H U2 O1 Polynomial Quadratic - Forrester High School ...U2+O1...H U2 O1 Polynomial Quadratic AutoSave 17 December 01, 2013 Exercise 7I, page 137 Questions 1, 2 and 6 1st November 2013

H U2 O1 Polynomial Quadratic

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December 01, 2013

6th May 2013 Unit 1 Outcome 1Starter

10QQ


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December 01, 20136th May 2013 Unit 2 Outcome 1

Completing the square

http://www.mathsisfun.com/algebra/completingsquare.html

2x2 - 20x + 10 = 0

x2 + 20x + 40 = 0x = -2.25, -17.75

4x2 + 4x -3 = 0x = 0.5, -1.5

x=9.47,0.53

http://www.mathsisfun.com/quadraticequationsolver.htmlshows graphs

• Tarsia• Worksheet number 2 - SolveQuad2.pdf

http://www.mathsisfun.com/algebra/completing-square.html

http://www.mathsisfun.com/quadratic-equation-solver.html


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December 01, 2013

11th June 2013

Starter

http://www.transum.org/Software/sw/Starter_of_the_day/Starter_June10.asp

Estimating from pie chart

http://www.transum.org/software/sw/starter_of_the_day/starter_june10.asp


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December 01, 2013

11th June 2013Completing the square

x2 + 4x -13 x2 - 8x +1


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December 01, 2013


• Make the coefficient of x2 = 1 by dividing.• Move constant to RHS.

• Divide coefficient of x by 2 and square it.

• Add that to both sides of the equation.

• Factor the perfect square.• Take square root of both sides.• Solve by finding + and - results.

2x2 - 20x + 10 = 0x2 - 10x + 5 = 0 (÷2)

x2 - 10x = -5

10 2 ( (2x2 - 10x + = -5 + 10

2 ( (210 2 ( (2

x2 - 10x + 52 = -5 + 52

(x - 5)2 = 20x - 5 = √20x - 5 = ±4.47

x = +4.47 + 5 x = -4.47 + 5x = 9.47 x = 0.53


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December 01, 2013


• Make the coefficient of x2 = 1 by dividing.• Move constant to RHS.

• Divide coefficient of x by 2 and square it.

• Add that to both sides of the equation.

• Factor the perfect square.• Take square root of both sides.• Solve by finding + and - results.

2x2 - 20x + 10 = 0x2 - 10x + 5 = 0 (÷2)

x2 - 10x = -5

10 2 ( (2x2 - 10x + = -5 + 10

2 ( (210 2 ( (2

x2 - 10x + 52 = -5 + 52

(x - 5)2 = 20x - 5 = √20x - 5 = ±4.47

x = +4.47 + 5 x = -4.47 + 5x = 9.47 x = 0.53


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December 01, 2013

http://studymaths.co.uk/topics/completingTheSquare.php

http://www.regentsprep.org/Regents/math/algtrig/ATE12/completesqlesson.htm

http://studymaths.co.uk/topics/completingthesquare.php

http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm


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December 01, 2013

29th October 2013

HOMEWORK

Due: Friday 1st

November

Unit 1 Extension Test


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December 01, 2013

29th October 2013 Unit 2 Outcome 1

Evaluating Polynomials

A polynomial is an expression containing terms of the form axn. Each term has a coefficient (a), multiplied by a variable (x) raised to a whole number power (n).A polynomial has a degree equal to its highest power.Examples of polynomials:

f(x) = 4x3 - 3x2 - 5x + 28 has degree 3f(x) = -7x5 has degree 5f(x) = 5 has degree 0, since 5=5x0

Not polynomials:f(x) = 5x-1

f(x) = ∜7x2


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December 01, 2013

Exercise 7B, page 127Question 2

29th October 2013 Unit 2 Outcome 1Evaluating PolynomialsAn easy way of evaluating polynomials is to use the nested form table.

Evaluate f(3) for f(x) = 4x3 - 3x2 - 5x + 28

Nested Form Table Method Old Method

Evaluate f(-2) for f(x) = 3x4 - 7x + 35


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December 01, 2013

30th October 2013Using Synthetic Division to find Roots (factors) of Polynomials

Remainder and Factor Theorem

• If a polynomial f(x) is divided by (x-h) the remainder is f(h).

• If f(h)=0 then (x-h) is a factor of f(x) or, if (x-h) is a factor of f(x) then f(h)=0

f(x) = (ax + b)Q(x) + f(h)

factor quotient remainder(from synthetic division)

In other words, if the result of synthetic division on a polynomial by h is zero, then h is a root of the polynomial, and (x – h) is a factor of it.

Compare this to non-algebraic division

35 ÷ 8 = 4 rem 3

OR 35 = 8 x 4 + 3

divisor quotientremainder


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December 01, 2013

Express (5x3-8x2-2x+11) ÷ (x-5) in the form f(x)=(ax+b)Q(x)+f(h)

(Do synthetic division)

(Express in correct form)

(Get root)

(Check)

(7D Q1)


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December 01, 2013

Show that (x-3) is a factor of x3+2x2-14x-3

(Do synthetic division)

(statement of f(h)=0)

(Get root)

(7E Q1)


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December 01, 2013

Factorise x3-2x2-11x+12 fully

(Do synthetic division on the factors)

(if f(h)=0 then this is a factor, rewrite f(x))

(Get factors of the constant)

(7E Q7a)

(factorise again if needed and state final, factorised f(x))


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December 01, 2013

31st October 2013Finding a Polynomial's Coefficients

Find the value of k for which (x + 3) is a factor of x3 - 3x2 + kx + 6


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December 01, 2013

Exercise 7F, page 132Questions 1 and 2

31st October 2013Finding a Polynomial's Coefficients

When f(x ) = px3 + qx2 −17x + 4q is divided by x − 2, the remainder is 6, and (x −1) is a factor of f(x) .Find p and q.


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December 01, 2013

Exercise 7I, page 137Questions 1, 2 and 6

1st November 2013 Curve Sketching• Find y-intercept - substitute for x=0 into f(x)• Find x-intercepts - that is, find the roots by factorising using synthetic division• Find coordinates of stationary points using f'(x), and their nature.• Sketch the curve and label with all the information found above.

Note that functions with repeated roots have stationary points that rest on the x-axis.

For example, f(x)=x2 - 2x + 1 factorises to f(x)=(x-1)2 and its graph looks like this. It should have 2 roots as it is x2 but it is a repeated root at x=1, resting on the x-axis

f(x) = 1 x3 - 2x2 - 3 x + 9 2 2Factorises to:f(x) = 1 (x+2)(x-3)2

2So it crosses the x-axis at -2 and has a repeated root at x=3, where the graph has a stationary point resting on the x-axis


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December 01, 2013

5th November 2013Finding the Equation of a Graph of a Polynomial

f(x) = k(x-a)(x-b)(x-c)Find k by substituting for (0,d)


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December 01, 2013

Find the equation of the cubic in this graph.

y=k(x+6)(x+3)(x-1)Use the point (0,-36)

The roots from the graph are:


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December 01, 2013

Exercise 7H, page 135/136

Questions 1-9

Find the equation of the cubic in this graph.

y=k(x+2)(x-3)(x-3)

Use the point (0,9)

The roots from the graph are:


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December 01, 2013

6th November 2013Completing the square to Sketch Graphs

f(x) = ax2 + bx + c

can be written in completed square form:

y = a(x+p)2 + q

• Axis symmetry of the quadratic: x=-p• Turning Point is located at: (-p,q)


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December 01, 2013

Complete the square:• Take the coefficient of x :

- half it- square it- add it- subtract it

• factorise the first three terms as a squared factor • Tidy up the constant (competed square form)

Use the method of completing the square to sketch and annotate the graph y = x2 - 3x + 2

y = x2 - 3x + - + 232

( )2 32

( )2

y = x - - + 232( 3

2()2 )2

y = x - - 32( 1

4)2

Axis of symmetry is x=32

TP is a minimum since x2 is positive:Minimum TP at 3

2-14

( ),

Calculate y-intercept:for x=0, y=2 which gives us point (0,2)

Sketch Graph


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December 01, 2013

By completing the square, Sketch the graph y = -2x2 - 8x + 9

Complete the square

Axis of symmetry

Coordinate of TP

y-intercept

Sketch

Complete the square:• Take the coefficient of x :

- half it- square it- add it- subtract it

• factorise the first three terms as a squared factor • Tidy up the constant (competed square form)


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December 01, 2013

y = -2x2 - 8x + 9y = -2 x2 + 4x - 2

9( )y = -2 x2 + 4x + - - 4

2( )2[ ]42( )2

29

42( )2y = -2 (x + )2 - - 4

2( )[ ]29

y = -2 (x+2)2 - 4 - ][ 29

][y = -2 (x+2)2 - 217

y = -2(x+2)2 + 17

axis of symm: x=-2Min TP at (-2,17)x=0, y= 9 (0,9)

(-2,17)(0,9)

-2


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December 01, 2013

7th November 2013Solving Quadratic Inequations

There are several methods to solve quadratic equations:• Using a graph• Factorising• Completing the square• Using the quadratic formula

To solve quadratic inequations:• find the roots of the function• sketch the graph• use the graph to answer the question.


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December 01, 2013

Exercise 8F, page 149Question 3

Solving Quadratic Inequations

Solve x2 + x - 12 > 0

Find the roots by factorising

Solve x2 + x - 12 < 0Find the roots by factorising


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December 01, 2013

8th November 2013

The Discriminant of the Quadratic Formula

Remember the quadratic formula, used to find the roots of a quadratic?

-b±√b2-4ac2a

x =

b2-4ac is called the discriminant.

if b2-4ac > 0, then the roots are real and distinct - 2 roots

if b2-4ac = 0, then the roots are real and equal (one, repeated root)

if b2-4ac < 0, then the roots are non-real (does not cross the x-axis)


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December 01, 2013

Determine the nature of the roots of the equation 4x(x - 3) = 9


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December 01, 2013

Find the value(s) of p given that 2x2 + 4x + p has real roots.


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December 01, 2013

Level C: Exercise 8H, questions 1, 2Exercise 8I, questions 1, 2, 5

Level A/B: Exercise 8I, questions 6, 8Exercise 8K, questions 10, 12

Find the value(s) of r given that x2 + (r - 3)x + r = 0 has no real roots.


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December 01, 2013

12th November 2013Testing for Tangency

To find if a line intersects a curve:• Substitute the equation of the line into the equation of the curve.• Use the discriminant to find how many times the line meets the curve.

if b2-4ac > 0, then the line and the curve intersect twice.

if b2-4ac = 0, then the line and the curve intersect once and the line is a tangent to the curve.

if b2-4ac < 0, then the line and the curve do not intersect.


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December 01, 2013Testing for Tangency

Show that the line y = 5x − 2 is a tangent to the parabola y = 2x2 + xand find the point of contact.

The line meets the curve when their equations are the same:

5x - 2 = 2x2 + x


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December 01, 2013Testing for Tangency

Find the equation of the tangent to y = x2 +1 that has a gradient of 3.


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December 01, 2013

Documents

H U2 O1 Polynomial Quadratic - Forrester High School ...U2+O1...H U2 O1 Polynomial Quadratic AutoSave 17 December 01, 2013 Exercise 7I, page 137 Questions 1, 2 and 6 1st November 2013