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CORE 1 Summary Notes
1 Linear Graphs and Equations
y = mx + c
gradient = increase in y increase in x
y intercept
Gradient Facts ß Lines that have the same gradient are PARALLEL
ß If 2 lines are PERPENDICULAR then m 1 ¥^ m 2 =^ – 1^ or^ m 2 = -^
e.g. 2y = 4x – 8 m^1 y = 2x – 4 gradient = 2 gradient of perpendicular line = -½
Finding the equation of a straight line e.g. Find the equation of the line which passes through (2,3) and (4,8) GRADIENT = y 1 – y 2 x 1 – x 2
GRADIENT = 3 – 8^ = – 5^ = 5
Method 1 y – y 1 = m(x – x 1 )
y – 3 =
(x – 2) 2
y =
x – 2 2 2y = 5x – 4
Using the point (2,3)
Method 2 y = mx + c
¥ 2 + c
c = – 2
y =
x – 2 2 2y = 5x – 4
Using the point (2,3)
Finding the Mid-Point
( x 1 y 1 ) and (x 2 y 2 ) the midpoint is Ê Ë
Á
x 1 + x 2 2
' y^1 +^ y^2 ˆ (^2) ¯
Given the points
Finding the point of Intersection Treat the equations of the graphs as simultaneous equations and solve Find the point of intersection of the graphs y = 2x – 7 and 5x + 3y = 45
Substituting y = 2x – 7 gives 5x + 3(2x –7) = 45 5x + 6x – 21 = 45 11x = 66 x = 6 y = 2 x 6 – 7 y = 5 Point of intersection = (6, 5)
2 Surds ß A root such as √5 that cannot be written exactly as a fraction is IRRATIONAL ß An expression that involves irrational roots is in SURD FORM e.g. 3 √
ß ab^ =^ a^ ¥^ b
ß
a b
= a b 75 – 12
= 5 ¥ 5 ¥ 3 – 2 ¥ 2 ¥ 3
= 5 3 – 2 3
= 3 3
e.g
ß RATIONALISING THE DENOMINATOR 3 + √2 and 3 √2 is called a pair of CONJUGATES
The product of any pair of conjugates is always a rational number e.g. (3 + √2)(3 √2) = 9 3√2+3√ = 7
Rationalise the denominator of
3. Quadratic Graphs and Equations
Solution of quadratic equations ß Factorisation
x =
x 2
- 3x – 4 = 0 (x + 1)(x – 4) = 0
- 1 or x = 4
4 Simultaneous Equations ß Simultaneous equations can be solved by substitution to eliminate one of the variables Solve the simultanoeus equations y = 2x – 7 and x^2 + xy + 2 = 0 y = 7 + 2x so x^2 + x(7 + 2x) + 2 = 0 3x^2 + 7x + 2 = 0 (3x + 1)(x + 2) = 0
x = –^1 3
y = 6^1 3
or x = – 2 y = 3
ß A pair of simultaneous equations can be represented as graphs and the solutions interpreted as points of intersection. If they lead to a quadratic equation then the DISCRIMINANT tells you the geometrical relationship between the graphs of the functions b^2 – 4ac < 0 no points of intersection b^2 – 4ac = 0 1 point of intersection b^2 – 4ac > 0 2 points of intersection
5 Inequalities Linear Inequality ß Can be solved like a linear equation except Multiplying or dividing by a negative value reverses the direction of the inequality sign e.g Solve^ – 3x^ +^10 £^4
x
- 3x + 10 £ 4
- 3x £ – 6 ≥ 2 Quadratic Inequality ß Can be solved by either a graphical or algebraic approach.
e.g. solve the inequality x^2 + 4x – 5 <
Algebraic x^2 + 4x – 5 < 0 factorising gives (x + 5)(x –1) < 0
Using a sign diagram x + 5 - - - 0 + + + + + + + x – 1 - - - - - - - 0 + + + (x + 5)(x – 1) + + + 0 - - - 0 + + +
The product is negative for – 5 < x < 1 y
- 7– 6 – 5 – 4 – 3 – 2– 1 1 2 x
2
Graphical
The curve lies below the x–axis for –5 < x < 1
6 Polynomials Translation of graphs To find the equation of a curve after a translation of (^) ˙ replace x with (x-p) and ˚
˘ Í Î
È p q replace y with (y - p)
e.g The graph of y = x^3 is translated by (^) ˙ ˚
˘ ÍÎ
È 3
The equation for the new graph is y=(x - 3)^3 -
Polynomial Functions
A polynomial is an expression which can be written in the form a + bx + cx^2 + dx^3 + ex^4 + fx^5 (a, b, c..are constants)
y
1 2 x
1
2
3
4
5
∑ Polynomials can be divided to give a QUOTIENT and REMAINDER x^2 -3x + 7 x + 2 x^3 - x^2 + x + 15 x^3 +2x^2 -3x^2 + x Qutoient -3x^2 - 6x 7x + 15 7x + 14 1 Remainder ∑ REMAINDER THEOREM When P(x) is divided by (x - a) the remainder is P(a)
∑ FACTOR THEOREM If P(a) = 0 then (x – a) is a factor of P(x)
e.g. The polynomial f(x) = hx^3 -10x^2 + kx + 26 has a factor of (x - 2) When the polynomial is divided by (x+1) the remainder is 15. Find the values of h and k.
Using the factor theorem f(2) = 0 8h -40 + 2k + 26 = 0 8h +2k = 14 Using the remainder theorem f(-1) = 15 -h -10 –k + 26 = 14 h + k = 2
Solving simultaneously k = 2 –h 8h + 2(2 –h) = 14 6h + 4 = 14
dx
y = x^3 + 4 x^2 – 3 x + 6 dy (^) = 3 x (^2) + 8 x – 3
9 Using Differentiation
∑ If the value of
dy dx
is positive at x = a, then y is increasing at x = a
∑ If the value of
dy dx
is negative at x = a, then y is decreasing at x = a
∑ Points where
dy dx
= 0 are called stationary points
Minimum and Maximum Points (Stationary Points)
Stationary points can be investigated
y
- 5 – 4 – 3 – 2 – 1 1 2 3 4 5 x
1
2
3
4
5
GRADIENT
GRADIENT
Local Maximum
∑ by calculating the gradients close to the point (see above)
∑ by differentiating again to find d 2 or f “(x)
(^2) y
dx
o d 2 > 0 then the point is a local minimum
(^2) y
dx
o d 2
(^2) y
dx
< 0 then the point is a local maximum
Optimisation Problems Optimisation means getting the best result. It might mean maximising (e.g. profit) or minimising (e.g. costs)
10 Integration
ß Integration is the reverse of differentiation
Ú
x
n
dx =
x
n + 1
n + 1
+ c
Constant of integration
e.g. Given that f '(x) = 8x 3
- 6x and that f(2) = 9 find f(x)
f(x) =
Ú
8x^3 – 6x dx
8x
4
4
2
2
= 2x^4 – 3x^2 + c
To find c use f(2) = 9
So f(x) = 2x^4 – 3x^2 –
32 – 12 + c = 9 c = – 11
11 Area Under a Graph
ß The are under the graph of y = f(x) between x= a and x = b is found by evaluating the definite integral
Ú
b
f ( x ) dx
a
e.g. Calculate the area under the graph of y = 4x – x^3 between the lines x = 0 and x = 2 (^) y
1
2
3
4
5
Ú 0
2 4 x – x^3 dx =
= 2 x^2 – x
4 4 = (8 – 4) – (0 – 0) = 4
ß An area BELOW the x–axis has a NEGATIVE VALUE
