P1 Chapter 4 :: Graphs & Transformations jfrost@tiffin.kingston.sch.uk www.drfrostmaths.com @DrFrostMaths Last modified: 14 th September 2017
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Chapter Overview There are a few new bits and pieces since GCSE! 1a:: Cubic Graphs Sketch the graph with equation: = 3 2 1b:: Quartic Graphs Sketch the graph with equation: = 1 2 + 1 2 NEW! to A Level 2017+ The old A Level onl included cubic graphs, not quartics. 1c:: Reciprocal Graphs Sketch the graph with equation = 3 2 NEW! to A Level 2017+ In addition to graphs of the form = a, ou now need to recognise the sketch of = a 2 2:: Points of Intersection Sketch the curves = 4 2 and = 2 1 on the same aes. Using our sketch, state, with a reason, the number of real solutions to the equation 4 1 4 = 0. 3:: Graph Transformations If f = 2 ( + 1), sketch the graph of = f( + a), indicating an intercepts with the aes. NEW! since GCSE The GCSE 2015+ sllabus included translations of graphs but not stretches.
Polnomial Graphs In Chapter 2 we briefl saw that a polnomial epression is of the form: a + b + c 2 + d 3 + e 3 + where a, b, c, d, e, are constants (which could be 0). The order of a polnomial is its highest power. Order Name 0 Constant (e.g. 4 ) 1 Linear (e.g. 2 1 ) 2 Quadratic (e.g. 2 + 3 ) 3 Cubic 4 Quartic 5 Quintic These are covered in Chapter 5. Chapter 2 eplored the graphs for these. We will cover these now. While these are technicall beond the A Level sllabus, we will look at how to sketch polnomials in general.
Polnomial Graphs Order: 2 What propert connects the order of the polnomial and the shape? The number of turns is one less than the order, e.g. a cubic has 2 turns, a quartic 3 turns. Bro Note: Actuall this is not strictl true, e.g. consider = 4, which has a U shape. But this is because multiple turns are being squashed into a single point. In Chapter 2 how did we tell what wa up a quadratic is, and wh does this work? 3 4 For a quadratic = a 2 + b + c, i.f. a > 0, we had a valle shape. This is because if was a large positive value, a 2 would be large and positive, thus the graph s value tends towards infinit. We would write: As, where means tends towards.
Polnomial Graphs e.g. If = 2 2 + 3, tr a large positive value like = 1000. We can see we d get a large positive value. Thus as, Resulting Equation If a > 0 Shape If a < 0 Resulting Shape = a 2 + b + c As, As, As, As, = a 3 + b 2 +c + d As, As, As, As, = a 4 + b 3 +c 2 + d + e As, As, As, As, = a 5 + b 4 + As, As, As, As, If a > 0, what therefore can we sa about the shape if: The order is odd: It goes uphill (from left to right) The order is even: The tails go upwards. (And we have the opposite if a < 0)
Cubics Sketch the curve with equation = ( 2)(1 )(1 + ) Sketch the curve with equation = 2 ( 1) Features ou must consider: Shape? If we epanded, 3 term would be negative, so downhill shape. Roots? If = 0, = 2, 1 or 1 -intercept? If = 0, = 2 Fro Tip: No need to epand out the whole thing. Just mentall consider the terms multiplied together. Shape? Roots? This is sort of because the curve crosses at 0 then immediatel crosses at 0 again! -intercept? 3 term is positive so uphill shape. = 0 or = 1 However as root of 0 is repeated (because the factor of appears twice), the curve touches at = 0. If = 0, = 0-1 1 2-2 Fro Tip: It s incredibl eas to forget to write in one of the intercepts. So don t! 1
Cubics Sketch the curve with equation = 2 + 1 2 Sketch the curve with equation = 4 3 Shape? Downhill. Shape? Uphill. Roots? = 2, or = 1 Curve crosses as 2, but touches at -1 (again, because of repeated root) Roots? = 4 But root is triple repeated. We have a point of inflection at = 4. -intercept? 2 -intercept? -64 2-1 2 4 Fro Eam Notes: The term point of inflection has been removed from the new A Level sllabus (bad idea!!). You might be able to see we get this shape at = 4 because as the root is triple repeated, the curve crosses at 4, then crosses again, then crosses again, hence ending up in the same direction and the line becoming momentaril horizontal. -64 A point of inflection is where the curve goes from conve to concave (or vice versa), i.e. curves in one direction before and curves in another direction after. You might have encountered these terms in Phsics.
Cubics with Limited Roots Sketch the curve with equation = + 1 2 + + 1 Shape? Roots? Uphill. Either + 1 = 0 (giving root of -1) or 2 + + 1 = 0. This does not have an solutions as the discriminant is -3. Thus -1 is the onl root. -intercept? 1 We don t have enough information to determine the eact shape. It could for eample have been: -1 1 1 However, in Chapter 12, we ll be able to work turning points using differentiation, and hence conclude that it doesn t have an! -1
Finding the equation ourself Edecel C1 Ma 2013(R) Q9 Figure 1 shows a sketch of the curve C with equation = f(). The curve C passes through the point ( 1, 0) and touches the -ais at the point (2, 0). The curve C has a maimum at the point (0, 4). The equation of the curve C can be written in the form. = 3 + a 2 + b + c where a, b and c are integers. (a) Calculate the values of a, b, c. If it crosses at 1,0 we must have ( + 1). If it touches at 2,0 we must have 2 2 = 2 2 + 1 = 2 4 + 4 + 1 = 3 + 2 4 2 4 + 4 + 4 = 3 3 2 + 4 a = 3, b = 0, c = 4
Test Your Understanding 1 Sketch the curve with equation = 3 2 3 A curve has this shape, touches the ais at 3 and crosses the ais at -2. Give a suitable equation for this graph. = 3 2 + 2 3 N Sketch the curve with equation = 2 2 1 + 1 3 Point of inflection Touches -ais Crosses -ais 2 Sketch the curve with equation = + 2 3-2 -8 (I took this question from m Riemann Zeta Club materials: www.drfrostmaths.com/rzc ) [end shameless plug]
Eercise 4A Pearson Pure Mathematics Year 1/AS Pages 62-63 1 Etension [MAT 2012 1E] Which one of the following equations could possibl have the graph given below? 2 [MAT 2011 1A] A sketch of the graph = 3 2 + 1 appears on which of the following ais? (a) (b) A) = 3 2 3 + 2 1 B) = 2 9 2 3 C) = 6 2 2 + 2 2 D) = 2 1 2 3 Solution: D (c) (d) Cubics can sometimes be factorised b pairing the terms: 2 1 1 1 = 2 1 1 = + 1 1 2 (c)
Recap If we sketched = a on the -ais at: b 2 c 3 what happens = a: The line crosses the ais. a = b: The line touches the ais. b = c: Point of inflection on the ais. c
Quartics If ou understand the principle of sketching polnomials in general, then sketching quartics shouldn t feel like anthing new. Recall that if the 4 term is positive, the tails both go upwards, otherwise downwards. Sketch the curve with equation = ( + 1)( 2)( 3) Shape: Tails upwards Roots: -1, 0, 2, 3 -intercept: 0 Sketch the curve with equation = 2 2 ( + 1)(3 ) Shape: Tails downwards Roots: -1, 2, 3 2 is repeated. -intercept: 4 1 3 = 12-1 2 3-1 2 3
Quartics Sketch the curve with equation Sketch the curve with equation = + 1 1 3 = 2 4-1 root onl appears once so line crosses at = 1 +1 root triple repeated so point of inflection at = 1 2 is a quadruple repeated root! Because the line effectivel crosses the ais 4 times all at -2, it ends up in the opposite direction, and hence looks like a touch point. 16-1 1 2-1
Test Your Understanding Sketch the curve with equation = 2 + 1 1 Sketch the curve with equation = ( + 1) 3 3 27-1 1-1 3
Eercise 4B Pearson Pure Mathematics Year 1/AS Pages 65-66 1 Etension [STEP I 2012 Q2a] a. Sketch = 4 6 2 + 9 b. For what values of b does the equation = 4 6 2 + b have the following number of distinct roots (i) 0, (ii) 1, (iii) 2, (iv) 3, (v) 4. a) B factorising, = 2 3 2. This is a b) quartic, where is alwas positive, and has repeated roots at = ± 3: B changing b, we shift the graph up and down. Then we can see that: i) 0 roots: When b > 9 ii) 1 root: Not possible. iii) 2 roots: When b = 9 or b < 0 iv) 3 roots: b = 0 v) 4 roots: 0 < b < 9
GCSE RECAP :: Reciprocal Graphs Sketch = 1 Sketch = 3 Fro Note: The scaling caused b the 3 isn t observable for this graph in isolation because the aes have no scale. This will onl be observation for multiple graphs on the same aes. Notice the distance between this line and the -ais (i.e. the line = 0) graduall decreases as the lines go off towards infinit. The line = 0 is known as an asmptote of the graph.! An asmptote is a line which the graph approaches but never reaches. Asmptotes of = a : = 0, = 0
Reciprocal Graphs This is new to the A Level 2017 Sketch = 3 Sketch = 4 2 sllabus. 2 Hint: Note that anthing squared will alwas be at least 0.
Reciprocal Graphs On the same aes, sketch = 1 and = 3 = 3 The value for = 3 will be 3 times greater than = 1 = 1
Eercise 4C Pearson Pure Mathematics Year 1/AS Page 67
Points of Intersection In the previous chapter we saw wh the points of intersection of two graphs gave the solutions to the simultaneous equations corresponding to these graphs. If = f() and = g(), then the values of the points of intersection can be found when f = g(). Eample: On the same diagram sketch the curves with equations = ( 3) and = 2 1. Find the coordinates of their points of intersection. = 2 1 1 3 3 = 2 1 2 3 = 2 3 3 3 = 0 2 3 = 0 = 0 or = 3 or = + 3 Substituting these values back into either equation, we obtain points: 3, 3 + 3 3, 0,0, 3, 3 3 3 Froflections: Cubics generall have 3 solutions. And this seems good news as we have 3 points of intersection. Fro Tip: A classic mistake is to divide b to get 2 3 = 0. NEVER divide an equation b a variable, because ou lose a solution. Alwas factorise.
Further eample involving unknown constants On the same diagram sketch the curves with equations = 2 3 a and = b, where a, b are positive constants. State, giving a reason, the number of real solutions to the equation 2 3 a b = 0 If the points of intersection are given b: 2 3 a = b a 3 then clearl: 2 3 a b = 0 There are 2 points of intersection, thus 2 solutions to this equation. If 2 3 a = 0 then = 0 or = a 3 We were told that a is positive, thus this latter root is positive. Fro Note: Note that the question is asking for the number of solutions, not the solutions themselves. We d have to solve a quartic, with roots in terms of a and b. While there is a quartic formula (like the quadratic formula), it is absolutel horrific.
Test Your Understanding On the same diagram sketch the curves with equations = ( 4) and = 2 2, and hence find the coordinates of an points of intersection. = 2 2 Looking at the diagram we epect that 0,0 will be the onl point of intersection (as the cubic will rise more rapidl than the quadratic). But we need to show this algebraicall. = 4 2 4 2 2 = 4 2 4 + 4 = 2 4 3 4 2 + 4 = 2 4 3 5 2 + 8 = 0 2 5 + 8 = 0 Thus = 0 giving (0,0). But the discriminant of 2 5 + 8 is -7, thus there are no further solutions to this equation. Hint: Remember ou can use the discriminant to reason about the number of solutions of a quadratic.
1 Eercise 4D Pearson Pure Mathematics Year 1/AS, Pages 69-71 Etension [MAT 2005 1B] The equation 2 + 1 10 = 2 2 2 A) has = 2 as a solution; B) has no real solutions; C) has an odd number of real solutions; D) has twent real solutions. To sketch = 2 + 1 10, sketch = 2 + 1, then realise that as 2 + 1 is alwas at least 1, raising it to a positive power (>1) makes it go up more steepl. 2 2 2 we can sketch b completing the square, where we realise it is alwas negative. The answer is B. 2 3 [MAT 2010 1A] The values of k for which the line = k intersects the parabola = 1 2 are precisel A) k 0 B) k 4 C) k 0 or k 4 D) 4 k 0 Equating: 1 2 = k 2 2 + 1 = k 2 + 2 k + 1 = 0 Discriminant: a = 1, b = 2 k, c = 1 k 2 + 4k + 4 4 0 k k + 4 0 k 4 or k 0 [MAT 2013 1D] Which of the following sketches is a graph of 4 2 = 2 + 1? 4 = 2 + 2 + 1 = + 1 2 2 = ±( + 1) i.e. = 2 1 or = 2 1 Answer is (b).
Transformations of Functions Suppose f = 2 Then f + 2 = + 2 2 Sketch = f : Sketch = f( + 2) We know = + 2 2 has a root of -2 where the graph touches. -2 What do ou notice about the relationship between the graphs of = f and = f + 2? The graph has been translated 2 b, i.e. we have 0 subtracted 2 from each value.
Transformations of Functions This is all ou need to remember when considering how transforming our function transforms our graph...! Change inside f( ) Change outside f( ) Affects which ais? What we epect or opposite? Opposite What we epect Therefore... = f 3 = f + 4 Translation b Translation b 3 0 0 4 = f 5 Stretch in -direction b scale factor 1 5 = 2f Stretch in -direction b scale factor 2
Sketching transformed graphs = 1 Sketch = 2 + 3 Sketch = 2 +1 If = 2, the +3 is outside the 0 squared function, so translation of 3. Imagine a sketch of = 2 and then do the translation, ensuring ou adjust an intercepts with the aes. 3 This looks like a reciprocal function = 2. The change of +1 is inside the reciprocal function, so we have a translation to the left b 1. 2 The transformation might result in new intercepts or roots. You can find these in the usual wa. Do not forget them! The asmptotes were previousl = 0 and = 0. The latter is unaffected but the former is now = 1. Draw asmptotes using a dotted line and write its equation on it.
More Eamples Sketch = ( + 2). On the same aes, sketch = a a + 2, where a > 2. The input has been replaced with a, i.e. a change inside the function. We translate right b a. The significance of a > 2 is that the original root of -2 will now be positive. Sketch = 2 4. On the same aes, sketch the graph with equation = 2 2 2 4 The input has been doubled to 2, again a change inside the function, so we do the opposite and halve the values. Ensure that 0 remains 0 and ou halve an roots. a a + 2-2 a 2 a 2 4 Note that our intercepts are in terms of a.
Reflections of Graphs If = ( + 2), sketch = f() and = f() on the same aes. You did this at GCSE. The minus is outside the function, so affects the output, i.e. the value. The values are negated, resulting in a reflection in the -ais. = f -2
Test Your Understanding If = ( + 1)( 2), sketch = f() and = f( ) on the same aes. 3 Sketch the graph of = 2 + 1, ensuring ou indicate an intercepts with the aes. = 1-3 -1 2 6-2 -2 To get this new root: 2 + 1 = 0 2 = 1 = 2 1 = 2
Eercise 4E/4F Pearson Pure Mathematics Year 1/AS Pages 74-75 (translations), 78 (stretches/reflections)
Effect of transformation on specific points Sometimes ou will not be given the original function, but will be given a sketch with specific points and features ou need to transform. Where would each of these points end up? = f 4, 3 1, 0 6, 4 = f + 1 3,3 0,0 5, 4 = f 2 2,3 0.5, 0 3, 4 = 3f 4,9 1,0 6, 12 = f 1 4,2 1, 1 6, 5 = f 4 12,3 4,0 24, 4 = f 4,3 1,0 6, 4 = f 4, 3 1,0 6,4
Test Your Understanding Edecel C1 Ma 2012 Q10
Eercise 4G Pearson Pure Mathematics Year 1/AS Pages 80-81