Maths Methods 1 MATHEMATICAL METHODS UNITS 3 AND 4.3 Sketching Polnomial Graphs ou are required to e ale to sketch the following graphs. 1. Linear functions. Eg. = ax + These graphs when drawn will form a straight line. The gradient of this line will = a and the intercept =. Eg. = x - 1 x intercept intercept When =, = x - 1 When x =, = - 1 x = 1 = -1 1 X Note that a linear function is of one degree, and cuts the x-axis at no more than one point. -1 The gradient of a straight line joining the two points (x 1, 1 ) and (x,, ) is given m = 1 x x From this ou get the equation of the straight line joining the two points - 1 = m(x x 1 ) 1 The gradient of the slope is = tangent of the slope = x x If two straight lines are perpendicular to each other the products of their gradients is 1. The distance etween points (A and B) is given AB = The midpoint of a straight line joining (x 1, 1 ) and (x,, ) is the point 1 x1) ( 1) x 1 x 1 ( x Graphing inequations Eg. x - 1 x intercept intercept When =, = x - 1 When x =, = - 1 x = 1 = -1, 1-1 1 X The shaded area is x - 1
Maths Methods 1. Quadratic Functions Eg. = ax + x + c These graphs when drawn will form a paraola. A quadratic function is a polnomial function of degree, and cuts the x-axis in no more than two points and has no more than one turning point. To sketch the graph of a quadratic function (called a paraola) use the following: If a >, the function has a minimum value If a <, the function has a maximum value The value of c gives the -intercept The equation of the axis of smmetr is x = -, this comes from writing the formula as a 4ac x = - a a Use x = -, to find the turning point a The x axis intercepts are determined solving the equation ax + x + c = The general form of the paraolas are: x x = ax = -ax a x (,a) (, -a) a x = (x a)(x ) = -(x a)(x ) Positive paraola Negative paraola A quadratic can e solved : Factorising Eg. x + 5x 1 = (x 3)(x + 4) = x = 3 or 4 Completing the square Eg x + x 4 = i.e. a = 1, =, c = -4 Add and sutract x + x +(1 1) 4 = (x + 1) 5 = to complete the square.
(x + 1) = 5 x + 1 = 5 x = -1 5 Using the general quadratic formula x = Eg -3x 1x 7 = x = ( 1) ( 1) ( 3) Maths Methods 1 4( 3)( 7) 6 15 x = 3 Using the discriminant of the quadratic function f(x) = ax + x + c If 4ac >, the graph of the function has two x-axis intercepts If 4ac =, the graph of the function touches the x-axis If 4ac <, the graph of the function does not intersect the x-axis a 4ac 4ac a 4ac a 4ac 4ac a a Example 1 Sketch the graph of f(x) = -3x 1x 7 using the quadratic formula to calculate the x-axis intercepts. Since c = -7 the -intercept is (, -7) Turning points coordinates: Axis of smmetr, x = - a ( 1) = - 3 = - (Use this to find the co-ordinates of the turning point) and f() = -3(-) 1(-) 7 = 5 Turning point coordinates (-, 5) - a
x = Maths Methods 1 To calculate the x-axis intercepts: -3x 1x 7 = a 4ac x = = ( 1) 1 6 6 ( 1) ( 3) = 6 15 = 3-3.9 or.71 4( 3)( 7) 1 15 6 6 3.87 3 (-, 5) 6 4 (-3.9, ) (-.71, ) 4-3 - -1 x - -4-6 (, -7) -8 Determining the rule for a function of a graph. All paraolas are in the form of = ax + x + c. It is possile to identif the values of a, and c from the graph. The general form of the paraolas are: (,5) x This is in the form = ax When x = =5 5 = a() a = 4 5 The rule is = 4 5 x
Maths Methods 1 (, 3) (-3,1) The rule is = - 9 x + 3 This is in the form = ax + c, where we know that a <. For (, 3) 3 = a() + c implies c = 3 For (-3, 1) 1 = a(-3) + 3 1 = 9a + 3 a = - 9 (-1, 8) 3 x This is in the form = ax + x For (3, ) = a(3) + 3 = 9a + 3 (1) For (-1, 8) 8 = a(-1) 8 = a () 8 + = a Sustitute into (1) = 9(8 + ) + 3 = 7 + 9 + s -7 = 1 = -6 a = The rule is = x 6x -1 1 x This is in the form = ax + x + c, where we know that a <. For (-1, ) = a + c (1) For (, ) = c () For (1, ) = a + + c (3) Sustitute c = in (1) and (3) = a +
Maths Methods 1 - = a (4) = a + + = a + (5) Sutract (5) from (4) - = - = 1 Sustitute =1 and c = into (1) = a 1 + a = -1 The quadratic rule is = -x + x + 3. Cuic functions Graphs of = ax 3 + x + cx + d A cuic function is a polnomial function of degree 3, and cuts the x-axis in no more than three points and has no more than two turning points. The shapes of cuic graphs var, ut in general the follow the following forms. x x = ax 3 = -ax 3 x - x = ax (x ) = -ax(x + ) (, ac) -a c x -a c x (, -ac) = (x + a)(x )(x c) = -(x + a)(x )(x c) In the equation = ax 3 + x + cx + d, the ax 3 term will dominate, so if a is positive, when x,
Maths Methods 1 Example Sketch the graph of f(x) = x 3 4x 11x + 3 Factorise the expression first using the Factor Theorem. Let x 3 4x 11x + 3 = P(1) = 1 4 11 + 3 P(-1) = -1 4 + 11 + 3 P() = 8 16 + 3 = x is a factor. Dividing x 3 4x 11x + 3 x gives P(x) = (x )(x x 15) = (x )(x 5)(x + 3) x = or x 5 = or x + 3 = x = x = 5 x = -3 (, 3) -3 5 x The turning points are left undefined at this stage. = (x + 3)(x )(x 5) Determining the rule for a function of a graph. Example 3 The graph shown is that of a cuic function. Find the rule for this cuic function. 4-3 1 4 x This graph is in the form = k(x 4)(x 1)(x + 3). For (, 4) 4 = k(-4)(-1)3 k = 3 1 The rule is = 3 1 (x 4)(x 1)(x + 3). Example 4 The graph shown is that of a cuic function. Find the rule for this cuic function. (, 9) -3 1 x
Maths Methods 1 This graph is in the form = k(x 1)(x + 3). Since it onl touches the axis at this point For (, 9) 9 = k(-1)(9) k = -1 The rule is = -(x 1)(x + 3) Example 5 The graph of a cuic function passes through the points with coordinates (, 1), (1, 4), (, 17) and (-1, ). Find the rule for this cuic function. The cuic function must e of the form = ax 3 + x + cx + d For the point (, 1) 1 = + + + d d = 1 For the point (1, 4) 4 = a + + c + 1 3 = a + + c (1) For the point (, 17) 17 = 8a + 4 + c + 1 16 = 8a + 4 + c 8 = 4a + + c () For the point (-1, ) = -a + - c + 1 1 = -a + c (3) Add (1) and (3) = 4 = Sustitute in (1) and () 1 = a + c (4) 8 = 8a + c 4 = 4a + c (5) (5) (4) 3 = 3a a = 1. From (4) c = The rule is = x 3 +x + 1 4. Quartic functions The general form for a quartic function is = ax 4 + x 3 + cx + dx + e The shapes of quartic graphs var, ut in general the follow the following forms. x x Flat ottom = ax 4 = -ax 4
Maths Methods 1 intercept is acd, ecause, here x = = a c d = acd (, acd) -a - c d x -a - c d x (, -acd) = (x + a)(x + )(x c)(x d) = -(x + a)(x )(x c)(x d) The graph will cross the -axis at the point a c d. It will ehave like = ax 4 when x is large. A repeated factor (x a) implies that the graph will touch the x-axis at a. A quartic function is of degree four and cuts the x-axis in, at most, four points and has, at most, three turning points. Example 6 Sketch the graph of the function = (x + )(x + 4)(x )(x 3) If =, then the graph crosses the x-axis at the points (-, ), (-4, ), (, ) and (3, ) If x =, the graph crosses the -axis at the point (, 4 - -3) (, 48) As x gets large the graph ehaves like ax 4. (, 48) -4-3 = (x + )(x + 4)(x )(x 3) Cutting, Touching and Inflexion A polnomial function of n degrees ma: Cut the x-axis in, at most, n points Have, at most, (n 1) turning points. When factorised, if the expression is (x + a) 1, then the function cuts the x-axis at -a. if the expression is (x + a), then the function touches the x-axis at -a. +ve grad if the expression is (x + a) 3, then the point -a will e a point of inflexion. (i.e. a point of zero gradient with out a sign change of the gradient.) zero grad +ve grad
Maths Methods 1 Example 7 Sketch the graph of p(x) = (x 1)(x )(1 x) Use our graphics calculator to do this. Often this is a multiple choice question with a variet of graph for ou to choose from. Example 8 The graph of the function f:r R, defined f(x) = x 4 4x, has A no x-intercepts D three x-intercepts B one x-intercept E four x-intercepts C two x-intercepts The technique to solve this tpe of multiple choice question is to ignore the answers and solve the prolem ourself, and then see which answer is correct ecause it agrees with ou. x 4 4x = x (x 4) = x x(x )(x + ) the factors are x, x, x +, the intercepts are,, -. the answer is D. Example 9 The graph sketched to the right represents a polnomial function of degree 4. The rule could e A = x(x 1) ( x) B = -x(x 1)(x ) C = x(x 1) (x ) D = x (x 1)(x ) E = -x(x + 1) (x + ) It is not a good idea to tr to plot all these graphs on our calculator to tr to find the correct answer, as this will take too much time on the exam. Tr to eliminate as man possiilities as ou can. The graph touches at 1, the (x 1) factor must e squared. This leaves onl A and C. Graph C can e eliminated ecause it is a positive quartic, which means that the graph must tend to + as x +. A
Maths Methods 1 Example 1 Express f(x) = x 3 + x 8x 4 as the product of linear factors. ou can use factor 7 on our calculator to find the factors. The slow wa is x 3 + x 8x 4 = x (x + 1) 4(x + 1) = (x + 1)(x 4) = (x + 1)(x )(x + ) Don t forget to do this last step, the factors must e linear. Example 11 Express f(x) = 6x 4 + 5x 3 6x as the product of linear factors. ou can use factor 7 on our calculator to find the factors. The slow wa is 6x 4 + 5x 3 6x = x (6x + 5x 6) = x (3x )(x + 3) = x x (3x )(x + 3) Don t forget to do this last step, the factors must e linear. Example 1 The linear factors of x 3 4x are A x, x 4 B x, x 4, x - 4 C x, x D x, x + E x, x, x + The technique to solve this tpe of multiple choice question is to ignore the answers and solve the prolem ourself, and then see which answer is correct ecause it agrees with ou. x 3 4x = x(x 4) = x(x )(x + ) the factors are x, x, x +, the answer is E. Example 13 Sketch the graph of f(x) = x (x 9) ou should quickl sketch this on our graphics calculator, to see the answer. When ou are sketching it ou must ensure that ou clearl lael all relevant points on the graph. This means all intercepts, and make sure that ou lael the axes. ou can also think aout the answer, to confirm what the calculator is showing. This is a quartic function so it must have one of the general shapes. From the equation the graph needs to touch at the point (, ), and cross at the solutions for x 9 =. it needs to cross at x = 3. It is also a positive quartic.
Maths Methods 1 Example 14 The graph of the function f: R R, defined f(x) = x 4 x, has A no x-intercepts D three x-intercepts B one x-intercept E four x-intercepts C two x-intercepts To find the intercepts, ou need to factorise the expression. x 4 x = x(x 3 1) = x(x 3 1) = x(x 1)(x + x + 1) check using 4ac = 1 4 = -3 no solutions This has two real solutions, x = and x = 1, for when the function is equal to. C Example 15 Sketch the graph of the function f: R R, f(x) = x ( x)(1 + x), clearl showing all important features. Use our graphics calculator to do this, ut think aout the answer first. If ou multipl the expressions out, ou end up with a negative x 4, as x, -. The x intercepts: Let f(x) =, x =, x =, and x = -1. These are the solutions when f(x) =. It must touch at x =, since it comes from x, so this is a turning point. So the graph looks like this. Example 16 The equation which defines the graph at right could e A = (x 1)(x + 1)(x 9) B = (x 1)(x 3) C = (x 1)(x + 1)(x 3)(x + 3) D = (x 1)(x + 9) E = (x 1)(x + 1)(x + 3) Since it crosses the axis at +1 and 1, the factors are (x 1)(x + 1) = (x 1). It also touches at x = 3, must have factors (x 3) B
Maths Methods 1 Sketching Graphs When ou are sketching graphs, ou must include the following information on ever graph. 1. The x and axes must e laelled. The axes need to e drawn with arrow heads in the positive direction. 3. The origin must e clearl laelled. 4. If the domain is specified in the question then the graph is a line. 5. If the domain is not specified in the question then the graph should have arrowheads on either end. 6. An open circle on the end of the graph is used to show < or >. 7. A closed circle is used to show either or 8. When asked for coordinates, ou must alwas give the answer in coordinate form eg. (x, ). Don t use expressions like, crosses the x axis at 3. 9. All intercepts must e shown with coordinates. Example 17 If f(x) = x, sketch f(x) for the following domains. {x: x R} {x : -3 x 3} x x {x: -3 < x < 3} {x : x } x x {x: x > } x