Flashcards · 0580 · Topic 2

Algebra and graphs flashcards

81 cards on Topic 2, subtopics 2.1 to 2.13. Pick a subtopic or study them all.

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2.1 Introduction to algebra

Letters as generalised numbers: if n is any integer, what can you say about 2n and 2n + 1?: 2n is always even and 2n + 1 is always odd, whatever integer n is.
A pen costs p cents and a ruler costs r cents. Write an expression for the cost of 3 pens and 2 rulers.: 3p + 2r cents
How do you substitute a negative number into a formula?: Put the negative number in brackets, then follow the order of operations. v = u + at with u = 12, a = −3, t = 5: v = 12 + (−3)(5) = −3.
Work out 2x² when x = −3: 2 × (−3)² = 2 × 9 = 18. Square first, then multiply (not (2 × −3)² = 36).

2.2 Algebraic manipulation

Simplify 5x + 3y − 2x + y: Collect like terms: 3x + 4y
Expand and simplify (x + 3)(x − 5): x² − 5x + 3x − 15 = x² − 2x − 15
Expand (a + b)²: a² + 2ab + b². Not a² + b²: the middle term 2ab is the common slip.
Factorise fully 6x² + 9x: 3x(2x + 3). "Factorise" means take out the highest common factor of the numbers and the letters.
Difference of two squares: factorise 2x² − 18Extended: Take out the common factor first: 2(x² − 9) = 2(x + 3)(x − 3). a² − b² = (a + b)(a − b)
Factorise x² + 5x + 6 and 2x² + 7x + 3Extended: x² + 5x + 6 = (x + 2)(x + 3) 2x² + 7x + 3 = (2x + 1)(x + 3)
Factorise by grouping: xy + 2x + 3y + 6Extended: x(y + 2) + 3(y + 2) = (x + 3)(y + 2)
Complete the square for x² + 6x + 5Extended: Halve the coefficient of x: (x + 3)² − 9 + 5 = (x + 3)² − 4

2.3 Algebraic fractions

Simplify (x² − 9)/(x² + 3x)Extended: Factorise top and bottom: (x + 3)(x − 3) / x(x + 3) Cancel the common factor (x + 3): (x − 3)/x
Exam trap: cancelling the x² in (x² + 3)/(x² − 1)Extended: Only whole factors cancel, not single terms. This fraction does not simplify by cancelling x².
Write 2/x + 3/(x + 1) as a single fractionExtended: Common denominator x(x + 1): (2(x + 1) + 3x) / x(x + 1) = (5x + 2) / x(x + 1)
Simplify (x + 1)/2 − (x − 3)/3Extended: (3(x + 1) − 2(x − 3))/6 = (3x + 3 − 2x + 6)/6 = (x + 9)/6 Keep the bracket round the subtracted numerator: −2(x − 3) = −2x + 6.
Simplify x/3 ÷ x²/6Extended: Multiply by the reciprocal: x/3 × 6/x² = 6x/(3x²) = 2/x

2.4 Indices II

Simplify (2x³)⁴: The power applies to every factor, numbers included: 2⁴x¹² = 16x¹²
Simplify 6x⁵ ÷ 2x²: Divide the numbers and subtract the indices: 3x³
Solve 2ˣ = 32: 32 = 2⁵, so x = 5.
Write x⁻² and 5x⁰ without negative or zero indices: x⁻² = 1/x² 5x⁰ = 5 × 1 = 5
Simplify (8x⁶)^(1/3)Extended: ∛8 × (x⁶)^(1/3) = 2x²
Solve 9ˣ⁺¹ = 27ˣExtended: Write both sides as powers of 3: 3^(2x + 2) = 3^(3x) 2x + 2 = 3x, so x = 2. Set the indices equal only once the bases match.

2.5 Equations

Solve 5x + 3 = 2x + 15: 3x + 3 = 15, so 3x = 12, x = 4. Do the same operation to both sides.
Solve the simultaneous equations 2x + y = 7 and x − y = 2: Add to eliminate y: 3x = 9, so x = 3. Substitute: 2(3) + y = 7, so y = 1.
Elimination: when do you add the equations and when do you subtract?: Once the coefficients of one letter match in size: if their signs are the same, subtract; if their signs are different, add.
Make t the subject of v = u + at: v − u = at, so t = (v − u)/a
The quadratic formulaExtended: For ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a)
Solve x² − 4x + 1 = 0, giving exact answersExtended: x = (4 ± √(16 − 4)) / 2 = (4 ± √12)/2 = 2 ± √3 (x ≈ 3.73 or 0.268)
Solve y = x + 1 and y = x² − 1 simultaneouslyExtended: Substitute the linear equation: x² − 1 = x + 1, so x² − x − 2 = 0, (x − 2)(x + 1) = 0. x = 2, y = 3 or x = −1, y = 0.
Solve x/3 + x/4 = 7Extended: Multiply every term by 12: 4x + 3x = 84, so 7x = 84 and x = 12.
Make x the subject of ax + 3 = bx + 7 (the subject appears twice)Extended: Collect the x terms on one side: ax − bx = 4 Factorise: x(a − b) = 4 Divide: x = 4/(a − b)

2.6 Inequalities

Number line: open circle or filled circle?: Open circle (○) for < or >: the end value is not included. Filled circle (●) for ≤ or ≥: the end value is included.
List the integers n with −2 < n ≤ 3: −1, 0, 1, 2, 3 (−2 is not included; 3 is).
Solve 4 − 2x ≥ 10Extended: −2x ≥ 6. Dividing by a negative number reverses the sign: x ≤ −3.
Solve −5 < 2x + 3 ≤ 11Extended: Subtract 3 from every part: −8 < 2x ≤ 8 Divide every part by 2: −4 < x ≤ 4
Regions for inequalities in two variables: line style and shadingExtended: Draw the boundary solid for ≤ or ≥ and broken (dashed) for < or >. Shade the UNWANTED side unless the question says otherwise. Test a point such as (0, 0) to choose the side.

2.7 Sequences

Term-to-term rule and nth term: Term-to-term rule: how to get from one term to the next (e.g. add 3). nth term: a formula for any term from its position n (e.g. 3n + 2).
Find the nth term of 5, 8, 11, 14, …: The difference is 3, so 3n + b. When n = 1: 3 + b = 5, so b = 2. nth term = 3n + 2
Find the nth term of 20, 17, 14, 11, …: The difference is −3, so −3n + b. When n = 1: −3 + b = 20, b = 23. nth term = 23 − 3n
How do the differences tell you the type of sequence?: Constant 1st difference: linear (an + b). Constant 2nd difference: quadratic (an² + bn + c). Constant 3rd difference: cubic.
Find the nth term of 2, 5, 10, 17, …: These are one more than the square numbers 1, 4, 9, 16: nth term = n² + 1.
Find the nth term of the quadratic sequence 3, 8, 15, 24, …Extended: 2nd difference 2, so the n² coefficient is 2 ÷ 2 = 1. Subtract n²: 2, 4, 6, 8, which is 2n. nth term = n² + 2n
Find the nth term of the exponential sequence 3, 6, 12, 24, …Extended: Each term is double the one before: nth term = 3 × 2ⁿ⁻¹

2.8 Proportion

y ∝ x and y ∝ 1/x as equationsExtended: Direct: y = kx Inverse: y = k/x k is the constant of proportionality. The same pattern gives y = kx², y = k√x, y = k/x², and so on.
How do you solve a proportion question?Extended: 1. Write the equation with k (e.g. y = kx²). 2. Substitute the known pair of values to find k. 3. Rewrite the equation with that k. 4. Use it to find the unknown.
y is proportional to x². y = 12 when x = 2. Find y when x = 5.Extended: 12 = k × 4, so k = 3 and y = 3x². x = 5: y = 3 × 25 = 75
y is inversely proportional to x². y = 5 when x = 4. Find y when x = 2.Extended: 5 = k/16, so k = 80 and y = 80/x². x = 2: y = 80/4 = 20
y = kx²: x is multiplied by 3. y = k/x³: x is doubled. What happens to y?Extended: y = kx²: y is multiplied by 3² = 9. y = k/x³: y is divided by 2³ = 8.

2.9 Graphs in practical situations

Distance–time graph: what do the gradient, a horizontal section and a downward slope mean?: Gradient: speed. Horizontal: stationary. Sloping down: travelling back towards the start.
Gradient of a straight-line graph in a practical situation: The rate of change of the vertical quantity with the horizontal one, e.g. cost per item, litres per minute, km per hour.
A van travels 36 km in 45 minutes. Find its speed in km/h.: 45 min = 0.75 h, so 36 ÷ 0.75 = 48 km/h (not 36 ÷ 45 = 0.8).
Conversion graph trap: a line that does not pass through the origin: You cannot scale up from one reading (e.g. a fixed charge plus a rate). Read each value from the graph, or use the equation of the line.
Speed–time graph: what do the gradient, the area under it and a horizontal section mean?Extended: Gradient: acceleration (a negative gradient is a deceleration). Area under the graph: distance travelled. Horizontal: constant speed (not stopped).
An object accelerates uniformly from rest to 12 m/s in 4 s. Find the acceleration and the distance travelled.Extended: Acceleration = 12 ÷ 4 = 3 m/s² Distance = area of triangle = ½ × 4 × 12 = 24 m
How do you estimate the rate of change at a point on a curved graph?Extended: Draw the tangent to the curve at that point, then find the gradient of the tangent using two points on the tangent (not on the curve).

2.10 Graphs of functions

Shapes of y = ax + b, y = x² + ax + b and y = −x² + ax + b: y = ax + b: a straight line. y = x² + ax + b: a U-shaped parabola (minimum). y = −x² + ax + b: an n-shaped parabola (maximum). Both parabolas are symmetrical.
Shape of the graph of y = a/x: Two separate branches that never touch the axes. For a > 0 they lie in the first and third quadrants; for a < 0, in the second and fourth. There is no point at x = 0.
Drawing a curve from a table of values: two common slips: Squaring a negative wrongly: (−3)² = 9, not −9 (a wrong value breaks the symmetry). Joining points with a ruler: draw a smooth curve.
How do you solve x² − 2x − 3 = 0 from the graph of y = x² − 2x − 3?: Read the x-coordinates where the curve crosses the x-axis (y = 0): x = −1 and x = 3.
How do you solve x² − 2x − 3 = x + 1 using the graph of y = x² − 2x − 3?: Draw the line y = x + 1 on the same axes and read the x-coordinates of the points where it meets the curve.
Which graphs does Extended add in 2.10?Extended: y = axⁿ for n = −2, −1, −½, 0, ½, 1, 2, 3 (and sums of up to three such terms), and exponential graphs y = abˣ + c.
Features of an exponential decay graphExtended: The value falls by the same factor in each equal time interval (a constant halving time), and the curve never reaches 0.

2.11 Sketching curves

Sketch y = x² − 2x − 8: what must you show?: U-shape (positive x²). Roots: (x − 4)(x + 2) = 0, so x = −2 and x = 4. y-intercept −8. Line of symmetry halfway between the roots, x = 1. Label the intercepts.
How do you sketch a straight line such as 2x + 5y = 10?Extended: Find the intercepts: x = 0 gives y = 2; y = 0 gives x = 5. Join (0, 2) and (5, 0) and extend the line.
Turning point of y = (x + p)² + q, and of y = (x + 1)² − 16Extended: y = (x + p)² + q has its turning point at (−p, q). y = (x + 1)² − 16 has its turning point at (−1, −16), not (1, −16).
Asymptotes of y = a/x + bExtended: Vertical asymptote x = 0 and horizontal asymptote y = b. The two branches never cross or touch them.
Key features of y = arˣ + bExtended: Horizontal asymptote y = b. Crosses the y-axis at (0, a + b). Growth if r > 1, decay if 0 < r < 1.
Sketching y = ax³ + bx² + cx: what do you know about the roots?Extended: Factorise as x(ax² + bx + c), so x = 0 is always a root. There can be up to three roots and two turning points.

2.12 Differentiation

Derivative of y = axⁿExtended: dy/dx = anxⁿ⁻¹ Multiply by the power, then reduce the power by 1. A constant term differentiates to 0.
Differentiate y = 4x² − 3x + 7Extended: dy/dx = 8x − 3
Find the gradient of y = x³ − 12x at x = 1Extended: dy/dx = 3x² − 12. At x = 1: 3 − 12 = −9.
How do you find the turning points of a curve?Extended: 1. Find dy/dx. 2. Solve dy/dx = 0 for x. 3. Substitute each x into the equation of the curve to find y.
Find the turning points of y = x³ − 12x and say which is a maximumExtended: 3x² − 12 = 0, so x = ±2. Points (2, −16) and (−2, 16). d²y/dx² = 6x: at x = 2 it is 12 > 0 (minimum); at x = −2 it is −12 < 0 (maximum).
Ways to tell a maximum from a minimumExtended: Second derivative: d²y/dx² > 0 minimum, < 0 maximum. Or the sign of dy/dx either side: + then − is a maximum; − then + is a minimum. Or an accurate sketch.

2.13 Functions

Domain and range of a functionExtended: Domain: the set of inputs the function is allowed to take. Range: the set of outputs it produces from that domain.
g(x) = 3x² + 1 and k(x) = 5/(x − 4). State the range of g and a value not in the domain of k.Extended: x² is never negative, so the range of g is g(x) ≥ 1. x = 4 is not in the domain of k (division by zero).
f(x) = x + 2 and g(x) = 3x. Find gf(x) and fg(x).Extended: gf(x): f first, then g: 3(x + 2) = 3x + 6 fg(x): g first, then f: 3x + 2
How do you find an inverse function f⁻¹(x)?Extended: 1. Write y = f(x). 2. Swap x and y. 3. Rearrange to make y the subject. That expression is f⁻¹(x).
Find f⁻¹(x) when f(x) = (2x + 1)/3Extended: x = (2y + 1)/3, so 3x = 2y + 1 and y = (3x − 1)/2. f⁻¹(x) = (3x − 1)/2
Exam trap: f⁻¹(x) and 1/f(x)Extended: They are different. f⁻¹(x) undoes f; 1/f(x) is the reciprocal of the output.