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# Quadratic functions my maths presentation

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### Quadratic functions my maths presentation

1. 1. x 2=16ax2+ bx + cax2+ bx + c
2. 2. HOW DO THEY LOOK?
3. 3. THEIR SHAPE IS CALLED APARABOLA
4. 4. WHY IS THIS SHAPE STUDIED?The parabola is used to make lights that have afocused beam as those used in motor vehicles. Itis also used in parabolic louvres and as lightfixtures inside buildings.
5. 5. What do we need to know?1. To factorize….2. To solve ….3. To draw the graph …4. To locate the roots of the equation.5. To locate the axis of symmetry …6. To state the maximum or minimum value …7. To locate and state the turning point …8. To use the graph to solve simple inequalities
6. 6. STEPS TO FACTORISE :To factorize a x2+ b x + c1. Find the product a,c2. Find two factors of a,c that will add to give the valueof b3. Replace b x by the two factors found in 2.4. Factorize in pairs.
7. 7. ExampleFactorize 2x2– x – 3a = 2 b = -1 c = -3ac = 2 x -3 = -6Two factors of -6 are -2 and 3 but -2 + 3 = 1which is not equal to b.Use the two factors 2 and -3( check that 2 + (-3 ) = -1 which is equal to b)The expression is now re-written as2x2+ 2x – 3x -3 (factorize by pairing)2x(x + 1) -3(x + 1)(x + 1)(2x – 3)
8. 8. Steps to solve a quadraticequation:1. Take all terms to one side of the equation.2. Factorize the equation.3. Put each factor equal to 0 and solve for x.4. ALL QUADRATC EQUATIONS YIELD TWOANSWERS! Perfect squares give two identical values.
9. 9. To solve an equation:To solve 2x2– x = 3(Take all terms to one side of the equation.)2x2– x – 3 = 0(Factorize the equation.)(x+1)(2x – 3) = 0(Put each factor equal to 0 and solve for x.)x + 1 = 0 and 2x – 3 = 0x = -1 x = 1.5The two solutions are x = -1 and 1.5
10. 10. TO DRAW THE GRAPH:1. Choose at least 5 values of x from a given range.2. Substitute each value into the equation and find thecorresponding values of y.3. Plot each point on a graph page.4. Draw a smooth curve connecting the points.
11. 11. Example graph of y = 2x2– x – 3The first two steps can be represented in the table as shownbelow. Working for the values of y do not need to be shown butif shown you will not lose any marks.x -2 -1 0 1 2 3y 7 0 -3 -2 3 12
12. 12. Graph of y = 2x2– x – 3
13. 13. Roots of the equationThe roots are the values of x for which the function isequal to zero.i.e. the answer you get when you solve:y = 2x2– x – 3 = 0Remember when we solved this equation we got: x = -1 or1.5.The axis of symmetry is the average of the roots. x =½ (-1 + 1.5) = 0.25
14. 14. Roots and axis of symmetry
15. 15. Turning Points: maximum or minimumIn this case the turning point is a minimum.The minimum point is the coordinates of the pointwhere the axis of symmetry crosses the curve.The minimum point for this graph is(0.25, -3.1)The minimum value is -3.1
16. 16. Graph showing the minimum point
17. 17. Curve with a maximum value
18. 18. Graphical solutions of otherequations and inequalitiesA graph that is already drawn can be used todetermine the solution to other equations orinequalities by drawing a suitable line on the curvethat was already drawn.For example to solve 2x2– x – 3 = 2 we simply needto1. draw the line y = 2 on the graph of 2x2– x – 3 = 0and2.state the values of x where the curve and the lineintersect.
19. 19. Graphical solution of :2x2– x – 3 = 2
20. 20. Graphical solution of :2x2– x – 3 = -x + 5
21. 21. Note:The solutions are the x coordinates of the pointswhere the curve and the line intersect.QUESTION:Do you think this method can be used to solve 2x2– x– 3 = x2+ 4x – 5?How would you do this?