Higher. Polynomials and Quadratics. Polynomials and Quadratics 1

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1 Higher Mathematics Polnomials and Quadratics Contents Polnomials and Quadratics 1 1 Quadratics EF 1 The Discriminant EF Completing the Square EF Sketching Paraolas EF 7 5 Determining the Equation of a Paraola RC 9 6 Solving Quadratic Inequalities RC 11 7 Intersections of Lines and Paraolas RC 1 8 Polnomials RC 1 9 Snthetic Division RC 15 1 Finding Unknown Coefficients RC Finding Intersections of Curves RC 1 1 Determining the Equation of a Curve RC This document was produced speciall for the HSN.uk.net wesite, and we require that an copies or derivative works attriute the work to Higher Still Notes. For more details aout the copright on these notes, please see

2 Polnomials and Quadratics Polnomials and Quadratics 1 Quadratics EF A quadratic has the form provided a. You should alread e familiar with the following. a + + c where a,, and c are an real numers, The graph of a quadratic is called a paraola. There are two possile shapes: concave up (if a > ) concave down (if a < ) This has a minimum turning point This has a maimum turning point To find the roots (i.e. solutions) of the quadratic equation we can use: factorisation; completing the square (see Section ); the quadratic formula: EXAMPLES 1. Find the roots of = ( + 1)( ) =. Solve + + =, a c ± ac = (this is not given in the eam). a =. + 1 = or = = 1 =. ( )( ) = = + + = + = or + = = =. Page 1

3 Polnomials and Quadratics. Find the roots of + 1=. We cannot factorise + 1, ut we can use the quadratic formula: ± 1 ( 1) = 1 ± 16 + = ± = 5 = ± = ± 5. Note If there are two distinct solutions, the curve intersects the -ais twice. If there is one repeated solution, the turning point lies on the -ais. If < when using the quadratic formula, there are no points where the curve intersects the -ais. Page

4 Polnomials and Quadratics The Discriminant EF Given a + + c, we call the discriminant. This is the part of the quadratic formula which determines the numer of real roots of the equation a + + c =. If If If >, the roots are real and unequal (distinct). =, the roots are real and equal (i.e. a repeated root). <, the roots are not real; the paraola does not cross the -ais. EXAMPLE 1. Find the nature of the roots of a = 9 = c = 16 Since =. = 9 16 = = =, the roots are real and equal.. Find the values of q such that Since a = 6 = 1 c = q 6 1 q 6 1 q + + = has real roots, ac q q 1 q q 1 q = has real roots. ac : two roots one root no real roots Page

5 Polnomials and Quadratics. Find the range of values of k for which the equation no real roots. k + 7= has For no real roots, we need a = k = c = 7 < : < k ( 7) < + 8k < 8k < k < 8 k <.. Show that ( k ) ( k ) ( k ) = has real roots for all real values of k. a = k+ = k+ c = k 1 7 ( k ) ( k )( k ) 9k 1k ( k )( k 8) = + + = = k + k+ k = + + k ( k ) 1k 6 = + 6. Since ( ) = k + 6, the roots are alwas real. Completing the Square EF The process of writing a c called completing the square. = + + in the form ( ) = a + p + q is Once in completed square form we can determine the turning point of an paraola, including those with no real roots. The ais of smmetr is = p and the turning point is ( pq, ). + p = + p + p. For eample, we The process relies on the fact that ( ) can write the epression + using the racket ( + ) since when multiplied out this gives the terms we want with an etra constant term. Page

6 Polnomials and Quadratics This means we can rewrite the epression gives us the correct and terms, with an etra constant. We will use this to help complete the square for Step 1 Make sure the equation is in the form = a + + c. Step Take out the -coefficient as a factor of the and terms. Step Replace the + k epression and compensate for the etra constant. + k using ( + k ) since this = + 1. = + 1. ( ) = +. = ( ( + ) ) = ( + ) 1. Step Collect together the constant terms. = ( + ) 15. Now that we have completed the square, we can see that the paraola with equation = + 1 has turning point (, 15). EXAMPLES 1. Write = in the form = ( + p) + q. = = ( + ) 9 5 = ( + ) 1. Note You can alwas check our answer epanding the rackets.. Write + ( ) ( ) + in the form ( + p) + q. = + 9 = + 5. Page 5

7 Polnomials and Quadratics. Write = + 8 in the form = ( + a) + and then state: (i) the ais of smmetr, and (ii) the minimum turning point of the paraola with this equation. = + 8 ( ) ( ) = + 16 = (i) The ais of smmetr is =. (ii) The minimum turning point is (, 19).. A paraola has equation = (a) Epress the equation in the form = ( + a) +. () State the turning point of the paraola and its nature. (a) 1 7 = + ( ) = + 7 (( 9 ) ) ( ) ( ) = + 7 = 9+ 7 =. () The turning point is ( ), and is a minimum. Rememer If the coefficient of is positive then the paraola is concave up. Page 6

8 Polnomials and Quadratics Sketching Paraolas EF The method used to sketch the curve with equation on how man times the curve intersects the -ais. = a + + c depends We have met curve sketching efore. However, when sketching paraolas, we do not need to use calculus. We know there is onl one turning point, and we have methods for finding it. Paraolas with one or two roots Find the -ais intercepts factorising or using the quadratic formula. Find the -ais intercept (i.e. where = ). The turning point is on the ais of smmetr: O The ais of smmetr is halfwa etween two distinct roots. O A repeated root lies on the ais of smmetr. Paraolas with no real roots There are no -ais intercepts. Find the -ais intercept (i.e. where = ). Find the turning point completing the square. EXAMPLES 1. Sketch the graph of = Since ac = ( 8 ) 1 7 >, the paraola crosses the -ais twice. The -ais intercept ( = ) : = ( ) 8 ( ) + 7 = 7 (, 7 ). The -ais intercepts ( = ) : 8 + 7= ( 1)( 7) = 1= = 1 or 7= = 7 ( 1, ) ( 7, ). The ais of smmetr lies halfwa etween = 1 and = 7, i.e. =, so the -coordinate of the turning point is. Page 7

9 Polnomials and Quadratics We can now find the -coordinate: ( ) ( ) = 8+ 7 = = 9. So the turning point is (, 9).. Sketch the paraola with equation Since ac ( 6) ( 1) ( 9) = 6 9. = =, there is a repeated root. The -ais intercept ( = ) : = ( ) 6 ( ) 9 = 9 (, 9 ). Since there is a repeated root, (, ) is the turning point.. Sketch the curve with equation The -ais intercept ( = ) : = 6 9 ( ) = ( + )( + ) = + = = = (, ). Since ac = ( 8 ) 1 <, there are no real roots. The -ais intercept ( = ) : = ( ) 8( ) + 1 = 1 (,1 ). So the turning point is (, 5 ). Complete the square: = O 9 = 6 9 ( ) = + 1 = ( ) O = ( ) + 5. = + = (, 9) 8 1 O (, 5) Page 8

10 Polnomials and Quadratics 5 Determining the Equation of a Paraola RC Given the equation of a paraola, we have seen how to sketch its graph. We will now consider the opposite prolem: finding an equation for a paraola ased on information aout its graph. We can find the equation given: the roots and another point, the turning point and another point. When we know the roots If a paraola has roots where k is some constant. = aand = then its equation is of the form = k( a)( ) If we know another point on the paraola, then we can find the value of k. EXAMPLES 1. A paraola passes through the points ( 1, ), ( 5, ) and (, ). Find the equation of the paraola. Since the paraola cuts the -ais where = 1 and = 5, the equation is of the form: = k( 1)( 5. ) To find k, we use the point (, ): = k( 1)( 5) = k ( 1)( 5) = 5k k = 5. So the equation of the paraola is: = ( 1)( 5) 5 5 ( 6 5) = + 18 = Page 9

11 Polnomials and Quadratics. Find the equation of the paraola shown elow. ( 1, 6) Since there is a repeated root, the equation is of the form: = k( + 5)( + 5) = k( + 5 ) Hence = ( + ). 5 O To find k, we use ( 1, 6 ) : = k( + 5) 6= k ( 1+ 5) k = 6 =. When we know the turning point Recall from Completing the Square that a paraola with turning point pq, has an equation of the form ( ) where a is some constant. = a( + p) + q If we know another point on the paraola, then we can find the value of a. EXAMPLE. Find the equation of the paraola shown elow. O (, ) 7 Since the turning point is (, ) the equation is of the form: = a( ). = Hence ( ), To find a, we use (, 7) : ( ) = a ( ) 7= a 16a = 5 a = Page 1

12 Polnomials and Quadratics 6 Solving Quadratic Inequalities RC The most efficient wa of solving a quadratic inequalit is making a rough sketch of the paraola. To do this we need to know: the shape concave up or concave down, the -ais intercepts. We can then solve the quadratic inequalit inspection of the sketch. EXAMPLES 1. Solve + 1 <. The paraola with equation The -ais intercepts are given : + 1 = ( + )( ) = + = or = = =. = + 1 is concave up. Make a sketch: = + 1 So + 1 < for < <.. Find the values of for which The paraola with equation = 6+ 7 is concave down. The -ais intercepts are given : 6+ 7 = Make a sketch: ( 7 6) = ( + )( ) = = = or = = =. So for. Page 11

13 Polnomials and Quadratics. Solve 5 >. The paraola with equation The -ais intercepts are given : 5 = ( 1)( ) = = 5 is concave up. Make a sketch: = 5 1 = or = = 1 =. 1 So > for < 1 and >. 5. Find the range of values of for which the curve is strictl increasing. We have d 5 d = +. The curve is strictl increasing where + 5>. + 5= Make a sketch: ( 1)( + 5) = 1 = or + 5= = 1 = = Rememer Strictl increasing means d d. = + 5 So the curve is strictl increasing for < 5 and > Find the values of q for which ( ) For no real roots, < : a = 1 = q 1 1 q = q = ( q )( q ) q c = 1 q = q 8q+ 16 q + q + 1 q = has no real roots. ( ) ( )( ) = q q We now need to solve the inequalit q 1q+ 16 <. The paraola with equation q q = is concave up. Page 1

14 Polnomials and Quadratics The -ais intercepts are given : q 1q+ 16 = ( q )( q ) 8 = q = or q 8 = Therefore q = q = 8. < for q 8 no real roots when < q < 8. Make a sketch: = q 1q < <, and so ( ) + q + 1 q = has 7 Intersections of Lines and Paraolas RC To determine how man times a line intersects a paraola, we sustitute the equation of the line into the equation of the paraola. We can then use the discriminant, or factorisation, to find the numer of intersections. q If >, the line and curve intersect twice. If =, the line and curve intersect once (i.e. the line is a tangent to the curve). If <, the line and the paraola do not intersect. EXAMPLES 1. Show that the line = 5 is a tangent to the paraola = + and find the point of contact. Sustitute = 5 into: = + 5 = + + = + 1= ( 1)( 1) =. Since there is a repeated root, the line is a tangent at = 1. To find the -coordinate, sustitute = 1 into the equation of the line: = 5 1 =. So the point of contact is ( 1, ). Page 1

15 Polnomials and Quadratics. Find the equation of the tangent to = + 1 that has gradient. The equation of the tangent is of the form = m + c, with m =, i.e. = + c. Sustitute this into c = + + c = =. Since the line is a tangent: ( ) ( 1 c ) = = 9 + c = c = 5 c = 5. Therefore the equation of the tangent is: = 5 5 =. 1 Note You could also do this question using methods from Differentiation. 8 Polnomials RC Polnomials are epressions with one or more terms added together, where each term has a numer (called the coefficient) followed a variale (such as ) raised to a whole numer power. For eample: + + or The degree of the polnomial is the value of its highest power, for eample: has degree has degree 18. Note that quadratics are polnomials of degree two. Also, constants are polnomials of degree zero (e.g. 6 is a polnomial, since 6= 6 ). Page 1