes a context exponential, with the same multiplier applied in every equal time step.
- Work out a value after n time steps by raising the constant multiplier to the power n, as in 2000 × 1.05ⁿ.
- Estimate a doubling time or half-life from a curve and explain why any rate taken from it belongs to that moment only.
- Explain the difference between a tangent, which gives the rate at one instant, and a chord, which gives an average rate.
- Calculate the gradient at a point by ruling a tangent, building a large triangle on it and dividing the rise by the run.
- Estimate the area under a velocity-time graph by splitting it into rectangles and trapeziums, then read it as a distance.
- Interpret a gradient taken from a distance-time graph as a speed, or from a financial graph in the units of its axes, such as pounds per month.
- Write down the equation of a circle centred on the origin with a given radius and test whether a point lies on it.
- Find a tangent equation by taking the negative reciprocal of the radius gradient and substituting the point of contact.
- Explain why the tangent is perpendicular to the radius at the point of contact and what that gives you to work with.
Graphs exam tips
Quick Revision Summary (Key Takeaway)
Graphs in AQA GCSE Mathematics covers plotting and interpreting linear, quadratic, cubic, reciprocal and exponential functions, as well as solving equations graphically and identifying key features such as roots, intercepts, turning points and asymptotes. Mastery requires accurate plotting, recognising shape families and using graphs to solve simultaneous equations and inequalities.
Topic Overview
Graphs is a fundamental topic in AQA GCSE Mathematics that involves representing relationships between variables visually. You will learn to plot and interpret linear, quadratic, cubic, reciprocal and exponential graphs, and use them to solve equations and inequalities. This topic connects algebra and geometry, and is essential for understanding functions and modelling real-world situations.
Graphs appear across both Foundation and Higher tiers, with Higher tier including more complex functions such as exponential graphs and transformations. Mastery of graphs is crucial for solving simultaneous equations, finding roots, and interpreting rates of change. It also provides a foundation for A-level Mathematics topics like calculus and functions.
Key Concepts
- →Linear graphs have the form y = mx + c, where m is the gradient and c is the y-intercept; they are straight lines.
- →Quadratic graphs have the form y = ax^2 + bx + c and are U-shaped (parabolas); the turning point and roots are key features.
- →Cubic graphs have the form y = ax^3 + bx^2 + cx + d and have an S-shape with up to three roots.
- →Reciprocal graphs have the form y = a/x and have two branches with asymptotes at x = 0 and y = 0.
- →Exponential graphs have the form y = k^x and show rapid growth or decay, passing through (0, 1) when k > 0.
Marking Points
- each point plotted in the correct region with both signs read the right way round
- the method of averaging the two horizontal values and the two vertical values when a midpoint is wanted, even if the final answer is wrong
- the answer written as an ordered pair in brackets, horizontal value first
- using the scale on each axis rather than counting one square as one unit
- rearranging an equation such as 2y + x = 6 to leave y on its own before substituting
- a correct table of values, with the substitution shown for at least one input
- at least two points plotted correctly from that table, even if the line is then drawn badly
- a single ruled straight line through the plotted points, covering the range asked for
- Correctly calculating the gradient from two points using the formula m = (y₂ − y₁) / (x₂ − x₁).
- Identifying that parallel lines have an equal gradient m.
- Substituting the gradient and the coordinates of one point correctly into y = mx + c to find the y-intercept c.
- Stating the final answer as a complete equation in the form y = mx + c or an equivalent form.
- the gradient of the given line, read off directly or found by rearranging first
- the perpendicular gradient as the negative reciprocal, even if everything after it is wrong
- substituting the given point to find the intercept
- the final equation, or for showing the two gradients multiply to −1 when a proof is asked for
- a correct rise and run taken from two points that genuinely lie on the line
- the gradient as a value with the correct sign, even if the interpretation is missing
- the intercept read where the line meets the vertical axis
- interpreting the gradient in the units of the context, such as cost per kilogram
- setting the expression to zero before solving
- a correct factorisation, even if the roots are then written with the wrong signs
- each root stated as a value of x
- the line of symmetry as the midpoint of the two roots, and one for the turning point written as a coordinate pair
- halving the coefficient of x correctly, including its sign
- the completed square arrangement, even if the turning point is then read off wrongly
- the turning point as a coordinate pair, with the sign inside the bracket reversed
- the least value stated as the constant outside the bracket when a minimum is asked for
- the correct overall shape, a straight line or a parabola opening the right way
- each axis crossing marked and labelled with its value
- a turning point in a sensible place on a quadratic sketch
- a smooth curve rather than a set of joined straight segments
- a table of values with negative inputs handled correctly, especially cubes of negative numbers
- all points plotted correctly within the tolerance of the grid
- a single smooth curve, with a reciprocal graph drawn as two separate branches
- identifying which sketch belongs to a cubic and which to a reciprocal in a matching question
- an exponential curve through (0, 1) with the correct growth or decay direction
- an exponential curve that approaches the horizontal axis without crossing it
- a sine or cosine curve of the right height and period, starting at the right place
- asymptotes drawn in the right places on a tangent curve, and one for using symmetry to give a further solution
- moving the curve in the correct direction for the transformation given
- a transformed curve that keeps the shape and width of the original
- the image of a named point given as a coordinate pair
- a reflection in the correct axis, even if a labelled point is then misread
- Plotting points accurately from a table and drawing a suitable line or curve through them.
- Correctly labelled axes with appropriate, linear scales and units.
- Calculating the gradient correctly by dividing the change in the vertical axis by the change in the horizontal axis.
- Estimating the area under a speed-time graph by splitting it into appropriate shapes (e.g., trapeziums) and summing the areas.
- A final answer with the correct units derived from the context, e.g., m/s for speed or m for distance.
- recognising the shape as a falling curve that levels off, not a straight line
- a correct calculation of the missing quantity, such as dividing a distance by a speed
- a reading taken from the curve at the stated position on the horizontal axis
- describing the flattening of the curve in the language of the context
- identifying the constant multiplier from the context, such as 1.05 for a rise of 5% a year
- raising that multiplier to the correct number of time steps
- a smooth curve that steepens for growth or flattens for decay
- a reading taken from the curve when a doubling time or half-life is asked for
- a tangent ruled so that it touches the curve at the point named
- a correct rise and run taken from that tangent, even if the division then goes wrong
- splitting the region into named shapes, or for stating what one square is worth
- the parts of the area added together, and one for interpreting the result with the correct units
- the equation of the circle written with the square of the radius on the right
- the gradient of the radius from the origin to the point of contact
- the negative reciprocal taken as the gradient of the tangent, even if the final equation is wrong
- substituting the point of contact to find the intercept, and one for the tangent equation
Examiner Tips
- 💡before you plot anything, work out what one square is worth on each axis, as the two scales can differ
- 💡mark a point with a small neat cross rather than a blob, so the exact position can be seen
- 💡if a shape is involved, plot every vertex before joining anything up
- 💡work out one extra value beyond those asked for; if it does not sit on the same straight line, one of your points is wrong
- 💡once y is on its own, check that your line crosses the vertical axis at the constant term, so y = 3x − 2 must cross at −2; for 2y + x = 6 the crossing is at 3, not 6
- 💡rearrange first if y is not already on its own, since substituting into a mixed-up equation invites errors
- 💡If a line is parallel to ax + by = d, first rearrange the equation into the form y = mx + c to find the gradient m correctly.
- 💡After finding the equation of a line passing through two points, substitute the coordinates of the *second* point into your final equation as a check. If the equation holds true, your answer is very likely correct.
- 💡multiply your two gradients together as a check; the result must be −1
- 💡get the given line into the arrangement with y on its own before you read any gradient from it
- 💡leave a fractional gradient as a fraction, since a rounded decimal such as 0.33 loses accuracy in the intercept
- 💡pick two points far apart on the line, because a small triangle magnifies any reading error
- 💡answer an interpretation question in a sentence that names the quantity and its unit
- 💡if the equation is not arranged with y on its own, rearrange it before reading the gradient and intercept
- 💡check a root by substituting it back into the expression; the result should be zero
- 💡the line of symmetry sits at the mean of the two roots, and the turning point shares that same horizontal value
- 💡read a turning point off a graph as a coordinate pair, using the scale on both axes
- 💡expand your completed square in your head to check it gives back the original expression
- 💡the horizontal value of the turning point is whatever makes the bracket zero, and its height is the constant outside
- 💡if a question asks for the least value of the expression, that is the constant alone, not the whole coordinate pair
- 💡label every axis crossing with its number, since a sketch earns marks for the labels as much as for the shape
- 💡read the sign of the squared term before you draw a single stroke
- 💡keep a sketch large, so the turning point is clearly away from the axes and can be seen
- 💡for a reciprocal graph, add extra inputs close to zero such as 0.5 and −0.5, to show how steep the curve becomes
- 💡put brackets round a negative number before cubing it on the calculator
- 💡lift your pencil between the two branches of a reciprocal graph rather than drawing straight through
- 💡mark the horizontal axis in steps of 90° before sketching a trigonometric curve, so the peaks and zeros land correctly
- 💡for an exponential curve, work out the height at zero and at one to fix its position before drawing
- 💡use the symmetry of the sine and cosine curves to find the second solution in a given interval
- 💡track the turning point or another labelled point through the transformation and plot that first
- 💡a change outside the function moves the curve up or down as you would expect; a change inside the bracket moves it sideways the opposite way
- 💡use the original curve as a template so the new one keeps exactly the same width
- 💡When plotting, choose a scale that uses at least half of the available grid. This makes your graph clearer and your readings more accurate. Always label axes with the quantity and its units.
- 💡For interpretation questions, show your working on the graph itself. Draw lines to show how you read a value or a triangle to show how you calculated a gradient. This can earn method marks.
- 💡where the two quantities have a constant product, such as speed and time over a fixed distance, multiply your reading back to check it
- 💡plot extra points near the left-hand end, where the curve changes fastest
- 💡give the units, and say that the answer is an estimate when it came from a reading
- 💡write the multiplier down before any powers, and check it is above one for growth and below one for decay
- 💡for a half-life, mark half the starting height on the vertical axis and work across to the curve first
- 💡keep the time steps and the units straight; a monthly rate needs the number of months in the power
- 💡let the tangent run well past the point of contact, so the triangle you build on it is large and easy to read
- 💡an estimate is expected here, so a value inside the tolerance band scores; do not chase the last decimal place
- 💡multiply the units of the two axes for an area, and divide them for a gradient
- 💡the radius always runs from the origin to the point of contact, so its gradient is the vertical coordinate over the horizontal one
- 💡check that the point of contact satisfies both the circle equation and your tangent equation
- 💡keep gradients as fractions, because a rounded decimal makes the intercept messy
- 💡Always label your axes and use a suitable scale; marks are awarded for correctly labelled axes and accurate plotting.
- 💡When drawing curves, use a sharp pencil and draw a smooth curve through the points; do not use a ruler for curves.
- 💡For graphical solutions, clearly state the coordinates of intersection points and check your answer by substitution.
Common Mistakes
- writing a point the wrong way round, so (2, 5) is plotted where (5, 2) belongs
- dropping a minus sign in the bottom left region and plotting (−4, −1) as (4, 1)
- assuming each square is one unit when an axis is marked in twos, fives or tens
- counting from the nearest labelled gridline rather than from the origin
- joining the points with short segments so the graph bends slightly instead of being one ruled line
- losing a sign when a negative input goes into a term such as −4x, so one point sits out of line
- stopping the line at the last plotted point when the grid carries on further
- reading the table row for x and plotting it as the vertical value
- Inverting the gradient calculation to be (change in x) / (change in y). Correction: The gradient is 'rise over run', so it must be calculated as (change in y) / (change in x).
- Making sign errors when calculating the gradient with negative coordinates. Correction: Use brackets and be careful with subtractions, remembering that a − (−b) is a + b.
- Finding the gradient correctly but then stopping, or forgetting to find the specific y-intercept c. Correction: After finding m, you must substitute a point's coordinates into y = mx + c to find the value of c for that specific line.
- changing the sign without inverting, so a gradient of 2 becomes −2 instead of −½
- inverting without changing the sign, so a gradient of 2 becomes ½
- reading the gradient off an equation such as 2y = 6x + 4 as 6 rather than 3, because it was not rearranged first
- treating perpendicular like parallel and copying the gradient across unchanged
- dividing the run by the rise, which gives the reciprocal of the gradient
- quoting a positive gradient for a line that slopes downwards
- counting squares rather than units, so a rise of five squares on an axis marked in tens is recorded as five
- giving the value where the line crosses the horizontal axis as the intercept
- taking the roots straight from the brackets, so (x − 4)(x + 1) is read as roots of −4 and 1 rather than 4 and −1
- giving a turning point as a single number instead of a pair of coordinates
- quoting the point where the curve meets the vertical axis as a root
- joining plotted points with straight segments, so the curve has a flat base instead of a smooth lowest point
- keeping the sign as it appears inside the bracket, so (x − 3)² − 4 is read as a turning point at (−3, −4)
- forgetting to subtract the square of the halved coefficient, leaving the constant unchanged
- not factoring out the coefficient of the squared term first, so the halving is done on the wrong number
- stopping at the completed square when the question asked for the coordinates of the turning point
- building a full table of values when a sketch was asked for, and running short of time
- drawing a parabola with a flat base or with its two arms at different heights
- sketching a curve opening upwards when the squared term is negative
- leaving a sketch bare, so there is nothing labelled for the crossings and roots to be credited on
- cubing a negative number and writing a positive result, so (−2)³ is given as 8 rather than −8
- joining the two branches of a reciprocal graph through the origin
- drawing a reciprocal curve so that it touches or crosses an axis
- using too few points, so the flat part of a cubic near the origin comes out as a sharp corner
- drawing an exponential curve that cuts the horizontal axis and continues below it
- swapping sine and cosine, so the sine curve is drawn starting at its highest point
- giving only the value the calculator returns when the curve shows more solutions in the range asked for
- using a period of 360° for the tangent curve instead of 180°
- shifting y = f(x + 4) four to the right instead of four to the left
- mixing up y = −f(x) and y = f(−x), so the reflection lands in the wrong axis
- stretching or squashing the curve while moving it, so its shape changes
- moving the axes rather than the curve
- Using a non-linear scale (e.g., 0, 1, 2, 4, 8) or joining points with a ruler when a smooth curve is required.
- Confusing the meaning of the gradient and the area under the graph, especially for speed-time graphs.
- Interpreting a horizontal line on a distance-time graph as constant speed instead of stationary.
- treating the relationship as a straight line, so halving one quantity is assumed to halve the other
- drawing the curve so that it touches or crosses the horizontal axis
- taking a careless reading from the steep left-hand part, giving an answer well outside the tolerance
- trying to use the value at zero, where the relationship has no meaning
- adding the same amount each step instead of multiplying, so the graph is drawn as a straight line
- using 0.15 rather than 0.85 as the multiplier for a loss of 15% a year
- reading the steepness across a wide interval and calling it the rate at a single instant
- expecting a decay curve to reach zero and drawing it meeting the axis
- ruling a line that cuts the curve at two points and calling it a tangent, which gives an average rate instead
- counting squares without working out what one square represents on the two scales
- leaving an area as a number of squares rather than converting it into a distance or an amount
- quoting an acceleration in the units of speed, or an area in the units of one axis alone
- writing the radius itself on the right of the equation instead of its square
- giving the radius as the number on the right without taking its square root
- using the gradient of the radius as the gradient of the tangent
- inverting the gradient but forgetting the sign, so 4/3 becomes 3/4 rather than −3/4
- Students often think that a quadratic graph is a straight line or that it can be drawn with straight segments; it must be a smooth curve.
- Students may confuse the gradient and y-intercept when identifying the equation of a straight line; remember y = mx + c, where m is gradient and c is y-intercept.
- Students sometimes believe that reciprocal graphs cross the axes; they do not, as they have asymptotes at x = 0 and y = 0.
Revision Plan
- 1Week 1: Revise plotting linear graphs using tables of values and identifying gradient and y-intercept. Practice drawing straight lines from equations.
- 2Week 1: Move on to quadratic graphs; practice completing tables of values, plotting points, and drawing smooth curves. Identify roots and turning points.
- 3Week 2: Study cubic and reciprocal graphs; focus on their shapes and key features. Practice sketching them from equations.
- 4Week 2: Learn to solve simultaneous equations graphically and use graphs to solve quadratic equations. Attempt past paper questions.
- 5Week 2: Review exponential graphs and transformations of graphs if studying Higher tier. Complete a mixed topic test.
Exam Question Types
- 📋Plotting a graph from a table of values: Ensure you plot points accurately and join with a smooth curve for non-linear graphs.
- 📋Solving simultaneous equations graphically: Draw both lines and read off the intersection point; give both coordinates.
- 📋Using a graph to solve an equation: Find the x-intercepts for equations of the form f(x) = 0, or draw a new line for f(x) = k.
- 📋Identifying key features of a graph: State roots, intercepts, turning points, and asymptotes as required.
Command Word Expectations (AQA)
Draw points accurately on a grid and join them appropriately (straight line for linear, smooth curve for non-linear).
Draw a rough graph showing the correct shape and key features, without needing exact plotting.
Find the value(s) that satisfy the equation, often by reading from a graph or using algebraic methods.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: Complete the table of values for y = x^2 - 2x - 3 for x = -2 to 3, then plot the graph and use it to solve x^2 - 2x - 3 = 0.
- 1.Step 1: Substitute each x value into y = x^2 - 2x - 3. For x = -2: y = 4 + 4 - 3 = 5; x = -1: y = 1 + 2 - 3 = 0; x = 0: y = -3; x = 1: y = 1 - 2 - 3 = -4; x = 2: y = 4 - 4 - 3 = -3; x = 3: y = 9 - 6 - 3 = 0.
- 2.Step 2: Plot the points (-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0) on a grid and join with a smooth curve.
- 3.Step 3: The solutions to x^2 - 2x - 3 = 0 are the x-intercepts where y = 0. From the graph, the curve crosses the x-axis at x = -1 and x = 3.
Question: Solve the simultaneous equations y = 2x + 1 and y = -x + 4 graphically.
- 1.Step 1: Create a table of values for each equation. For y = 2x + 1: when x = 0, y = 1; when x = 2, y = 5. For y = -x + 4: when x = 0, y = 4; when x = 3, y = 1.
- 2.Step 2: Plot both straight lines on the same axes using the points from the tables.
- 3.Step 3: Identify the point where the two lines intersect. The lines cross at (1, 3). Check: 2(1) + 1 = 3 and -(1) + 4 = 3.