(Y1 ONLY) algebra and functions whole topic

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65 Terms

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polynomials

finite expressions with positive whole number indices

polynomials: 5, 2x+3, 8xy-x+6

not polynomials: √5, x^-2, 4/x, 5/3

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dividing polynomials

the use of long division to divide polynomials by (x±k)

-ensure the polynomial is arranged in descending order of degree of power

-divide the first term of the polynomial by x, append this to the solution

-multiply the result by (x±k)

-subtract the result from the original polynomial

-repeat this until the original polynomial is reduced to 0 or an integer remainder which is appended to the solution

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expanding brackets

when mutiple polynomials in brackets are multiplied together, each term must be multiplied by each term in the other bracket(s)

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factor theorem

for a polynomial f(x), if f(p) = 0, then (x-p) is a factor of f(x)

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factorising polynomials

you can factorise a cubic polynomial by finding p where f(p) = 0 and dividing f(x) by (x-p)

the result of this should be factorised further if possible

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quadratic equation

ax^2 + bx + c = 0

a, b and c are real constants and a != 0

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quadratic roots

the solutions to the quadratic equation

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discriminant

b^2 - 4ac

the value indicates how many roots the quadratic equation has:

if b^2 - 4ac > 0, there are two distinct roots

if b^2 - 4ac = 0, there is one repeated real root

if b^2 - 4ac < 0, there are no real roots

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factorising

-the quadratic equation is formatted into (nx ± a)(mx ± b) = 0

-each bracket is treated as if it equals zero

-the values of x equal the values of the roots

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completing the square

-the quadratic equation is formatted into a(x+d)2 ± e = 0

where d = b/2a

and e = c - b2/4a (aka c - d²)

-this can be solved giving x = ±√(e/a) - b/2a (aka ±√(e/a) - d)

-gives two values of x which equal the values of the roots

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additional uses of completing the square

-to find the turning point on a quadratic graph

-to prove and/or show results using the fact that a squared term is always >= 0 i.e. k(x±a)^2 ± b always gives a result >= b

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quadratic formula

(-b ± (b^2 - 4ac))/2a

gives the values of the roots

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functions

functions take inputs, apply mathematical operations to it and output the value

usually denoted as f(x), or g(x) if f(x) is already specified

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roots of functions

the values of x which cause an output of 0

commonly determined by solving the function's equation as quadratic(it may need to be converted first)

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quadratic graphs

graphs plotted from quadratic functions, the line takes the shape of a parabola

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parabola shape and QE

if a is positive the parabola will be u shaped, if it is negative it will be n shaped

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x-intercept and QE

the roots of the quadratic equation equal the co-ordinates at which the parabola intersects the x axis

the number of roots determines whether the parabola crosses the x axis, merely touches it or does not touch it

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y-intercept and QE

the y intercept of the parabola is equal to c

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turning point and QE

the turning point of the parabola can be found by completing the square to a(x+d)2 ± e = 0 where the co-ordinates equal (-d, e)

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sketching polynomials

sketching polynomials requires taking into account the following and connecting them through a smooth curve:

-y intercept

-roots

-turning points (exact values don't need to be known)

-shape

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finding the y intercept of polynomials

set x to 0 where y = f(x)

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finding the roots of polynomials

find values which give a y of 0 where y = f(x)

the maximum number of roots is equal to the highest power in the polynomial

odd-degree polynomials must have at least one root

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turning points in polynomials

the number of turning points is equal to the highest power in the polynomial minus one

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shapes of polynomials

where f(x) = ax^n + bx^n-1 ...

a positive value of a means the left end of the graph goes up for even degree polynomials and down for odd degree polynomials while a negative value means the opposite

for even-degree polynomials the right end goes in the same direction as the left end, while for odd-degree polynomials it goes in the opposite

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reciprocal functions

fractions with an x term on the denominator

e.x. 1/x, 3/x^2, -2/x

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reciprocal graphs

graphs of reciprocal functions- contain two asymptotes in different quadrants which never touch the x or y axis but get infinitely close at the ends

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shapes of reciprocal graphs

for a/x the asymptotes are in diagonally opposite quadrants

where a is positive the asymptotes are in the bottom left and top right quadrants and where it is negative they are in the other two

for a/x^2 the asymptotes are in horizontally opposite quadrants

where a is positive the asymptotes are above y=0 and where it is negative they are below

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shapes of asymptotes

the further a is from zero, the further the asymptotes are from the origin, and the closer it is the more L-shaped they are

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proportional relationships

connections between two variables, can be direct or inverse

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direct proportionality

y = kx

as one increases or decreases by a factor of k, so does the other

if plotted graphically, k determines the gradient of the line

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inverse proportionality

y = k/x

as one increases the other decreases and vice versa

if plotted graphically, k determines the closeness of the asymptote to zero

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translating graphs

f(x) + a causes the graph to move vertically by a

f(x + a) causes the graph to move horizontally by -a

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stretching graphs

af(x) causes the graph to stretch vertically by a

f(ax) causes the graph to stretch horizontally by 1/a

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reflecting graphs

-f(x) reflects the graph in the x axis

f(-x) reflects the graph in the y axis

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linear simultaneous equations

simultaneous equations with two unknowns, both to the power of one, with one pair of solutions

can be solved by elimination or substitution

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quadratic simultaneous equations

simultaneous equations where one is linear and one is quadratic, with up to two pairs of solutions

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solving simultaneous equations by elimination

-multiply one or both of the equations so that the multiple of one of the unknowns matches

-subtract one equation from another to eliminate the equal unknown

-calculate the value of the second unknown using the result

-calculate the value of the first unknown by subtracting the second unknown from one of the equations

-check the results by substituting into the original equations

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solving simultaneous equations by substitution

-rearrange one of the equations to make one of the unknowns the subject

-substitute this equation into the other equation

-solve the second equation to find the value of one of the unknowns

-substitute this unknown into the first equation and use that to figure out the value of the other unknown

-check the results by substituting into the oriignal equations

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solving quadratic simultaneous equations

-rearrange the linear equation to make one of the unknowns the subject

-substitute this equation into the quadratic equation

-solve the quadratic equation to find the value(s) of one of the unknowns

-substitute the known unknowns into the rearranged linear equation to find the value of the other unknown

-check the results by substituting into the original equations

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simultaneous equations on graphs

simultaneous equations can be plotted on graphs- linear equations will be straight lines while quadratic equations will be parabolas

the solution(s) are equal to the point(s) of intersection

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linear inequalities

similar to linear equations, but with inequality signs instead therefore the solutions take ranges of values

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quadratic inequalities

similar to quadratic equations, but with inequality signs instead therefore the solutions take ranges of values

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solving linear inequalities

linear inequalities are solved the same way as other linear equations, but the solutions take a range using inequality signs

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solving quadratic inequalities

quadratic inequalities are solved the same as other quadratic equations, but the solutions take a range

the range of solutions either take the area between the x-intercepts or outside of them, or above or below the curve

which of these ranges it takes depends on the value of a and the inequality sign, it can be determined by sketching the quadratic equation as a graph

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representing solutions to inequalities

inequalities can be:

-drawn on number lines

-written normally ie 5 < x < 7

-written using set notation ie {x: x < 5} ∪ {x: x > 10}

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number lines

can be used to represent solutions to inequalities

filled in dots represent <= and >= while empty dots represent < and >

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set notation

can be used to represent solutions to inequalities e.x. {x: x < 3}

useful for non-graphically representing values outside of a range rather than in it ie

{x: x < 5} ∪ {x: x > 10} (x is smaller than 5 or bigger than 10, not between them)

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inequalities on graphs

inequalities can be plotted on graphs

solid lines represent < and > while dotted lines represent <= and >=

the region of the graph which satisfies the inequalities is shaded

this region is determined using the fact that the solution for each line is on one side of the line for straight lines, or within a range inside or outside of the curve for parabolas

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a^m x a^n

a^(m+n)

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a^m / a^n

a^(m-n)

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(a^m)^n

a^(mn)

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(ab)^n

(a^n)(b^n)

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a^-m

1/(a^m)

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a^(m/n)

n√(a^m)

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surds

irrational roots of integers and their multiples

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√ab

√a x √b

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√(a/b)

√a / √b

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simplifying surds

split surds of non-prime numbers into their prime factors and multiply together any equivalent surds to produce whole numbers, the result is the sum of these

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rationalising n/√a

multiply by √a

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rationalising n/(√a + √b)

multiply by √a - √b

also applies if the plus and minus signs are reversed

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proof by contradiction

not in the test, add later

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algebraic division with no improper fractions

where F(x) and G(x) are polynomials:

F(x)/G(x) = Q(x) + r/G(x)

where Q(x) is the quotient and r is the remainder (of F(x)/G(x))

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partial fractions

a fraction that has more than one linear(ax + b) factor in the denominator, that can therefore be split up into separate fractions

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finding and converting split fractions

-factorise the polynomial in the denominator

-split it into the sum of proper fractions with linear denominators and algebraic placeholders for the numerators

-equate the original numerator to the sum of the algebraic placeholders multiplied by all the denominators of all other partial fractions

-choose a simple value of x to make finding the values of the algebraic placeholders easier

-done

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repeated linear factors

if a squared expression appears in the factors, the unsquared expression and the squared expression will both be denominators in the split fractions as they are both factors