All reasonable efforts have been made to make sure the notes are accurate. The author cannot be held responsible for an damages arising from the use of these notes in an fashion. Quadratic equationsgraphs Quadratic graphs and equations notes luvis 11/19/01
These graphs follow the following equation How do we sketch quadratic graphs? Quadratic Equations/Graphs a b c, where a, b and c are real numbers Method-1 Method- Method-3 Method-4 Method-5 Just selected a variet of values of and find the corresponding value of and plot them. This takes time but at least it is fail safe. Use a graphic calculator and it will do it for ou. Good method but in an eam where ou are not allowed to use a graphics calculator it will not help. Factorize the equation if it is possible and find the intercept should give a rough sketch of the quadratic graph. Complete the square as this will give us the turning point and again give us a good wa to sketch the graph. Use the formula to find the roots of the quadratic equation and sketch it The are a little bit trickier than linear function but there are a variet of trusted methods Let us review each of the methods above Method-1-Sketching graph using values gives us a graph as follows: 10 5 4 + 3 Intercept ( 0, 3 ) -10-5 5 10 Intercept ( 1, 0 ) Intercept ( 3, 0 ) -5-10 Page 1 of 8
Method--Using the Casio classpad calculator Sketch 4 3 Step in using Casio classpad Step-1- Start classpad Step-- Press Main and ou will get the following Step-3- Enter the equation into the classpad Make sure ou have using the bold as the stand for variables Step-6 Now ou will see the following Step-4 Now click on the graph tab Step-5-Now highlight the epression and drag it to the bottom of the screen Page of 8
Step7 Now the rezize option on the bottom of the scrren to get full look at the graph Step-8 Now press analsis we can find the answers to various questions Press G-solve Step-9 After pressing G-solve we can find the root( intercepts),min,ma etc Step-10 Pressing G-Solve we can find answers to various questions. Press the navigation button ou will obtain all the intercepts. Roots =1 and = 3 Turning points are = = -1 Y intercept = 3 So what are the main steps in graphing and solving equations using the classpad? 1. Enter the equation in the main screen. Make sure ou press enter 3. Then click on the graph tab 4. Drag the equation into the graph section 5. Resize it and then ou can use Analsis tab to solve various points Method-3-Factorizing / Using the formula This method does not alwas work so it actuall pas to find the determinant first to see if the equation can be factorized. Remember the general equation is the following: a b c b b 4ac Now the general solution is given b the following equation a The determinant of a quadratic is what is inside the square root brackets and is given a special smbol b 4ac Now looking at the table below we find the determinant has a few special conditions Page 3 of 8
Meaning Graphical meaning 0 One solution eist Graph touches ais once 0 Two solutions eists Graph crosses the ais twice 0 No solution eists Graph does not cross ais at all So let us return back to the original graph and see it stacks up this time using the formula 4 3 What I normall do is compare it to the standard equation and note the values of a, b and c Y Y a b c 1 4 3 a 1 b 4 c 3 Now find to see if solutions actuall eists (that does not mean we cannot sketch it if no solutions eists, just that it does not cross the -ais!) b 4 ac ( 4) 4(1)(3) 16 1 4 and since the determinant is greater than zero it means it crosses the ais twice. b b 4ac Now lets us use the big formula, to find the values of, we need to be careful here. a We have alread found the determinant which is the part in the square root sign so we need to put it into the formula and proceed as normal b b 4 ac ( 4) 4 4 4 4 a (1) So we can two values for, namel the following 4 4 or 6 or 3 or 1 This gives us the roots where the graph crosses the -ais. In fact we can actuall factorise it also as manipulating these two solutions gives us the following: ( 3)( 1) 4 3 This method does not give us the turning point but because of smmetr ou can guess that the turning point is located alwas half wa between the two roots. b A simple formula is given as follows,, this gives us the turning point of the parabola a 4a b ( 4) 4 4 4 a 4a (1) 4(1) 4 Testing the formula ields the following results:,,,, 1 what we epected which is eactl Page 4 of 8
Various other facts about quadratic equations Facts about quadratic equations General equation of a quadratic equation If a > 0 then parabola opens upwards If a < 0 then parabola opens downwards The coefficient c controls the height of the parabola more specificall it is the point where it crosses the ais The coefficient b alone is the declivit of the parabola as it crosses the -ais Finding the roots of a quadratic equation using the formula on the left a b c The coefficient a controls the speed on increase or decrease of the quadratic function from the verte Bigger positive a makes the function increase faster and the graph appear more closed b b 4ac a b b 4ac b a a Finding the determinant, b 4ac Complete the square- It is done to find the turning point of a quadratic equation b hand if ou are not allowed to use a calculator! Let s look at the process of completing the square Start with the original quadratic 67 0 Move the loose number over to the other side 6 7 Take half the term ( that is divide it b two, and do not forget the sign). Square the answer Add this to both sides of the equation 6 6 6 7 6 3 7 3 69 7 9 69 16 Convert the left hand side to the squared form Now square root both sides and remember to include Now put it into simpler forms showing the two equations Now solve for and simplif the equations We could epress this as a equation and complete the square ( 3) 16 ( 3) 16 ( 3) 4 3 4 and 3 4 34 41 3 and 6 7 3 4 41 5 6 6 6 7 6 979 Page 5 of 8
This is the verte form of a quadratic equation It actuall shows the turning point which is located at 3 and 16 Let s us plot the graph to have a look at it ( 3) 16 0 + 6 7 10-0 -10 Intercept Intercept 10 0 ( -7, 0 ) ( 1, 0 ) -10 Intercept ( 0, -7 ) Local Minimum ( -3, -16 ) -0 What happens when we have a number in front of the original equation? Eample 4 5 Let us see if we can actuall solve this equation without using the formula, that is using the determinant and the big formula. Start with the original quadratic 4 5 0 Take out the 4 from both sides 5 4( ) 0 4 4 Focus on the middle inside the brackets Take half the term ( that is divide it b two, and do not forget the sign). Square the answer Add this to both sides of the equation 5 0 4 4 1 5 0 4 1 5 4 Page 6 of 8
1 1 1 5 4 1 1 5 1 4 4 4 1 1 1 5 16 4 16 1 0 1 4 16 16 1 1 4 16 Now square root both sides and remember to include Now put it into simpler forms showing the two equations and we can find real answers to both of these equations 1 1 4 16 1 1 4 4 1 1 4 4 1 1 1 1 or 4 4 4 4 We could have put the equation into verte form and epressed it in such a wa so to see the turning point of this parabola 4 5 Minimum: Y min=-5.5 for = 0.5 Root: = -0.895643937 Root: = 1.39564394 Turning point can be seen from the equation in verte form 0.5 and 5.5 4 5 4 4 5 4 4 1 5 4 1 1 5 1 16 4 16 4 1 0 1 4 4 16 16 1 1 4 4 16 1 84 4 4 16 1 4 5.5 4 Page 7 of 8
10 4 5 5-10 -5 5 10 Intercept ( -0.895644, 0 ) Intercept ( 1.395644, 0 ) -5 Local Minimum ( 0.5, -5.5 ) -10 Page 8 of 8