# MathsMethodsU1-2FormulaNotes

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Downloaded from – www.vcehelp.com.au Created by VCE student Nicholas Buttigieg for Maths Methods Unit 1/2 Here is an article for all of the functions and graphs used in Maths Methods Units 1 & 2. You can use these as study notes if you like. To look for any diagrams on the shape of graphs, refer to your textbook. Linear Graphs ã The form is y = mx + c, where m is the gradient and c is the y-intercept. ã ã ã ã ã ã ã ã ã ã The gradient between two points is equal to m = = Two lines with the same g
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## Trigonometric Functions

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Get more free VCE resources atwww.vcehelp.com.au  Downloaded from –www.vcehelp.com.au Created by VCE student Nicholas Buttigieg for Maths Methods Unit 1/2Here is an article for all of the functions and graphs used in Maths Methods Units 1 & 2. Youcan use these as study notes if you like. To look for any diagrams on the shape of graphs,refer to your textbook. Linear Graphs ã The form is y  = mx  + c  , where m is the gradient and c  is the y  -intercept. ã The gradient between two points is equal to m = = ã Two lines with the same gradient are parallel to each other and do not intersect. ã To find the equation of a straight line given two points, use the formula: m (  x  -  x  1 ) = y  - y  1 , where m is the gradient. ã When given equations of the form ax  + by  + c  = 0, then transpose to y  = mx  + c    ã Vertical lines are of the form  x  = a , where the  x  -axis intercept is given as ( a , 0). ã Horizontal lines are of the form y  = c  , where the y  -axis intercept is given as (0, c  ). ã Perpendicular lines can be calculated using the formula: m 1   m 2 = -1. ã To calculate the tangent of the angle of slope, calculate the gradient, then substituteit into tan  to get the angle. ã The distance between two points is given as ã The midpoint of a line segment is given as Parabolas ã The form is y  = ax  2 + bx  + c  , which is known as polynomial form. ã The form is y  = a (  x – b) 2 + c  , which is known as turning point form. ã Parabolas of the form y  = ax  2 , in which the axis of symmetry is  x  = 0, and the turningpoint is (0,0). If  a is greater than 1, the graph will be narrower than y  =  x  2 . If it isbetween 0 and 1, the graph will be wider. If  a is negative, then the graph is invertedor reflected in the  x  -axis. ã Parabolas of the form y  = ax  2 + c  involve a translation up or down the y  -axis. Theaxis of symmetry is  x  = 0 and the turning point is (0, c  ), which correlates to the y  -intercept. ã Parabolas of the form y  = a (  x   – b ) 2 involve a translation left or right along the  x  -axis.The axis of symmetry is  x  = - b and the turning point is (- b , 0), which correlates to the  x  -intercept. ã The turning point of a parabola is (- b , c  ) from the turning point form. ã The turning point of a parabola from polynomial form can be found by completing thesquare so it is in turning point form. ã To find  x  -intercepts, either factorise the equation or use the quadratic formula:  x  = ã The axis of symmetry is  x  = - b from the turning point form. ã The axis of symmetry is  x  = from the polynomial form. ã The discriminant is equal to  Δ = b 2 – 4 ac  .  Get more free VCE resources atwww.vcehelp.com.au  ã If   Δ < 0, then there are no real solutions. ã If   Δ = 0, then there is one rational solution. ã If   Δ > 0, then there are two solutions. If it results in a perfect square, there are tworational solutions. If it is not a perfect square, there are two irrational solutions. Rectangular Hyperbolas ã The form is y  = a / (  x  - h ) + k  . ã If  a is greater than 1, then the graph is dilated from the  x  -axis, or moves further awayfrom the  x  -axis, compared to the graph of  y  = 1 /  x  . If  a is between 0 and 1, then thegraph is dilated to the  x  -axis, or moves closer to the  x  -axis. If  a is negative, then thegraph is inverted or reflected in the y  -axis. ã The asymptotes are given as  x  = - h and y  = k  . ã The  x  -intercepts can be calculated by substituting y  = 0. ã The y  -intercepts can be calculated by substituting  x  = 0. The Truncus ã The form is y  = a / (  x  – h ) 2 + k  . ã If  a is greater than 1, then the graph is dilated from the  x  -axis, or moves further awayfrom the  x  -axis, compared to the graph of  y  = 1 /  x  2 . If  a is between 0 and 1, then thegraph is dilated to the  x  -axis, or moves closer to the  x  -axis. If  a is negative, then thegraph is inverted or reflected in the  x  -axis. ã The asymptotes are given as  x  = - h and y  = k  . ã The  x  -intercepts can be calculated by substituting y  = 0. ã The y  -intercepts can be calculated by substituting  x  = 0. The Square Root Function ã The form is y  = a √  - (x  – h ) + k    ã If  a is positive, then the top half of the graph will be sketched. If  a is negative, thenthe bottom half of the graph will be sketched. ã The stationary point, or turning point, is determined by the coordinates (- h , k  ). ã If the expression under the radical is negative, then the graph is reflected in the y  -axis. ã If the entire equation is negative, then the graph is reflected in the  x  -axis. ã The  x  -intercepts can be calculated by substituting y  = 0. ã The y  -intercepts can be calculated by substituting  x  = 0. Circles and Semicircles ã The form is  x  2 + y  2 = r  2 for circles with a centre at (0,0). ã The form is (  x  – h ) 2 + ( y  – k  ) 2 = r  2 for circles not centred at the srcin. This is equal tothe form  x  2 + y  2 – 2 hx  – 2 ky  + c  = 0, where c  = h 2 + k  2 – r  2 . ã The centre of the circle is denoted as (- h ,- k  ). ã The radius of the circle is equal to r  . ã Circles of the form  x  2 + y  2 – 2 hx  – 2 ky  + c  = 0 can be sketched by completing thesquare, so it is in the form (  x  – h ) 2 + ( y  – k  ) 2 = r  2 .  Get more free VCE resources atwww.vcehelp.com.au  ã When sketching semicircles, draw the graph of the circle. If there is a positive sign infront of the radical, then draw the top half of the circle. If there is a negative sign,then draw the bottom half of the circle. ã The  x  -intercepts can be calculated by substituting y  = 0. ã The y  -intercepts can be calculated by substituting  x  = 0. Cubic Graphs ã The form is y  = ax  3 + bx  2 + cx  + d  , which is in polynomial or factor form. ã The form is y  = a (  x  – h ) 3 + k  , which is the basic form. ã The point of inflexion from the basic form is given as (- h , k  ). This only goes for graphs that resemble this form. ã Cubic graphs are said to take two distinct patterns: the basic form and the factor form, plus their negative forms. ã To find the  x  -intercepts, divide the polynomial so it is in factor form. Then use theNull Factor Law to find the  x  -intercepts. ã If there is a repeated factor in the equation, then the graph will only have two  x  -intercepts. ã If, when factorised, one linear factor is produced, then there is one  x  -intercept. Thisis because the other factors cannot be simplified another further. ã Before sketching the graph, draw a sign diagram that helps you to sketch the graphmore accurately. ã The y  -intercept can be found by substituting  x  = 0. Quartic Graphs ã The form is y  = ax  4 + bx  3 + cx  2 + dx  + e , which is in polynomial or factor form. ã The form is y  = a (  x  – h ) 4 + k  , which is the basic form. ã The turning point from the basic form is given as (- h , k  ). This only goes for graphsthat resemble this form. ã Quartic graphs are said to take 6 distinct patterns, which are:- y  = ax  4  - y  = ax  4 + cx  2  - y  = ax  2 (  x  – b )(  x  – c  )- y  = a (  x  – b ) 2 (  x  – c  ) 2  - y  = a (  x  – b )(  x  – c  ) 3  - y  = a (  x  – b )(  x  – c  )(  x  – d  )(  x  – e ) ã To find the  x  -intercepts, divide the polynomial so it is in factor form. Then use theNull Factor Law to find the  x  -intercepts. ã When factorised, the number of factors is the same as the number of   x  -intercepts. ã Before sketching the graph, draw a sign diagram that helps you to sketch the graphmore accurately. ã The y  -intercept can be found by substituting  x  = 0. Exponential and Logarithmic Functions ã The form is y  = k  × a bx  + c  . ã For simple graphs of the form y  = a  x  , the only asymptote is y  = 0 or the  x  -axis. The y  -intercept will always be equal to 1.  Get more free VCE resources atwww.vcehelp.com.au  ã If the value of  a is greater than 1, the graph has a positive gradient. This means thatwhen  x  approaches ∞ , then y  will increase exponentially. If the value of  a is between0 and 1, then this will produce a negative index. Therefore, this is the same as thegraph with the positive index, but is reflected in the y  -axis. ã For graphs of the form y  = k  × a bx  + c  , the following apply:- The horizontal asymptote is equal to y  = c  .- The y  -intercept is equal to (0, k  + c  )- If  b is a number other than 1, then every  x  value must be divided by b to get thepoints of the transformed graph.- If  k  is a number other than 1, then every y  value must be multiplied by k  to get thepoints of the transformed graph.- If  k  is negative, then the graph of its positive form is reflected in the horizontalasymptote. ã Logarithmic functions are the inverse of exponential functions. This means that theyare exponential functions reflected in the line y  =  x  . ã For simple graphs of the form y  = log a  x  , the only asymptote is  x  = 0 or the y  -axis.The  x  -intercept will always be equal to 1. ã If the value of  a is greater than 1, the graph has a positive gradient. This means thatwhen  x  approaches ∞ , then y  will increase logarithmically. If the value of  a isbetween 0 and 1, then this will produce a negative logarithm. Therefore, this is thesame as the graph with the positive index, but is reflected in the  x  -axis. ã For graphs of the form y  = log a (  x  – c  ), the following apply:- The vertical asymptote is equal to  x  = c  .- The  x  -intercept is equal to (1 + c  , 0). Sine and Cosine Functions ã The form is y  = a sin n (  x  ± b ) + c  or  y  = a cos n (  x  ± b ) + c  . ã The basic graph is y  = sin  x  . It has the following characteristics:- The graph has a wavelength or period of 2 π units. It is a periodic function.- The maximum and minimum values are y  = 1 and y  = -1. This means that theamplitude is equal to 1.- The y  -intercept is equal to (0, 0). ã The other basic graph is y  = cos  x  . It has the following characteristics:- The graph has a wavelength or period of 2 π units. It is a periodic function.- The maximum and minimum values are y  = 1 and y  = -1. This means that theamplitude is equal to 1.- The y  -intercept is equal to (0, 1). ã In the graphs of the form y  = a sin ( nx  ) and y  = a cos ( nx  ), the followingcharacteristics apply:- The graph has a wavelength or period of 2 π / n .- The amplitude is always equal to a .- The maximal domain is equal to R  .- The range of each function is [- a , a ].- If  a is negative, then the graph of the positive form is reflected in the  x  -axis. ã In graphs of the form y  = a sin n (  x  ± b ) and y  = a cos n (  x  ± b ), a horizontaltranslation along the t  -axis occurs, with the graph moving b units to the left or right.
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