Printable · GCSE Foundation · ages 14-16
Recognising and sketching graphs worksheet — GCSE Foundation
Fifteen questions on "recognising and sketching graphs" — DfE statement A12. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Recognising and sketching graphs worksheet — GCSE Foundation
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- 1.A graph has equation y = 3x + 2. Which word or phrase best describes its shape?y = 3x + 2
- 2.A table of values is being drawn for the graph of y = x³. Work out the value of y when x = −2.y = x
- 3.The diagram shows two curves drawn on the same grid. Curve A is the graph of . Which equation could represent curve B?
- 4.Which of these equations gives a straight-line graph when plotted?
- 5.The diagram shows a straight line passing through the origin, drawn on a numbered grid. Which of these points lies on the line?
- 6.The diagram shows the graph of a straight line, drawn on a grid numbered from −5 to 5 on both axes. Which equation could represent the line?
- 7.Which of these equations gives a graph shaped like a U, symmetrical about a vertical line?
- 8.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
- 9.The diagram shows the graph of a quadratic function. Write down the values of x where the graph crosses the x-axis.
- 10.The diagram shows the graph of , together with the horizontal line . Use the graphs to estimate, to 1 decimal place, the two solutions of .
- 11.The diagram shows the graph of a function. Which equation could it represent?
- 12.Which of these equations gives a graph with two separate curved branches — one where x and y are both positive, and one where x and y are both negative?
- 13.A graph has equation y = −2x² + 5. Which statement about its shape is correct?y = -2x² + 5
- 14.The diagram shows the graph of a quadratic function. Which equation could it represent?
- 15.The diagram shows part of the graph of a reciprocal function of the form , passing through the labelled point. Work out the value of k.
Answer key
- (b) A straight line. — y = 3x + 2 is a linear function, since the highest power of x is 1, so its graph is a straight line with gradient 3 and y-intercept 2. Saying it is a U-shaped curve confuses a linear graph with a quadratic graph, which has an x² term. Saying it decreases then increases describes a curve with a turning point, which a straight line does not have. Saying it gets closer to an axis but never reaches it describes a reciprocal graph, y = k/x, not a linear one.
- (c) −8 — (−2)³ = (−2) × (−2) × (−2) = −8, since multiplying three negative numbers gives a negative result. A candidate who forgets the sign of a negative number when cubing it might treat (−2)³ as if it were 2³ = 8. A candidate who multiplies −2 by 3 instead of cubing it might get −2 × 3 = −6. A candidate who combines both mistakes — multiplying by 3 and dropping the sign — might get 2 × 3 = 6.
- (d) $y = x^2 + 3$ — Method: a point lies on a curve only if substituting its x-coordinate into the equation gives back its y-coordinate, so read one or two points off curve B and test each equation. Working: curve B crosses the y-axis at (0, 3), and its lowest point is also (0, 3); substituting x = 0 into y = x² + 3 gives 0² + 3 = 3, which matches. Checking a second point: at x = 2 curve B is at y = 7, and 2² + 3 = 4 + 3 = 7, which matches as well. Answer: curve B has equation y = x² + 3. Distractor refutation: y = x² − 3 comes from reading the 3 as a move down instead of a move up; substituting x = 0 gives −3, so that curve would cross the y-axis three squares below the origin, while curve B crosses it three squares above. y = (x − 3)² comes from putting the 3 inside the brackets; substituting x = 0 gives (−3)² = 9, and that curve's lowest point is at (3, 0), three squares to the right along the x-axis, whereas curve B has its lowest point on the y-axis. y = x² + 3x comes from attaching the 3 to the x term instead of writing it on its own; substituting x = 0 gives 0² + 3 × 0 = 0, so that curve passes through the origin, and curve B does not pass through the origin.
- (b) y = 2x − 3 — A straight-line graph has an equation of the form y = mx + c, where x appears only to the power 1. Of the four equations, y = 2x − 3 fits this form, so it gives a straight line. y = x² − 3 has x squared, so it gives a curved (quadratic) graph, not a straight line. y = x³ − 3 has x cubed, so it gives a cubic curve. y = 3/x has x on the bottom of a fraction, so it gives a reciprocal curve with two branches.
- (c) (2, 6) — Method: substitute the x-coordinate of each point into the rule for the line (y = 3 × x) and compare it with the point's y-coordinate. Working: the line passes through the origin and rises 3 squares for every 1 square across, so at x = 2 the line's y-value is 3 × 2 = 6, giving the point (2, 6). Answer: (2, 6). Distractor refutation: (2, 3) comes from counting only 3 squares up in total between the origin and x = 2, instead of 3 squares up for every 1 square across, halving the true rise. (3, 2) comes from swapping the x-coordinate and the y-coordinate round. (2, 5) comes from a miscounted gridline, landing one square below the line.
- (d) $y = 2x - 1$ — Method: pick two points the line passes through, work out the gradient as vertical change ÷ horizontal change, and read the y-intercept from where the line crosses the y-axis. Working: the line passes through (0, −1) and (1, 1), so the gradient is (1 − (−1)) ÷ (1 − 0) = 2, and it crosses the y-axis at −1. Answer: y = 2x − 1. Distractor refutation: y = 2x + 1 comes from reading the y-intercept as +1 instead of −1, misreading which side of the origin the line crosses. y = x − 1 comes from taking the gradient as 1, counting the same number of squares across and up instead of checking the rise is twice the run. y = −2x − 1 comes from a sign error on the gradient, treating the line as sloping downward from left to right when it actually rises.
- (a) y = x² + 2 — A quadratic equation, with x², gives a graph shaped like a symmetrical U (or an upside-down U). Of the four equations, y = x² + 2 is quadratic, so it gives this U-shape. y = x³ + 2 is cubic, giving an S-shaped curve rather than a symmetrical U. y = 2x + 2 is linear, giving a straight line. y = 2/x is reciprocal, giving two separate curved branches rather than a single U-shape.
- (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
- (a) −2 and 2 — Method: the graph crosses the x-axis where y = 0, so read the two crossing points straight off the curve. Working: the curve meets the horizontal axis two squares to the left of the origin and two squares to the right, at x = −2 and x = 2. Answer: −2 and 2. Distractor refutation: −4 and 4 comes from reading the value where the curve crosses the y-axis, −4, and using that value and its positive partner as the x-axis crossings instead. −2 and 3 comes from misreading the right-hand crossing point one square out along the grid, taking it where the curve is already above the axis. 0 and 2 comes from confusing the lowest point of the curve, which sits on the y-axis at x = 0, with one of the crossing points.
- (a) x = −1.8 and x = 2.8 — Method: the solutions of x² − x − 2 = 3 are the x-coordinates of the points where the curve y = x² − x − 2 meets the line y = 3, read off the grid to 1 decimal place. Working: the curve meets the line y = 3 at approximately x = −1.8 and at x = 2.8. Answer: x ≈ −1.8 and x ≈ 2.8. Distractor refutation: x = −1.0 and x = 2.0 comes from reading where the curve crosses the x-axis (y = 0), solving x² − x − 2 = 0, instead of where it meets the line y = 3. x = −3.0 and x = 4.0 comes from taking the two ends of the drawn curve as the intersection points, instead of finding where it actually crosses the line y = 3. x = 1.8 and x = 2.8 comes from a sign slip on the left-hand intersection, reading it as positive instead of negative.
- (b) $y = x^3 - 4x$ — Method: count how many times the curve crosses the x-axis and check whether it is a cubic (an S-shape with up to three crossing points) rather than a lower power, and note which way it runs overall from bottom-left to top-right or the reverse. Working: the curve crosses the x-axis at three points, x = −2, 0 and 2, and runs from bottom-left to top-right, which matches y = x³ − 4x = x(x − 2)(x + 2). Answer: y = x³ − 4x. Distractor refutation: y = −x³ + 4x comes from a sign error on every term, which would flip the curve so it ran from top-left to bottom-right instead. y = x³ + 4x comes from a sign error on the x term only, which removes two of the three crossing points, since x(x² + 4) has only x = 0 as a real root. y = x² − 4x comes from dropping the cubic term altogether, giving a parabola with only two crossing points instead of an S-shaped curve with three.
- (b) y = 1/x — A graph with two separate curved branches must have a value of x that is excluded from it, creating a break in the curve. For y = 1/x, x = 0 is excluded, so the graph splits into one branch where x and y are both positive, and one where x and y are both negative. y = x² is a single U-shaped curve with no break, even though it has two 'arms'. y = x³ is one continuous curve that passes through the origin without any break. y = 2x + 1 is a single straight line; even though it passes through regions where x and y share the same sign, it is one continuous line, not two separate branches.
- (d) It is n-shaped, since the x² coefficient is negative. — The coefficient of x² is −2, which is negative, so the quadratic curve opens downward — shaped like an n, with a maximum turning point. Saying it is U-shaped focuses only on x² being non-negative and ignores that the −2 in front of it flips the whole curve to open downward. Saying it is a straight line confuses having a constant term with being linear — any equation with an x² term is a curve, not a line. Saying it repeatedly rises and falls like a wave describes a trigonometric graph such as y = sin x, not a quadratic.
- (d) $y = x^2 - 2x - 3$ — Method: read the two x-intercepts (roots) from the graph, write the quadratic as the product of the corresponding factors, then expand. Working: the curve crosses the x-axis at x = −1 and x = 3, so the equation factorises as (x + 1)(x − 3), which expands to x² − 2x − 3. Answer: y = x² − 2x − 3. Distractor refutation: y = x² − x − 6 comes from misreading the left-hand crossing point as x = −2 instead of x = −1, giving factors (x + 2)(x − 3). y = x² − x − 2 comes from misreading the right-hand crossing point as x = 2 instead of x = 3, giving factors (x + 1)(x − 2). y = x² + 2x − 3 comes from writing the factors as (x − 1)(x + 3), swapping which root gets the plus sign and which gets the minus sign, giving the wrong sign on the x term.
- (b) 6 — Method: for any point that lies on y = k/x, the value of k is found by multiplying the x-coordinate and the y-coordinate together, since k = x × y. Working: k = 2 × 3 = 6. Answer: k = 6. Distractor refutation: 1.5 comes from dividing the y-coordinate by the x-coordinate instead of multiplying them. 5 comes from adding the two coordinates instead of multiplying them. 9 comes from misreading the point's x-coordinate as 3 instead of 2, then multiplying 3 × 3.
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