AMB111F Notes 3 Quadratic Equations, Inequalities and their Graphs

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1 AMB111F Notes 3 Quadratic Equations, Inequalities and their Graphs The eqn y = a +b+c is a quadratic eqn and its graph is called a parabola. If a > 0, the parabola is concave up, while if a < 0, the parabola is concave down as shown in the diagrams below: (i) Case a > 0: (a, b) (ii) Case a < 0: (a, b) In each case the point (a, b) is a turning or critical point. The vertical line = a through (a, b) is the ais of symmetry of the parabola. Now consider the eqn y = a + b + c of the parabola. c is the y-intercept. For - intercepts, solve the eqn 0 = a +b+c. The ais of symmetry is given by the eqn = b/a, thus the turning point has coordinates ( b/a, f( b/a)) = ( b/a, ac (b /4a)) where y = f(). The y-coordinate is found by completing the square, see later. To solve 0 = a + b + c you may use the formula = b± b 4ac or factorise a the epression and then solve easily. Eample 9: Graph the parabola y = 5 + 6, showing intercepts, ais of symmetry and turning points. Soln. y-int: y = 6. -int: Solve = 0; thus = 5± 5 4(1)(6) = 3 or 1

2 . Therefore the -intercepts are and 3. Ais of symmetry: = b/a = 5/. So the turning point is (5/, 1/4). Since the coefficient of is 1 and thus positive, then the parabola is concave up. Hence its graph is as shown below: y 3 (5/, 1/4) Eample 10: Consider the quadratic eqn = ay + by + c. It is like the previous quadratic eqn with and y interchanged. Thus it also represents a parabola which is concave right if a > 0, and concave left if a < 0. Its ais of symmetry is given by y = b/a and its turning point is (f( b/a), b/a) where = f(y). For eample draw the graph of = y + 5y 6. Soln. -int: = 6. y-int: solve y + 5y 6 = 0; thus y-intercepts are and 3. Ais of symmetry: y = b/a = 5/. Thus the turning point is ( 1/4, 5/). Hence the following graph:

3 y 3 ( 1/4, 5/) Finding an Eqn of a Parabola given some points Competing the square for y = a + b + c, we may write the equation of the parabola as y = a[( + b/a) + c b /4a ]. Eample 11: A concave up parabola has a turning point (5/, 1/4) and one -intercept at 3. Find its eqn with an aid of a diagram. y 3 (5/, 1/4) Soln. By symmetry, = is another - intercept. The general eqn of this parabola is y = a + b + c. From this, construct eqns using the intercepts and the turning point. Thus we solve simultaneously 0 = 4a + b + c 0 = 9a + 3b + c. Subtracting, we have b = 5a. Using the turning point we have 1/4 = 5/4a + 5/b + c. Subtracting this from the first eqn, we have 3

4 1/4 = 11/4a+1/b. Using b = 5a we get a = 1 and b = 5. Substitute these into the first eqn to get c = 6. Thus the eqn of the parabola is y = 5+6. SECOND METHOD: y = a + b + c = a[( + b/a) + c b /4a ] and b/a = 5/ implying b = 5a. Also when = b/a, y = f( b/a) = ac b /4a = a[c b /4a ] = a[c (5/) ] = 1/4. When = 3, y = 0 = a[( + b/a) + c b /4a ], implying [( + b/a) + c b /4a ] = 0 = [(3 +( 5/)) + c (5/4)]. Thus c = 6. Using a[c (5/) ] = 1/4, we have a = 1. Thus b = 5a = 5. Hence the eqn of the parabola is y = Other quadratics Eample 1: Draw the graph of the eqn = y + 1 Soln. Squaring, we have y = 1, 0. Thus this is the right half of the parabola y = 1 as shown in the graph below: y 1 1 Eample 13: Draw the graph of the eqn y = + 1. Soln. Squaring, we have = y 1 = (y 1)(y + 1); y 0. Thus the graph is the upper half of the parabola = y 1. The sketch is shown below: 4

5 y 1 1 Quadratic Inequalities In this section we solve inequalities of form a + b + c 0, or may also be replaced by any inequality sign. We discuss two methods of solving these inequalities, viz the parabola method and the table of signs. Parabola Method : Assume we want to solve the inequality a + b + c 0. First draw the graph of the eqn y = a + b + c. Suppose the -intercepts are 0 and 1 and that the graph is concave up. y 0 1 Since y 0, we consider the part of the graph lying above the -ais. The soln is the set of all -coordinates of the points on the parabola that lie above the -ais. Clearly these points are on the left side of 0 and on the right side of 1. We then write the soln as 0 or 1. On the number line the 5

6 soln is as follows: 0 1 Suppose in the above discussion is replaced with <, then using the same graph but taking points below the -ais, we see that the soln is 0 < < 1. Its number line is as follows: 0 1 Eample 14: Solve the inequality > 0. Soln. Let y = +5 6 = ( )( 3). Thus the parabola is as shown below. 6

7 y 3 We want all the -coordinates of points on the graph that lie above the -ais since y > 0. Clearly these -coordinates are real numbers between and 3. Thus the solution is < < 3. Table of signsmethod Eample 15 : Solve Soln. First solve = 0 and find the roots and 3. This means ( )( 3) 0. Then draw up a table of signs (negative or positive) of ( ) and ( 3) as follows: (-)(-3) is -ve for all < while 3 is -ve for all less than 3. Thus ( )( 3) is -ve for all between and 3. To find the required soln, we look for the region in which ( )( 3) 0, noting that less than zero essentially means negative. Clearly then the soln is 3, i.e. all real numbers from to 3. Note that the same table gives soln to ( )( 3) > 0 as < or > 3. Eample 16: Draw the region y a + b + c. Soln. First draw the parabola y = a + b + c, then shade the region above the parabola. (For, shade below the parabola). 7

8 y 3 Inequalities involving Absolute Values The symbol means equivalent. Two statements are equivalent when they have the same meaning. For eample the statements 3 < 5 and 5 > 3 are equivalent. (1) We claim that a a a Proof. Suppose a. Case 0. Then = a. Thus a 0 a. Case < 0. Then = a implies a. Thus a a. Suppose a a. Case 0. The = a. Case < 0. Then = a. () Similarly a a or a Eample 17: Solve 3 < 1. Soln. By (1) above, 1 < 3 < 1, thus < < 4. Eample 18: Solve 3 1. Soln. By () above, 3 1 or 3 1. Thus 4 or. Squaring Method for solving absolute value inequalities Eample 19: Solve 3 < 1. Soln. Square both sides to get < 1, implying ( )( 4) < 0. Using the parabola method we get < < 4. Eample 0: Solve 3+1 Soln. Squaring we get (3+1) 4( ). Simplifying, we have 5 + 8

9 15 0. Using the parabola method, the soln is 5 3/ /5 Eample 1: Solve 3+1. Soln. Note that there is no absolute value here. Thus we cannot square. We use common sense. Case > 0. Then 3+1 ( ) = 4. This implies 5. But our assumption implies >. This conflicts with 5. Hence we cannot have > 0. Thus < 0, implying <. Therefore 3+1 ( ) = 4 giving 5. Using <, we arrive at the soln 5 <. 9