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Quadratic Functions What you should know from Intermediate 2 A quadratic is a function of the form ax2 + bx + c (a ≠ 0) The graph of a quadratic is called a parabola. Turning Points If a > 0, the graph of the function If a < 0, the graph of the function has a minimum turning point has a maximum turning point For a quadratic expressed in the form y = a(x - b)2 + c, the axis of symmetry is x = b and the turning point is (b,c). e.g. y = 2(x – 1)2 – 3 Minimum turning point at (1, -3) Axis of symmetry x = 1 y = -(x + 4)2 + 2 y = -(x – (-4))2 + 2 Maximum turning point at (-4, 2) Axis of symmetry x = -4 To complete the square o take the coefficient of x2 out as a factor of the quadratic o Work out what would have to be squared to give x2 and x terms o Subtract the term that would come from the constant squared o Multiply by factor originally removed e.g. 5 3x 2 x 2 2 x 2 3 x 5 3 5 2 ( x 2 x ) 2 2 3 9 5 2 2 x 4 16 2 2 3 9 2 x 5 4 8 2 3 9 40 2 x 4 8 8 2 3 31 2 x 4 8 3 31 which would have a maximum turning point at , 4 8 Roots The roots of a quadratic are where the graph cuts the x-axis (in other words, the values of x such that f(x) = 0). The x-value of the turning point is always half way between the roots. To find the roots (if they exist), factorise, set each factor equal to zero and solve for x or use the b b 2 4ac quadratic formula x where f(x) = ax2 + bx + c. 2a e.g. x2 -2x – 15 = 0 (x – 5)(x + 3) = 0 x–5=0 or x+3=0 x=5 x = -3 Graph cuts x-axis at -3 and 5 x2 + 5x – 9 = 0 a = 1, b = 5, c = -9 b b 2 4ac x 2a 5 5 2 4 1 (9) 2 1 5 25 36 2 5 61 2 5 61 5 61 or 2 2 6.405...or1.405... Graph cuts x-axis at -6.41 and 1.41 Now try Ex 8B p143 (25 mins) Ex 8C p144 Q1,2 (25 mins) Ex 8D p146 Q1, 2(a), (b), (c) (25 mins) Finding the equation of a quadratic If you are given points on the graph of f(x), it is possible to find the equation. e.g. Find the equation of the quadratic, f(x), which passes through the points (-1, 0), (3, 0) and (0, -6). Note that we need to Since x = -1 and x = 3 are roots, (x + 1) and (x – 3) are factors of f(x) introduce a f(x) = k(x + 1)(x – 3) coefficient k as there is a family of parabolas which have We also know that f(0) = -6 roots at -1 and 3. -6 = k(0 + 1)(0 – 3) If we are told that the -6 = k(1)(-3) quadratic is of the k=2 form x2 + bx + c then f(x) = 2(x + 1)(x – 3) we would not need k. Now try Ex8C p144 Q3(a),(b), 5 (25 mins) More About The Quadratic Formula There are 3 possible scenarios when using the quadratic formula: Formula gives 2 roots Formula gives 1 root Formula gives no roots (real roots) (real, equal roots) (no real roots) y x Now try Ex8G p150 Q1, 3 (30 mins) The Discriminant You will have seen that when o b2 – 4ac > 0 there are 2 real roots o b2 – 4ac = 0 there are real, equal roots o b2 – 4ac < 0 there are no real roots (since we can’t take the square root of a negative) b2 – 4ac is called the discriminant. It is enough to use this to find the nature of the roots, but notice it doesn’t find where they are if they do exist. Now try Ex 8H p151 (25 mins) Ex 8I p152 Q1(a),(b),(c) (10 mins) Condition for Tangency to a Parabola There are 3 scenarios for how a line meets a parabola: Intersects at 2 points Intersects at one point Does not intersect (tangent) y x For a quadratic y = ax2 + bx + c and a line y = mx + d, points of intersection occur when ax2 + bx + c = mx + d this can then be rearranged to give a quadratic equal to zero. When the discriminant is used on this, if: o b2 – 4ac > 0 there are 2 points of intersection o b2 – 4ac = 0 there is one point of intersection so line is a tangent to the curve o b2 – 4ac < 0 there are no points of intersection Again, the discriminant only tells us the type of intersection, not where it intersects. To find this, you must solve by factorising or using the quadratic formula. e.g. Find the point(s) of intersection, if they exist, between y = x2 – 8x + 6 and y = 2x – 19. Points of intersection occur when x2 – 8x + 6 = 2x – 19 x2 – 10x + 25 = 0 a = 1, b = -10, c = 25 b2 – 4ac = (-10)2 – 4 x 1 x 25 = 100 – 100 =0 line is a tangent to parabola x2 – 10x + 25 = 0 (x – 5)2 = 0 x=5 when x = 5, y = 2 x 5 -19 = 10 – 19 = -9 Point of intersection between tangent and parabola is (5, -9). Now try Ex 8J p155 Q1, 2, 4 (30 mins). Ex 8K (Mixed Questions) These notes should be used in conjunction with the examples given in your textbook. The times given are approximations of the amount of time you should spend on each exercise in class. You should also do extra work at home. I will expect you to have done all the questions listed with the exception of Ex 8K. Once you have completed the questions, if you have any spare time in class you can use this time to revise for the Unit 1 A/B Test.