UNCORRECTED. To recognise the rules of a number of common algebraic relations: y = x 1 y 2 = x

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1 5A galler of graphs Objectives To recognise the rules of a number of common algebraic relations: = = = (rectangular hperbola) + = (circle). To be able to sketch the graphs of these relations. To be able to sketch the graphs of simple transformations of these relations. To find the ke features of the graphs of these relations. To determine the rules of relations of these tpes given sufficient information. In Chapter, we looked at linear graphs, sketching them and determining their rules given sufficient information. All linear graphs can be considered as transformations of =. The features we concentrated on for linear graphs were the -ais intercept, the -ais intercept and the gradient. In Chapter, we considered quadratics written in turning point form and sketched their graphs b using transformations of the graph of the basic quadratic =. The features we concentrated on for graphs of quadratic polnomials were the -ais intercepts, the -ais intercept and the coordinates of the turning point (verte). In this chapter, we stud some other common algebraic relations, and develop methods similar to those used in Chapter to sketch the graphs of these relations. The relations in this chapter have different tpes of ke features. For eample, we introduce asmptotes for graphs of rectangular hperbolas, and the coordinates of the centre and the length of the radius are ke features in the stud of circles. Chapter 5 Knowledge check See the online test of required knowledge, with links to revision lessons. Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

2 66 Chapter 5: A galler of graphs 5A Rectangular hperbolas Consider the rule = = for We can construct a table of values for = for values of between 4 and 4 as follows: We can plot these points and then connect the dots to produce a continuous curve. A graph of this tpe is an eample of a rectangular hperbola. Note that is undefined when =, and that there is no -value that will produce the value =. Asmptotes There are two lines associated with this graph that help to describe its shape. Horizontal asmptote 4 = 4 From the graph we see that, as approaches infinit in either direction, the value of approaches zero. The following notation will be used to state this: As, +. This is read: As approaches infinit, approaches from the positive side. As,. This is read: As approaches negative infinit, approaches from the negative side. The graph approaches the -ais (the line = ) but does not cross this line. The line = is a horizontal asmptote. Vertical asmptote As approaches zero from either direction, the magnitude of becomes ver large. The following notation will be used to state this: As +,. This is read: As approaches zero from the positive side, approaches infinit. As,. This is read: As approaches zero from the negative side, approaches negative infinit. The graph approaches the -ais (the line = ) but does not cross this line. The line = is a vertical asmptote. 4 Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

3 Dilations from an ais The diagram on the right shows the graphs of =, = and = The asmptotes are the -ais and the -ais, and the have equations = and = respectivel. As can be seen from the diagram, the graphs of = and = have the same shape and asmptotes as the graph of =, but the have been stretched. = 4 5A Rectangular hperbolas 67 (, ) (, ), 4, (, ) = (, ) The transformation that takes the graph of = to the graph of = is called the dilation of factor from the -ais. For eample, the point (, ) on the graph of = is taken to the point (, ) on the graph of =. Dilations will be considered formall in Chapter 7. Reflection in the -ais When the graph of = is reflected in the -ais, the result is the graph of =. The asmptotes are still the two aes, that is, the lines = and =. Similarl, = is the reflection of = in the -ais. Reflecting in the -ais gives the same result for these two graphs. Translations Now let us consider the graph of = +. The basic graph of = has been translated unit to the right and units up. Asmptotes The equation of the vertical asmptote is now =, and the equation of the horizontal asmptote is now =. Intercepts with the aes The graph now has -ais and -ais intercepts. These can be calculated in the usual wa to add further detail to the graph. = = + (, ) = (, ) = Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

4 68 Chapter 5: A galler of graphs Sketching rectangular hperbolas Using dilations, reflections and translations, we are now able to sketch the graphs of all rectangular hperbolas of the form = a h + k. Eample Sketch the graph of = +. = 4 Eample Sketch the graph of = 4 6, =. Eplanation The graph of = has been translated unit to the left and units down. The asmptotes have equations = and =. When =, = + =. the -ais intercept is. When =, = + = + ( + ) = = the -ais intercept is. The graph of = is obtained from the graph of = b reflection in the -ais. This graph is then translated unit to the right to obtain the graph of =. = = = (, ) = Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

5 5A 5A Rectangular hperbolas 69 Section summar Eample, For a >, a dilation of factor a from the -ais transforms the graph of = to the graph of = a. A reflection in the -ais transforms the graph of = a to the graph of = a. For h, k, a translation of h to the right and k upwards transforms the graph of = a to the graph of = a h + k. A rectangular hperbola with rule of the form = a h + k has: vertical asmptote = h horizontal asmptote = k. Eercise 5A Sketch the graphs of the following, showing all important features of the graphs: a = b = c = d = e = + f = g = 4 h = + 5 i = j = = + k + + l = 4 Write down the equations of the asmptotes for each of the graphs in Question. a We can write = + 6 as = ( + ). Sketch the graph of = and hence the graph of = + 6. b We can write = + 4 as = ( + ). Sketch the graph of = and hence the graph of = + 4. c We can write = + 4 as = ( + ). Sketch the graph of = and hence the graph of = + 4. d We can write = + as = ( + ). Sketch the graph of = and hence the graph of = +. 4 Sketch the graphs of the following, showing all important features of the graphs: a = = + b + c = + d = + e = + 4 f = + + g = + = h + 4 SF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

6 7 Chapter 5: A galler of graphs 5A 5 Show that + = and hence sketch the graph of =. 6 Show that + + = + + and hence sketch the graph of = Show that = and hence sketch the graph of =. 5B The graph of = Now consider the rule =. We can construct a table of values for between 4 and 4: We plot these points with against and then connect the dots to produce a continuous curve. The graph of = is a parabola. It can be obtained from the graph of = b a reflection in the line =. The verte of the parabola is at (, ), and the ais of smmetr is the -ais. 4 4 = The transformations considered in the previous section can be applied to the graph of =. All graphs of the form ( k) = a ( h) will have the same basic parabola shape. The verte of the parabola will be at the point (h, k), and the ais of smmetr will be the line = k. Eample Sketch the graph of: a ( 4) = + b + = + a (, 4) Eplanation The graph of = is translated units to the left and 4 units up. The verte is (, 4). When =, ( 4) = 4 = ± When =, 6 = + = = 4 ± CF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

7 5B 5C The graph of = 7 Eample b Complete the square: (, ) + = = + 4 ( + ) = ( + ) Eercise 5B The graph of ( + ) = ( + ) is obtained from the graph of = b a dilation of factor from the -ais and then a translation units to the left and unit down. The verte has coordinates (, ). When =, ( + ) = 4 + = ± When =, = + 4 = = ± = or = Sketch the graph of each of the following relations, showing all important features: a ( ) = b ( + ) = + 4 d = ( + 5) e ( 4) = ( + ) c f = ( + 4) = g ( + ) = 4 h = i + 4 = + 4 j = k + = l = m = n + = 5C The graph of = The rule = = for corresponds to the upper part of the graph shown opposite. It is one arm of the parabola =. = Coordinates of points on the graph of = include (, ), (, ), (4, ) and (9, ). All graphs of the form = a h + k will have the same basic shape as the graph of =. (, ) (4, ) = = SF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

8 7 Chapter 5: A galler of graphs Eample 4 Sketch the graph of = + 6. (, 6) = Ö + 6 (, ) When =, = (, ) When =, + 6 = Eample 5 + = 6 + = + = 4 = Sketch the graph of = +. When = : (, ) Square both sides:, 4 + = = 4( ) = 9 Eplanation The graph is formed b dilating the graph of = from the -ais b factor and then translating unit to the left and 6 units down. The rule is defined for. The set of values the rule can take (the range) is all numbers greater than or equal to 6, i.e. 6. Eplanation The graph is formed b dilating the graph of = from the -ais b factor, reflecting this in the -ais and then translating it unit to the right and units up. The rule is defined for. The set of values the rule can take (the range) is all numbers less than or equal to, i.e.. Therefore = = 4 Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

9 5C 5C The graph of = 7 The graph of = The rule = for ields a graph which is the reflection of the graph of = in the -ais. All graphs of the form = a ( h) + k will have the same basic shape as the graph of =. Eample 6 Sketch the graph of = +. Note: = ( ) When =, = +. Section summar (, + ) (, ) Eplanation = We can write the rule as = ( ) + The rule is defined for. The set of values the rule can take (the range) is all numbers greater than or equal to, i.e.. All graphs of the form = a h + k will have the same basic shape as the graph of =. The graph will have endpoint (h, k). Eample 4 The graph of = is the reflection in the -ais of the graph of =. Eercise 5C For each of the following rules, sketch the corresponding graph, giving the ais intercepts when the eist, the set of -values for which the rule is defined and the set of -values which the rule takes: a = + b = + c = d = + + e = + + f = + SF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

10 74 Chapter 5: A galler of graphs 5C Eample 5, 6 For each of the following rules, sketch the corresponding graph, giving the ais intercepts when the eist, the set of -values for which the rule is defined and the set of -values which the rule takes: a = + b = ( 4) c = ( + 4) d = e = f = 4 4 For each of the following rules, sketch the corresponding graph, giving the ais intercepts when the eist, the set of -values for which the rule is defined and the set of -values which the rule takes: 5D Circles a = b = ( ) c = d = ( ) e = 4( ) f = 4 ( ) 4 Consider a circle in the coordinate plane with centre the origin and radius r. If P(, ) is a point on the circle, its distance from the origin is r and so b Pthagoras theorem + = r. Conversel, if a point P(, ) in the plane satisfies the equation + = r, its distance from the origin is r, so it lies on a circle with centre the origin and radius r. To the right is the graph of the circle with equation + =. All circles can be considered as being transformations of this basic graph. As has been seen with other graphs, the basic graph ma be translated horizontall and verticall. The equation for a circle is ( h) + ( k) = r where the centre of the circle is the point (h, k) and the radius is r. If the radius and the coordinates of the centre of the circle are given, the equation of the circle can be determined. SF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

11 5D Circles 75 Eample 7 Write down the equation of the circle with centre (, 5) and radius. If the radius is and the centre is the point (, 5), then the equation will be ( ( )) + ( 5) = 4 ( + ) + ( 5) = 4 If the equation of the circle is given, the radius and the centre of the circle can be determined and the graph sketched. Eample 8 Find the centre and radius of the circle ( ) + ( ) = 4. The equation ( ) + ( ) = 4 defines a circle of radius with centre at (, ). Eample 9 Eplanation We can sketch the circle with a little etra work. When =, + ( ) = 4 ( ) = Hence = ± Sketch the graph of the circle ( + ) + ( + 4) = 9. When =, + ( + 4) = 9 ( + 4) = 8 Hence = 4 ± 8 = 4 ± 4 (, 4 + ) 4 (, 4 ) (, ) 4 (, + ) Eplanation The circle has radius and centre (, 4). The -ais intercepts can be found in the usual wa. 4 Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

12 76 Chapter 5: A galler of graphs The equation of a circle ma not alwas be written in the form ( h) + ( k) = r. Epanding the general equation of a circle gives ( h) + ( k) = r h + h + k + k = r + h k + h + k r = Let c = h + k r. Then we obtain an alternative form for the equation of a circle: The general form for the equation of a circle is + h k + c = You will note that there is some similarit with the general form of a straight line, a + b + c =. Notice that in the general form of the circle equation, the coefficients of and are both and there is no term. In order to sketch a circle with equation epressed in this form, the equation can be converted to the centre radius form b completing the square for both and. Eample Find the radius and the coordinates of the centre of the circle with equation = and hence sketch the graph. B completing the square for both and we have = ( 6 + 9) 9 + ( ) 4 = ( 6 + 9) + ( ) = 5 ( ) + ( + ) = 5 The radius is 5 and the centre is at (, ). 8 7 Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

13 Semicircles 5D Circles 77 Transposing the general equation of the circle + = r to make the subject, we have = r = ± r We can now consider two separate rules = + r and = r which correspond to the top half and bottom half of the circle respectivel. Similarl, solving for will give ou the semicircles to the left and right of the -ais: = ± r Eample Sketch the graphs of: a = + 4 b = 4 c = 4 d = + 4 a = + 4 b = 4 c = 4 d = + 4 Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

14 78 Chapter 5: A galler of graphs Eample Sketch the graph of = + 49 ( ). 5 When =, ( 5, ) (, ) (9, ) = + 45 = + 5 When =, + 49 ( ) = 49 ( ) = 49 ( ) = 4 Section summar ( ) = 45 = ± 5 Eplanation The equation of a circle with centre (h, k) and radius r is ( h) + ( k) = r The general form for the equation of a circle is + h k + c = It is a semicircle of the circle ( ) + ( + ) = 49 The centre is at the point (, ) and the radius is 7. It is the semicircle = 49 translated units the right and units down. In the usual wa, we find the -ais intercepts and the -ais intercept. The two separate rules for semicircles with their base on the -ais are = + r and = r The correspond to the top half and bottom half of the circle respectivel. The two separate rules for semicircles with their base on the -ais are = + r and = r The correspond to the right half and left half of the circle respectivel. Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

15 5D 5D Circles 79 Skillsheet Eercise 5D Write down the equation of each of the following circles, with centre at C(h, k) and radius r: Eample 7 a C(, ), r = b C(, ), r = 4 c C(, ), r = 5 Eample 8 Eample 9 Eample Eample Eample Eample d C(, 4), r = e C(, 4), r = 5 f C( 5, 6), r = 4.6 Find the centre, C, and the radius, r, of the following circles: a ( ) + ( ) = 4 b ( ) + ( + 4) = 5 c ( + ) + ( ) = 9 d ( + 5) + ( 4) = 8 Sketch the graphs of each of the following: a + = 64 b + ( 4) = 9 c ( + ) + = 5 d ( + ) + ( 4) 69 = e ( ) + ( 5) = 6 f ( + 5) + ( 5) = 6 Find the centre, C, and the radius, r, of the following circles: a = b = c = d = e = f = Sketch the graphs of each of the following: a = b = c = Sketch the graphs of each of the following: a = + 9 b = + 9 c = 6 d = 5 = 5 e 49 f = 4 Sketch the graphs of each of the following: a = 6 ( ) b = 4 ( + ) 8 The graph of + 9 is as shown. Note that (, ) satisfies + 9. The coordinates of ever point in the shaded region satisf the inequalit. Sketch the graphs of each of the following. Use a dotted line to indicate that the boundar is not included. a + 4 b + > c + 5 d + > 9 e + 6 f + < 8 (, ) SF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

16 8 Chapter 5: A galler of graphs 5E Determining rules In Chapters and we looked at some sufficient conditions for determining the rules for straight lines and parabolas. For straight lines these included: the coordinates of two points the gradient and a point. For parabolas these included: the coordinates of three points the coordinates of the verte and the coordinates of one other point. In this section we are looking at some sufficient conditions for determining the rules for the graphs of this chapter. Eample a The rectangular hperbola = a + 8 passes through the point (, 6). Find the value of a. b The rectangular hperbola = a + k passes through the points (, 7) and (, ). Find the values of a and k. a When =, = 6. Hence 6 = a + 8 = a a = 4 The equation is = b When =, = 7. When =, =. So we have the equations 7 = a + k () = a + k () Subtract () from (): 6 = a + a () Multipl both sides of equation () b : = a + a a = 4 From equation (): k = 5. The equation is = Eplanation The general technique is to substitute the given values into the general equation = a h + k In this case h = and k = 8. The general technique is to substitute the given values into the general equation = a h + k In this case h = and the values of a and k are unknown. Simultaneous equations need to be formed and then solved. Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

17 5E Determining rules 8 Eample 4 A graph which has rule = a h passes through the points (4, ) and (7, 4). Find the values of a and h. When = 4, =. When = 7, = 4. We have the equations = a 4 h () 4 = a 7 h () Divide () b (): 7 h = () 4 h Multipl both sides of equation () b 4 h: 4 h = 7 h Square both sides of the equation: 4(4 h) = 7 h 6 4h = 7 h h = 9 h = Substitute in () to find a =. The required equation is =. Eample 5 Eplanation The general technique is to substitute the given values into the general equation = a h + k In this case k = and the values of a and h are unknown. Simultaneous equations need to be formed and then solved. Note that h 4 from equation (). Find the equation of the circle whose centre is at the point (, ) and which passes through the point (4, ). Let r be the length of the radius. Then r = (4 ) + ( ( )) = + 4 = 5 Hence the equation of the circle is ( ) + ( + ) = 5 Eplanation We use the centre radius form for the equation of a circle: ( h) + ( k) = r The centre is (, ). We need to find the radius. Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

18 8 Chapter 5: A galler of graphs 5E Skillsheet Eample a Eample b Eample 4 Eample 5 Eercise 5E The rectangular hperbola = a + passes through the point (, 8). Find the value of a. A rectangular hperbola with rule of the form = a h + k has vertical asmptote =, horizontal asmptote = 4 and passes through the point (, 6). Find the values of a, h and k. The rectangular hperbola = a + k passes through the points (, 8) and (, 7). Find the values of a and k. 4 A rectangular hperbola with rule of the form = a h + k has vertical asmptote =, horizontal asmptote = 4 and passes through the point (, 4). Find the values of a, h and k. 5 A graph which has rule = a passes through the point (, 8). Find the value of a. 6 7 A graph which has rule = a h passes through the points (, ) and (, 4). Find the values of a and h. Find the equation of the circle whose centre is at the point (, ) and which passes through the point (4, ). 8 Find the equation of the circle whose centre is at the point (, ) and which passes through the point (, ). 9 Find the equation of the circle whose centre is at the point (, ) and which passes through the point (, ). Find the equation of the circle with centre (, ) which touches the -ais. Find the equation of the circle whose centre lies on the line = 4 and which passes through the points (, ) and (6, ). Find the equations of the circles which touch the -ais, have radius 5 and pass through the point (, 8). Find the equation of a circle which passes through the points A(, ), B(, ) and C( 4, ). SF CF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

19 5E 5E Determining rules 8 4 Find the rule for each of the following graphs. The general form of the rule is given for each graph. a ( h) + ( k) = r ( 5, ) (, ) Centre at (, ) b = a h + k (, ) (, ) c = a h + k d = a h + k = = (, ) = = e = a h + k f ( k) = b( h) (, ) (, ) (, ) ( 9, ) (, 5) CF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

20 84 Chapter 5: A galler of graphs Review AS Nrich Chapter summar The standard graphs: = Rectangular hperbola = + = Circle = Dilations of these graphs: = (, ) = (, ) = = = = = + = + = 4 Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

21 Chapter 5 review 85 Reflections in the aes: = = Translations of graphs: = + = = (, ) Equation for a circle with centre at (h, k) and radius r: ( h) + ( k) = r Alternative form: + h k + c = where c = h + k r. Technolog-free questions = = ( ) + ( ) = 4 Sketch the graphs of each of the following: a = b = = c + + d = e = 4 + f = g ( ) = + h = + i = + j = + + B completing the square, write each of the following equations in the form ( a) + ( b) = r : a = b = c = d = e + = 6( + ) f + = 4 6 For the circle = 4, find the equation of the diameter which passes through the origin. SF CF Review Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

22 86 Chapter 5: A galler of graphs Review 4 For the circle + + = 6, find the equation of the diameter which cuts the -ais at an angle of Find the equation of the circle with centre C and radius r for each of the following and sketch the graph: a C(, 4), r = 5 b C(, ), r = c C(4, 4), r = d C(, ), r = 6 6 The equation of a circle is =. Find the centre and radius. 7 Find the length cut off on the -ais and -ais b the circle + 4 =. 8 Sketch the graphs of the following semicircles: a = 9 b = 6 ( + ) c = d + = 4 ( + ) Multiple-choice questions The circle with equation ( a) + ( b) = 6 has its centre on the -ais and passes through the point with coordinates (6, 6). The values of a and b are A a = and b = 6 B a = and b = C a = and b = D a = 6 and b = E a = 6 and b = The equations of the asmptotes of the graph of = 5 5 are A = 5, = B = 5, = 5 C = 5, = 5 5 D = 5, = E = 5, = 5 5 For the rule = 5 +, when = a, = A 5 a + D 5 a + E a + B 5a + a + C a 4 If the -ais is an ais of smmetr and the circle passes through the origin and (, 4), the equation of the circle shown is A + ( ) = 4 B ( ) + = C ( + ) + = 4 D + ( + ) = 4 E + = 4 4 CF SF CF SF Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

23 Chapter 5 review 87 5 The equations of the asmptotes of the graph of = + 4 are A =, = 4 B =, = 4 C =, = D =, = 4 E = 4, = 6 For the circle with equation ( 5) + ( + ) = 9, the coordinates of the centre and the radius are A ( 5, ) and B ( 5, ) and 9 C (5, ) and 9 D (5, ) and E (, 5) and 7 For the rule = +, where, the range of possible -values is A B > C D E > 8 The equation of the circle which has a diameter with endpoints at (, 8) and (6, 8) is A ( + ) + ( 8) = 6 B ( + ) + ( + 8) = 64 C ( ) + ( 8) = 6 D ( ) + ( + 8) = 4 E ( ) + ( + 8) = 6 9 Which of the following is the equation for the graph of a circle? A = 6 B = 6 + C + = 6 D = 6 E = 6 The equation of the semicircle shown is A + ( ) = 9 B = 9 + C = 9 + D = 9 E = 9 + Etended-response questions The following questions also involve techniques developed in Chapters and. The line with equation = m is tangent to the circle with centre (, ) and radius 5 at the point P(, ). a Find the equation of the circle. b Show that the -coordinate of the point P satisfies the equation ( + m ) + 75 =. c Use the discriminant for this equation to find the eact value of m. d Find the coordinates of P. (There are two such points.) e Find the distance of P from the origin. 6 CU Review Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4

24 88 Chapter 5: A galler of graphs Review A circle has its centre at the origin and radius 4. a Find the equation of this circle. b Two lines which pass through the point (8, ) are tangents to this circle. i Show that the equations of these tangents are of the form = m 8m. ii Use techniques similar to those used in Question to find the value of m and, hence, the equations of the tangents. A circle has centre at the origin and radius 5. The point P(, 4) lies on the circle. a Find the gradient of the line segment OP, where O is the origin. b Find the gradient of the tangent to the circle at P. c Find the equation of the tangent at P. d If the tangent crosses the -ais at A and the -ais at B, find the length of line segment AB. 4 Let P(, ) be a point on the circle with equation + = a. a i Give the gradient of the line segment OP, where O is the origin. ii Give the gradient of the tangent to the circle at P. b Show that the equation of the tangent at P(, ) is + = a. c If = and a = 4, find the equations of the possible tangents. 5 An equilateral triangle ABC circumscribes the circle with equation + = a. The side BC of the triangle has equation = a. a Find the equations of AB and AC. b Find the equation of the circle circumscribing triangle ABC. 6 Consider the curve with equation = b + c. a Show that if the curve meets the line with equation = at the point (a, a), then a satisfies the equation a (c + )a + c + b =. b i If the line with equation = is a tangent to the curve, show that c = 4b. 4 ii Sketch the graph of = 4 and find the coordinates of the point on the graph at which the line with equation = is a tangent. c Find the values of k for which the line with equation = + k: i meets the curve with equation = 4 twice ii meets the curve with equation = 4 once iii does not meet the curve with equation = 4. 7 For the curve with equation = and the straight line with equation = k, find the values of k such that: a the line meets the curve twice b the line meets the curve once. CU Uncorrected rd sample pages Cambridge Universit Press Evans, et al Ph 867 4