Review Eercise a Use m m a a, so a a a Use c c 6 5 ( a ) 5 a First find Use a 5 m n m n m a m ( a ) or ( a) 5 5 65 m n m a n m a m a a n m or m n (Use a a a ) cancelling y 6 ecause n n ( 5) ( 5)( 5) ( 5) 5( 5) 6 5 5+ 5 5 and c ( ) a ( + )( ) ( ) + ( ) 6 + 6( ) + ( + )( ) 6( ) 6 ( ) and ( a ) 5 a mean k + + 5 k + k + 6 k + k Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.
5 range + 5 k ( k) + 6 k 6 a ( ) y 5 5 5 5 5y 5 5 5( 5 ) 7 Area h(a + ) + + 5+ ( ) + + The area of the trapezium is + cm. p+ q p q ( ) + ( ) ( ) ( ) 5 (5 ) (+ ) ( ) (+ ) 5+ 5 6 + + ( m, n ) 9 a + 6 ( )( ) Let y y ( y ) + 6 ( y )( y ) So y or y y or y a 6 Complete the square for 9 9 6 9 ( ) ( a and 5) 9 ( ) 5 5 Use the result from part a: ( ) 5 Take the square root of oth sides: ± 5 ± 5 5 9 5 5 since ( a) a Roots are ± 5 c and d ± f(a) a(a ) and g(a) a + 5 a(a ) a + 5 a a a 5 a a 5 Using the quadratic formula: ± ( ) ()( 5) a () ± 9.9 or.9 As a >, a.9 ( s.f.) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.
a The height of the athlete's shoulder aove the ground is.7 m.7 + t 5t Using the quadratic formula when a 5, and c.7 For, y P is the point (, ). The graph of y 6+ is a shape. For y a + + c, if a >, the shape is. ± t ( 5) ±.6 or.6 As t >, t.6 s ( s.f.) () ( 5)(.7) c.7 + t 5t.7 5(t t).7 5((t ) ).7 5(t ) + 5 6.7 5(t ) A 6.7, B 5 and C d Maimum when (t ), t s and maimum height 6.7 m f 6+ a 6 9 Complete the square for 6+ 6+ 9 + ( ) + 9 a and 9 y 6 + y ( ) + 9 ( ) Squaring a numer cannot give a negative result. The minimum value of ( ) is, when. So the minimum value of y is + 9, when. Q is the point (, 9). The curve crosses the y-ais where. Use the information aout P and Q to sketch the curve, so the part where < is not needed. c y + 9 Put y into the equation of C. ( ) + 9 Sutract 9 from oth sides. ( ) Take the square root of oth sides. ± ± 6 using ( a) a ± -coordinate of R is + The other value is which is less than, so is not needed. a Using the discriminant ac for equal roots ( ) ()( k) k k Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.
y + + ( + ) When y, When, y ( + ) 5 a g() 9 7 6 ( 6 7 ) To factorise 6 7, let y y 7y (y + )(y ) So g() ( + )( ) a, 6 c For equal roots, ac k k k 5 d The graph of + + 5 is a shape. For y a + + c, if a >, the shape is. : y + + 5 5 Meets y-ais at (, 5). y : + + 5 ( + 5)( + 5) 5 Meets -ais at ( 5, ). g() ( + )( ), or, or 6 a + + 6 + ( + 5) 5 Complete the square for + + 6 + + 6 5 5 + 6 ( ) + 5 + a 5 and + + 6 ( + 5) + Hence implied in part a must e used ( + 5) A real numer squared cannot e negative. There are no real roots. c + + k a,, c k 7 a The graph meets the -ais at just one point, so it touches the -ais. + + + ( + ) Complete the square for + + + + ( + ) + ( + ) + a and The graph of y + + is a shape. For y a + + c, if a >, the shape is. : y + + Put to find the intersection with the y-ais: Meets y-ais at (, ). Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.
7 Put y to find the intersection with the -ais: y : + + ( + ) + ( + ) A real numer squared cannot e negative, therefore, no real roots, so not intersection with the -ais. c + + a,, c ac Since the discriminant is negative, the equation has no real roots, so the graph does not cross the -ais. d + k + a, k, c For no real roots, < ac k < k < k+ k < This is a quadratic inequality with critical values and. 7 d The surds can e simplified using ( a) a < k < a ( ) + + + + + ( ) ( ) + + + ± ± ± a and Using y : y ± 6± Solution: ± y 6± 9 a ( ) + > 5 6+ > 5 6+ + > 5 > > Critical values: k, k < k < 7+ ( )( ) ( ) or Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 5
9 c 7+ > where < or > The equation in part a could e written as y 5 and y 6. Therefore, the solution to 6 5 are the same as for part a. These are the critical values for 6 > 5 : and. 6 > 5 6 5 + > > < < or > ( + ) 5 + 5 + + Using the discriminant ac ( ) ()() 7 As ac <, there are no real roots. Hence there is no value of for which p() q(). a y 5 5 6 5 + 6 ( )( ) 7 +, : y 5 7 : y 5 + 6 < or > a k ( k ) + + + a, k, c k+ > ac k > k+ k k > k+ k > k k ( k+ )( k 6) k, k 6 Solution, y and, y k k k k > where < or > 6 Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 6
6 + 5 < Multiply oth sides y ( + 5) 6( + 5) < ( + 5) 6 + < + + 5 + + > Solve the quadratic to find the critical values. + + ( + 7 + ) ( + 5)( + ) 5 or a 5 a ( ) ( )( ) + The solution is < 5 or >. a 9 ( + )( ) or When, y 9 To work out the points of intersection, solve the equations simultaneously. 9 6 6 + 5 ( 5)( ) or 5 Curve crosses the -ais where y +,, When, y When, y When, y When, y When 5, y 6 Let 6 7 6 The line crosses the -ais at 7,. c Crosses the y-ais at (, ). Crosses the -ais at (, ), (, ). y y This is a translation of + in the -direction. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 7
5 c 7 6 a Crosses the -ais at (, ), (, ) and (, ). c Meets the y-ais at (, 6). Meets the -ais at (, ), (, ). 6 Crosses the -ais at (, ), (, ). Image of P is (, ). a Meets the y-ais at (, ). Meets the -ais at (, ), (, ). 7 a Crosses the -ais at (, ), (, ).,. Image of P is Meets the y-ais at (, ). Meets the -ais at (, ), (, ). y is an asymptote. is an asymptote. The graph does not cross the y-ais (see sketch in part a). Crosses the -ais where y : +,, Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.
9 a ( 5 + )( 5 + ) For 5 + Using the quadratic formula: 5 ± ( 5) ()() ` () 5 ± 7.56 or. For 5 + ( )( ) or.,, or.56 When, y a The graphs intersect at point, so have point of intersection. a y is a translation y of a y f() is a reflection in the -ais of y f(), so P is transformed to (6, ). a y f( ) is a translation units to the right of y f(), so P is transformed to (9, ). c y f() is y f() which is a vertical stretch scale factor of y f(), so P is transformed to (6, ). a k y is the curve y, k < y ( ) is a translation, units to the right of the curve y When, y When y, When y ( + k) passes through the origin, and y. So k k k ± Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 9
Challenge a + 9 ( )( 9) or 9 ( ) + Multiply y + 9 Let y y y + 9 Using your answers from part a y or 9 or 9 or Let and y e the length and width of the rectangle. Area y 6 Perimeter + y y y Solving simultaneously ( ) 6 + 6 Using the quadratic formula ± ( ) ()(6) ± ± or Solving simultaneously + ( )( + ) + ( ) + + + ( + + ) The discriminant of + + ac ()() < so there are no real solutions for + + The only solution is when at (, ). f() ( + )( + ) ( + 5)( )( + )( ) when f() 5,, or g( k) ( k + 5)( k ) ( k + )( k ) When k, g( ) ( + )( 7)( )( ) ( + ), ( ) and ( ) match When k, g( + ) ( + )( )( + 5)( + ) ( ), ( + 5) and ( + ) match So k or. When, y When, y The dimensions of the rectangle are cm and cm. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.