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An in-depth exploration of linear and quadratic equations and their corresponding graphs. Topics covered include finding the equations of lines, the midpoint and point of intersection, surds, quadratic equations and their solutions, and simultaneous equations. Additionally, the document discusses the concepts of rationalizing the denominator, completing the square, and using the quadratic formula.
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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
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
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
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
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
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
1
2
3
4
5
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
n
n + 1
Constant of integration
e.g. Given that f '(x) = 8x 3
f(x) =
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
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
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