Unit 3 Notes Mathematical Methods

Unit 3 Notes Mathematical Methods Foundational Knowledge Created b Triumph Tutoring

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Polnomials Basic Polnomials This section covers concepts from Units and that are considered presumed knowledge for Unit 3 Mathematical Methods. Without a good prior knowledge of basic Unit and concepts, it will be difficult to build on skills required in Units 3 and 4. A polnomial in, for eample, is an epression that consists of terms which have non-negative integer powers of onl. That is, an epression must onl have positive whole number powers of to be classed as a polnomial. eg. 4 3 + is a polnomial whereas is not. The degree of a polnomial refers to the highest power of. eg. p() = 4 + 5 3 +, the degree of p() is 4. Eample Question If p() = 5 + a 4 + 3 + b, p() = 9 and p() = 8, find the values of a and b. Division and Factorisation of Polnomials Division of polnomials must be completed in order to factorise cubic or higher order graphs. When one polnomial p() is divided b another d(), the result can be epressed as p() d() = q() + r() d() where q() is the quotient, r() is the remainder and d() is the divisor. At this stage, the divisor will alwas be a simple linear epression. The linear epression is usuall found b using the remainder theorem, or through analsis to find an -intercept. eg. using the remainder theorem find a factor of p() = 3 + 5 6. Look at factors of the constant 6:,, 3, 6. Substitute the factors in until the epression equals 0. p() = () 3 + () 5() 6 p() = + 5 6 p() = 8 0 p( ) = ( ) 3 + ( ) 5( ) 6 p( ) = + + 5 6 p( ) = 0 Therefore, + is a factor.

3 eg. divide 4 3 3 + 8 b + Image sampled from Maths Quest 0 If the remainder is 0 (which can be verified using the remainder theorem), then the new epression for the original function can be represented b: quotient divisor. If the quotient is not a linear epression, then it can likel be factorised again (if it is a quadratic ou ma use simpler methods to achieve this). Note: not all functions can be factorised into linear terms. Often times this will be due to a quadratic factor of the function. If this occurs, ou can verif this b using the discriminant. The discriminant of a + b + c can be found using the following formula: = b 4ac If < 0, then we know that there are no solutions (or that it cannot be factorised further).

4 Linear Graphs The gradient of a straight line joining two points is m =, also known as rise over run. The general equation of a straight line is = m+c. m represents the gradient and c represents the vertical translation (also the -intercept). The equation of a straight line passing through the point (, ) and having a gradient of m is: = m( ). TIP: use this equation when evaluating straight lines, instead of = m + c. Parallel lines have the same gradient. Perpendicular lines, however, have gradients such that m m = (or m = m ). Domain and Range The domain of a function is the set of values of for which the function is defined. The range of a function is the set of values of for which the function is defined. Notation If the domain or range is unrestricted it can be denoted as R (all real numbers) or (, ). Circular brackets ( or ) are used to denote that the limiting number is not included. eg. (0, ) means from zero to infinit not including 0 Square brackets [ or ] are used to denote that the limiting number is included. eg. [0, ) means from zero to infinit including 0 Quadratics The general form of quadratic functions is = a + b + c. For a > 0, the graph has a minimum value (smile). For a < 0, the graph has a maimum value (mountain). Both of these tpes are called turning points. The -intercept is c. (Thinking: at the intercept =0. If =0 all terms equal zero, therefore onl c is left). The equation of the ais of smmetr and the -value of the turning point is = b a The -intercepts are found b solving the equation a + b + c = 0. You can factorise and then use the null factor law or use the quadratic formula. Quadratic formula: = b ± b 4ac a Note: Alwas check solvabilit using the discriminant.

5 The discriminant ( ) is the value underneath the square root sign in the quadratic formula. Discriminant: = b 4ac. If > 0, there are two solutions to the equation and hence two -intercepts on the graph. If = 0, the two solutions are equal and hence there is onl one -intercept on the graph. That is, the graph has a turning point on the ais. If < 0, there are no solutions to the equation and hence no -intercepts on the graph. The power form of quadratic functions looks like = a( b) + c, where a = dilation factor. Dilates the graph in the direction. b = horizontal translation. Translates the graph horizontall to either the left or right. c = vertical translation. Translates the graph verticall upwards or downwards. The turning point of the graph occurs at (b, c) To change from the general form to power form we need to complete the square. eg. B completing the square, factorise = + 0 + 3 into turning point form. = + 0 + 3. Half the coefficient of, square the result, add and subtract it like the eample. = ( + 0 + 5) + 3 5. Factorise the first three epressions. = ( + 5) Now that we have it in power form we know that: The graph is translated 5 units left The graph is translated units down The turning point is at ( 5, ) 5 0 5 0 5 5 0 5 5 5 0 5 0 5 0 ( 5, ) 5

6 Cubics Cubic functions are polnomials of degree 3. The general form of a cubic function is = a 3 + b + c + d. The power form of a cubic function is ver similar to that of a quadratic function: = a( b) 3 + c. The translations of a cubic graph are the same as a quadratic: a = dilation factor. Dilates the graph in the direction. b = horizontal translation. Translates the graph horizontall to either the left or right. c = vertical translation. Translates the graph verticall upwards or downwards. The point of inflection of the graph occurs at (b, c) For eample, the graph of = ( 3) 3 + 5 is the graph of = 3 translated 3 units right, 5 units up and dilated b a factor of from the -ais. 5 0 5 0 5 (3, 5) 5 0 5 5 5 0 5 0 5 0 5

Eponential and Logarithmic Equations The inde laws 7 A number in inde form has two parts, the base and the inde, power, eponent or logarithm. A number in inde form is represented like this: a where a is our base and is known as the inde, power or eponent The inde laws are as follows a a = a + a a = a (a ) = a a 0 = (a 0) a = a (a 0) a = a ( 0) a = a ( 0) (ab) = a b ( a b ) = a b (b 0) The logarithm laws Logarithms are an alternative wa of epressing a number in inde form. log a () =. For eample, 3 = 8 can be written as log (8) = 3. If = a, then The logarithm laws are as follows: since a 0, log a (0) is undefined a 0 = log a () = 0 a = a log a (a) = log a (m) + log a (n) = log a (mn) log a (m) log a (n) = log a ( m) n log a (m p ) = p log a (m)

8 Practicing eponential equations eg. simplif ( 3 ) 3 3( 4 ) 6 4 4. Remove the brackets b multipling out the indices ( 3 ) 3 3( 4 ) = 3 6 9 3 8 6 4 4 6 4 4. Add the indices of and add the indices of. Simplif 3 to 8 and. multipl the whole numbers together. 3 6 9 3 8 = 48 7 6 4 4 5 4 3. Subtract the indices of and. Divide the whole numbers. 4 8 7 5 = 4 3 3 eg. write 3 0.4 in simplest form. Write as a fraction with a positive inde 3 0.4 = 3 0.4. Change 0.4 into a fraction and simplif 3 = 0.4 3 5 3. Rewrite using the inde laws = 5 3 3 5 4. Simplif 5 3 = = 4 eg. solve 3 3 + 7 = 0 for. Write 3 as (3 ) (3 ) 3 + 7 = 0. Let 3 = a to create a simper quadratic equation to solve a a + 7 = 0 where a = 3 3. Factorise and solve for a (a 3)(a 9) = 0 a 3 = 0 or a 9 = 0 a = 3 or a = 9 4. Substitute back a = 3 and solve for 3 = 3 or 3 = 9 3 = 3 or 3 = 3 = or =

9 Practicing logarithmic equations eg. solve log 5 ( ) = for. Rewrite using the law a = log a () = 5 =. Solve for 5 = = 6 eg. solve log 0 () + log 0 ( 3) = log 0 (4) for. Simplif the left hand side using the law log a (m) + log a (n) = log a (mn) log 0 (( 3)) = log 0 (4). Equate the logs ( 3) = 4 3. Epand and rearrange the equation 3 4 = 0 4. Solve the equation ( 4)( + ) = 0 = 4 or = = 4 (as > 0) Euler s number Euler s number (e) is a number that is ver important, used in problems involving natural growth and natural deca. It is ver similar to in that it is irrational and has to be approimated: e =.783. The number e can be use like an other number. The same inde and logarithm laws appl to it. However, certain calculus rules and calculations can onl be performed on indices and d logarithms involving e. For eample, (log d e() =. This cannot be performed with an other base logarithm. eg. Write log e () + = log e () with as the subject. Use the laws log a (m) p = p log a (m) and log a (a) = to make each term a log log e ( ) + log e (e) = log e (). Use the law log a (m) + log a (n) = log a (mn) to create one logarithm on each side log e (e ) = log e () 3. Equate the insides of the logs = e

0 Unit & Trigonometr Revision Radians and the Unit Circle Angles are measured in degrees or radians. One radian is written as c c = 80 Degrees Radians: multipl angle b eg. 60 80 = 60 80 = 3 80 Radians Degrees: multipl angle b 80 eg. 6 80 = 80 6 = 30 The unit circle is a circle with a radius of one unit. This circle can be used to define radians, trigonometric ratios and other relationships. T tan(θ) The unit circle has a circumference of units, and the angle at the centre of the circle is radians, or 360. As shown in the first unit circle to the right: - the horizontal distance is defined as cos(θ) sin(θ) θ cos(θ) - the vertical distance is defined as sin(θ) - the point the diagonal hits on the line ST is defined as tan(θ) According to the CAST notation on the second unit circle to the right: - Cos(θ) is positive in the fourth quadrant - All are positive in the first quadrant - Sin(θ) is positive in the second quadrant θ θ S - Tan(θ) is positive in the third quadrant S T A C + θ θ

Eact Values and Smmetr In a right-angled triangle like the one to the right, we know that sin(θ) = O H O H cos(θ) = A H θ tan(θ) = O A A With this in mind, we can use the following triangles to create a table of values to use. 30 3 45 60 According to the above triangles, the following table is created. 30 45 60 sin 3 cos 3 tan 3 3 3 a Tr to memorise this table. Practicing questions that involve these values will help cement the knowledge and reall speed up our equation-solving skills.

The following two tables define complementar angles (angles which add up to 90 ) as part of the unit circle. First Quadrant Second Quadrant sin( θ) = cos(θ) sin( + θ) = cos(θ) cos( θ) = sin(θ) cos( + θ) = sin(θ) a In the first two quadrants, we can see that the complementar angles revolve around (90 ) to translate between the quadrants. Third Quadrant Fourth Quadrant sin( 3 θ) = cos(θ) sin( 3 cos( 3 θ) = sin(θ) cos( 3 + θ) = cos(θ) + θ) = sin(θ) a In the last two quadrants, we can see that the complementar angles revolve around 3 (70 ) to translate between the quadrants. eg. find the eact value for sin(0 ) without a calculator Method. Rewrite the angle in terms of 80 0 = (80 60). Identif the base angle 0 = (80 60) base angle: 60 3. Identif the quadrant and whether the value will be positive or negative nd Quadrant (S of CAST) - sin(0 ) will be positive 4. Reference the eact values table sin(60 ) = 3 5. Consolidate the information from steps 3 and 4 sin(0 ) = sin(60 ) = 3 Method. Reference the complementar angles table sin(0 ) = sin(90 + 30 ) = cos(30 ). Reference the eact values table cos(30 ) = 3 sin(0 ) = 3 This method can be followed to find the eact values of man different angles.

3 Pthagorean Formula One of the most important formulas that ou need to follow is this: (sin(θ)) + (cos(θ)) = This formula is derived from Pthagoras Theorem a + b = c. As seen in the unit circle on page 9, a triangle is created with the centre angle. This triangle has a hpotenuse of unit, and a = cos(θ) and b = sin(θ). Solutions to trigonometric equations From the general equation equation sin() = a, we can find an infinite number of solutions. An eample of this general equation is sin() = 3. A solution of this equation is = 60 or 3, because sin(60 ) = 3 or sin( 3 ) = 3. We also know that sin( ) = 3, because sin is positive in the second quadrant. 3 Therefore, for this particular equation there are two solutions between 0 and. These solutions are 3 and 3. Sin is negative in quadrants three and four, so there are no solutions in those areas. We can go around the unit circle as man time as we need to in order to find more solutions based on the required domain. For eample, since sin( + ) = 3 and sin(3 ) = 3 in the domain [0, 4] there are 4 3 3 solutions:,, 7, 8. 3 3 3 3 eg. Find all solutions to the equation cos() = in the domain [0, ].. Write out the equation. cos() =. Find the base angle b looking at the eact values table (ignore the negative) base angle: 4 3. Identif the quadrants in which cos is negative nd and 3rd Quadrants (ST from CAST) 4. Find the angle in these quadrants using the formulas on page 9 Second Quadrant: = 4 = 3 4 Third Quadrant: = + 4 = 5 4 = 3 4, 5 4

4 Plotting circular functions Now that we know how to find the solutions to an equation, we can combine this with a few tricks to plot the entire graph on a cartesian plane. To begin with, we need to recognise a normal sin, cos and tan graph. 3 3 5 one period 3 3 5 one period 3 one period These basic shaped graphs can be modified with transformations and dilations in the same wa that an other graph can be. The basic form these translations take is: where: = a sin(b( c)) + d = a cos(b( c)) + d = a tan(b( c)) + d a affects the amplitude of the graph b affects the period c affects the horizontal translation (in the -direction) d affects the vertical translation (in the -direction) The important translations to note to begin with for circular function graphing is a and b (amplitude and period respectivel).

5 The amplitude (a) affects how high and low the graph goes (the range). For eample, a standard sin (and cos) graph s range is [, ]. If we graph 3 sin() we are changing the graphs range, stretching the graph in the -ais as seen below. 3 3 3 5 3 As we can now see, changing the amplitude to 3 adjusted the highest and lowest points of the graph. The period affects how often the shape of a graph repeats itself. The period of a standard sin graph is as this is one full revolution of the unit circle. Changing b will change this domain, and thus have a big effect on the results. The period of a sin or cos graph is b. The period of a tan graph is b. We can use the period to know where the important points on our graph are. This includes maimums, minimums and inflection points. We know a graph has 4 of these important points ever period. Therefore a graph with a period of will have either an inflection, maimum or minimum ever units. 8 (, ) 8 ( 5 8, ) ( 4, 0) (, 0) ( 3 4, 0) (, 0) ( 3 8, ) ( 7 8, )