QUADRATIC GRAPHS ALGEBRA 2. Dr Adrian Jannetta MIMA CMath FRAS INU0114/514 (MATHS 1) Quadratic Graphs 1/ 16 Adrian Jannetta
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1 QUADRATIC GRAPHS ALGEBRA 2 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS Quadratic Graphs 1/ 16 Adrian Jannetta
2 Objectives Be able to sketch the graph of a quadratic function Recognise the shape of the parabola from the function. Use method completing the square to find turning point and lines of smmetr. Find intersection points with the coordinate aes. Understand wh sketching is better than plotting. Quadratic Graphs 2/ 16 Adrian Jannetta
3 In our previous education it is likel that ou were asked to plot the graph of a function - perhaps a straight line, a quadratic or something else. Plotting a quadratic Plot the graph of = Make a table of values for and Plot the points on a graph. Then draw a smooth curve through the points. An advantage of this method is that it s eas. However: important features (the intersections and the minimum point) are missed because we onl plotted certain values Quadratic Graphs 3/ 16 Adrian Jannetta
4 Sketching quadratic functions Given the general quadratic function = a 2 + b+c The graph of the function can be sketched b considering the following: 1 The shape of the curve. 2 The point where the curve intersects with the -ais. 3 Does the curve intersect with the -ais. If so, what are the coordinates of the points of intersection. 4 The location of the verte (turning point) of the curve. Quadratic Graphs 4/ 16 Adrian Jannetta
5 Sketching quadratic functions Shape of the curve Given the general quadratic function = a 2 + b+c There are two possible shapes for the graph of the quadratic curve. a>0 a<0 If a>0 then the quadratic curve is u-shaped or concave up. If a<0 then the quadratic curve is n-shaped or concave down. Quadratic Graphs 5/ 16 Adrian Jannetta
6 Sketching quadratic functions Intercept with the -ais Given the general quadratic function When =0 then = c. = a 2 + b+c c = a 2 + b+c That means the curve crosses the -ais at the point(0, c). Quadratic Graphs 6/ 16 Adrian Jannetta
7 Quadratic When <0 Graphs then there are no intersection 7/ 16 points. Adrian Jannetta Sketching quadratic functions Intercept with the -ais Given the general quadratic function = a 2 + b+c Provided the discriminant is nonnegative 0, then we can calculate the roots of the equation a 2 + b+c=0 to find intersections with the -ais. >0 =0 <0
8 Sketching quadratic functions Turning point and line of smmetr We ll use the method of completing the square to help sketch quadratics. =a 2 + b+c where p, q and r are real numbers. "complete the square" = p(+q)2 + r r ( q, r) = q The parabola has a line of smmetr at = q. The turning point of the parabola has coordinates( q, r). Quadratic Graphs 8/ 16 Adrian Jannetta
9 Sketching a quadratic Sketch the graph of =( 2) This quadratic function is alread in the useful completed square form. Epanding the brackets would give a positive 2 term; the curve is u-shaped. Line of smmetr at =2. 5 Turning point (minimum) at (2, 1). (The curve is completel above the -ais) Epand the brackets and simplif: = The -intercept is at(0, 5). 1 (2, 1) = 2 Quadratic Graphs 9/ 16 Adrian Jannetta
10 Sketching a quadratic Sketch the graph of = 2 6 Positive 2 term; the curve is u-shaped. The -intercept at(0, 6). Factorise =(+2)( 3) = 1 2 Intersections at = 3, = 2. Complete the square: =( 1 2 ) Line of smmetr: = 1 2 Minimum point:( , 4 ). 6 ( , 4 ) Quadratic Graphs 10/ 16 Adrian Jannetta
11 Sketching a quadratic Sketch the graph of = 4 2 We have a negative 2 term; the curve is n-shaped. The -intercept is at(0, 4). 4 Factorise the function: =(2 )(2+) Intersection points with the -ais at(2, 0) and( 2, 0). Line of smmetr is = 0 Maimum point is at(0, 4) Now sketch the curve! 2 2 Quadratic Graphs 11/ 16 Adrian Jannetta
12 Sketching a quadratic Sketch the graph of = We see a negative 2 term; the curve is n-shaped. The -intercept is at(0, 3). Epress the function in completed square form: = 2( 1) (1, 5) Line of smmetr at =1 Maimum point at(1, 5) Solve the equation = 0 to locate the roots: = 1± , 0.6 (1 D.P.) =1 Quadratic Graphs 12/ 16 Adrian Jannetta
13 Sketching a quadratic Sketch the graph of = We see a positive 2 term; the curve is u-shaped. The -intercept is at(0, 1). Epress the function in completed square form: = 0.8( 1 4 ) = 1 4 Line of smmetr at = 1 4 Minimum point at( 1 4, 1.05) Solve the equation =0 to locate the roots: = 1 4 ± , 1.4 (1 D.P.) ( 1 4, 1.05) Compare this to how we plotted the first eample: this time we found all the important points on the graph. Quadratic Graphs 13/ 16 Adrian Jannetta
14 Quadratic inequalities Although there are algebraic methods for solving inequalities, it is often quicker to visualise the solution using the quadratic graph. Consider the inequalit 2 4>0 Here is the corresponding graph. 2 2 Solutions to the inequalit are those places where the curve is above the -ais. It is eas to see that the solutions must be: < 2 and >2 4 Quadratic Graphs 14/ 16 Adrian Jannetta
15 Discriminant and roots Consider the quadratic equation 2 p+2p=0 Find the value of p for which the equation has comple roots. In this case we have a=1, b= p and c=2p. The condition on the coefficients to give comple roots is b 2 4ac<0. Therefore ( p) 2 4(1)(2p) < 0 p 2 8p < 0 p(p 8) < 0 Now we must solve the inequalit to find the values of p. This is quickl done using a sketch. Quadratic Graphs 15/ 16 Adrian Jannetta
16 The graph of = p 2 8p=p(p 8) is shown below p 16 To solve the inequalit p 2 8p< 0 we look that part of the curve which is less than zero below the ais. Clearl, this happens when p is between 0 and 8. Therefore the solutions to are comple when 0<p<8. 2 p+2p=0 Quadratic Graphs 16/ 16 Adrian Jannetta
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