(Suitable for CAS extension)
This activity involves a curve "with history"; an excellent example of combining geometry and algebra. This function, the so-called "Witch of Agnesi" is defined by a geometric description. After implementing the construction, students are then challenged to find the equation of the constructed curve. This equation, in turn, allows further investigations and generalisations, including some from the field of analysis - an alternative to conventional curve sketching.
The first and second parts of the task are suitable for students of secondary school age with knowledge of the theorems of intersecting lines and the laws of similarity, as well as the Pythagorean theorem. Methods of differential calculus are required only for the final task.
http://compasstech.com.au/TNSINTRO/TI-NspireCD/Exemplary_Activities_PDF/Act10_WitchofAgnesi.pdf
Tuesday, December 1, 2009
Activity 10: The Witch of Agnesi
Activity 9: Playing Rugby
What is the optimal position for a rugby player attempting a conversion?
The aim of this activity is to determine this optimal position and then to study the variations in the angle q obtained depending on the position where the try was scored.
This task enables us to evaluate the advantage gained by positioning oneself as close as possible to the posts before scoring the try.
Activity 8: Random Rectangles
http://compasstech.com.au/TNSINTRO/TI-NspireCD/Exemplary_Activities_PDF/Act8_RandomRectangles.pdf
Activity 7: Algebra Tools
This activity explores the equivalence of algebraic expressions in expanded and factored form, using patterning with CAS to expose commonly held student misconceptions.
The big algebraic mathematical ideas this activity explores are equivalence and symbol sense. More precisely, the activity speaks to the following curriculum expectations: expand and simplify second-degree polynomial expressions involving one variable that consist of the product of two binomials [e.g., (2x + 3)(x + 4)] or the square of a binomial [e.g., (x + 3)2 ], using a variety of tools (e.g., algebra tiles, diagrams, computer algebra systems, paper and pencil) and strategies (e.g. patterning)
http://compasstech.com.au/TNSINTRO/TI-NspireCD/Exemplary_Activities_PDF/Act7_AlgebraTools.pdf
Activity 6: Exploring the Parabola
This activity explores the key features of the parabola, both geometrically and algebraically. A variety of interactive representations support student learning as they build their understanding of this important curve and its real world applications.
The primary objective in the study of parabolas in many high school curricula, tend to be algebraic, moving quickly to the study of the quadratic function. Key defining features of this function are geometric in nature. Students often misrepresent other curves as 'parabolic' simply because they have a similar appearance. It is therefore important for students to understand some of the properties of a parabola, features that make this curve both unique and important. This activity supports students in actively linking some of the geometric and algebraic properties of a parabola.
Activity 5: Meeting a Friend
A version of this problem was set as the final question for the 2005 New South Wales Higher School Certificate examination in Mathematics. Copyright is held by the New South Wales Board of Studies.
http://compasstech.com.au/TNSINTRO/TI-NspireCD/Exemplary_Activities_PDF/Act5_MeetaFriend.pdf
Activity 4: The Diminishing Square
Study the diagram provided. A smaller square has been constructed inside a larger square, as shown.
A point x is located on the base of the larger square. (As shown) The smaller square is constructed using similar points on each of the remaining sides of the larger square. If x is the midpoint of the base, what is the ratio between the area of the larger square and the smaller square?
Explore the relationship between the position of this point and the area of the smaller square.
http://compasstech.com.au/TNSINTRO/TI-NspireCD/Exemplary_Activities_PDF/Act4_DiminishingSquare.pdf
Activity 3: The Beach Race
This beach race begins from a point 4 kilometres out to sea from one end of a 6 kilometre beach, and finishes at the opposite end. Contestants must swim to a point along the beach, and then run to reach the finish line first. I can swim at 4 km/h and run at 10 km/h - where should I aim to land on the beach so as to minimize my total time for the race?
http://compasstech.com.au/TNSINTRO/TI-NspireCD/Exemplary_Activities_PDF/Act3_BeachRace.pdf
Activity 2: Birthday Buddies
What is the chance of sharing a birthday with someone in your class? This simple question offers a rich context for mathematical modeling, which is potentially accessible to students from the early years of secondary school to seniors. Using TI-Nspire CAS, students are offered the tools by which they can investigate the problem and build a meaningful model, which will deepen their understanding of the problem, and help them to further appreciate the applications of mathematics to their world.
Activity 1: The Falling Ladder
What does it feel like to be at the top of a ladder as the bottom begins to slide away? Do you fall at a steady rate? If not, then what is the nature of your motion - and when are you falling fastest?
This modelling problem is suitable for students across the secondary school, from consolidation of work on Pythagoras' Theorem in the early years, to optimization using differential calculus in the senior years. At all levels, it is a realistic and valuable task, which links a variety of mathematical skills and understandings with a practical real-world context.
Practice Problem 6
Solve this on paper, preferably without a calculator:
2x2-18x+36 = 0
After you are done, you can click this button to see the solution, to check if you got it right.
Actual Graph of y = 2 x2-18 x+36
x = 6, 3 (sometimes solutions may be close approximations of the actual solutions)
Expression factored: 2 x2-18 x+36 = 2 (x-6)(x-3)
Solution ExplainedEquation:
a = 2 b = -18c = 36
Discriminant: b2-4ac = 18 2-4*2*36 = 36
Discriminant (36) is greater than zero. The equation has two solutions.
or
or
x1,2 = (18 ± 6) / 2*2
or
x1 = 24 / 4 = 6
x2 = 12 / 4 = 3
or
x1,2 = 6, 3
Equation factored:
2 (x-6)(x-3)
Review Quadratic Equations
Overview:
An exercise where you are asked to solve a quadratic equation is mostly a test of your attention span. If you can keep yourself focused for 5 minutes and perform all the steps that are clearly given to you, you win. You don't have to be smart or creative.
A quadratic equation is an equation like ax2 + bx + c = 0 , where a, b, and c are some numbers. These numbers are called coefficients. What you are typically required to do is find the values of x at which the left side of the equation (expression ax2 + bx + c) is equal to zero.
Example: for equation x2-1 = 0, it so happens that the values of x that make x2-1 equal to zero are +1 and -1. This means that when you replace x with 1, the expression would be 12-1 = 0. And it's obvious that this equation is true. When you use value -1 for x, you would get (-1)2-1 = 0, or after noting that -1 squared is 1, you again have 1-1 = 0, a true equation.
A frequent mistake that students make is not getting the "coefficients" right. They often forget a sign or confuse b with c. They they apply the correct formula to wrong numbers and get wrong results. Don't let this happen to you.
Standard Form
Any equation that equates a second order polynomial to zero is a quadratic equation. But it has to be in a standard form in order for the standard formula to apply.The standard form is always, no exceptions, as follows:
ax2+bx+c = 0
Note that a, b, and c can be positive as well as negative. The right part of the equation should always be zero.
The +- sign used above is actually a shortcut: it means that there are two solutions, one with the sign + and another is with the sign -.
Discussing discriminant
If you look at the formula above, you will notice the really imprtant part of it inside the square root sign: it is called a discriminant. It is usually denoted as d:
d = b2 - 4ac
If d is positive, that means that the expression under the square root is positive and therefore there are two distinct solutions.
If d is zero, this means that the expression under the square root sign is zero and therefore the value of the square root is zero too. This means that regardless of whether you add that zero or subtract it, you get the same result. So there is only one root.
If d is less than zero: as you know, there is no real number whose square is negative. Therefore, in such cases there are no real solutions. .
Summary: Steps to Solve Quadratics
You should always perform these steps when solving a quadratic equation. Note how these steps are followed in our quadratic equation calculator [in the sidebar]. If you miss even one of them, a mistake is almost guaranteed.
Remember that this is an exercise for your attention span.
Make sure that the equation is written in form ax2+bx+c = 0. Make sure that the right part is zero, for example.
Write down a, b, c. Make sure that you get the signs right.
Calculate the discriminant, d = b2-4ac. Again, do not forget about signs
If the discriminant is negative, stop. There is no real solution.
If the discriminant is not negative, calculate the solution according to the formula:
Review Graphing
Graphs that you may encounter:
Quadratic Function (Parabola)
Two Linear Functions
2.5*sin(x2)
How to make a nice graph:
Even someone as messy and uncoordinated as Patrick Star can make a nice graph with there few things:
Use a ruler.
Use a pencil, not a pen, it is easier to erase.
Plot with small strokes of your pencil. Do not try to plot the entire graph in one move of your hand.
If hand drawn graphs are not your forte, use a graphing calculator!
Understand the graph:
Before Spongebob cooks patties, he must understand his love for cooking at the Krusty Krab. Therefore, to make a graph you must understand its nature.
You can expect graphs of the following type:
Linear Functions. They are represented by straight lines. They are given by equations such as
y=3x-6
Their graphs are always represented by a straight line. You have to use a ruler to plot them. They look like this:
Quadratic Functions. They are represented by parabolas, which are curvy (not straight) lines sort of like cow horns. By the way, if you throw a rock, its line of fall is represented by a parabola (with horns obviously pointing down). Try it where windows and people are not in danger. Parabolas represent quadratic equations such as
y=x2+4x+3
Because parabolas are curvy, you cannot use a ruler to plot them. They look like this:
Absolute Values. In simple cases, they are represented by jagged lines composed of straigt segments. In more complicated cases, where the absolute value is mixed in with other functions, the lines may be not straight, however they have still some points where the graphs are not smooth and where the tangent angle changes abruptly. An example of a simple graph involving absolute value is
y=x-1
You can use a ruler to plot simple absolute value graphs, however you have to find where the straight lines break. The graphs look like this:
Rules to graphing:
BY HAND:
1. A good mathematician knows that to graph, you need coordinates.
(x, y)
The x value is the number on the horizontal axis, while the y value is on the vertical axis. Follow the values of both variables until their continuous paths meet. That meeting point is your first graphed coordinate.
2. You can trace the values of a coordinate by means of a table of values with a pattern
3. Use hand drawn graphs as a way of helping you sort through your logic. Don't depend on them to be accurate.
BY CALCULATOR:
*Refer to TI NSPIRE Graph or TI 84 Graph
Practice Problem 5
Solve this on paper, preferably without a calculator:
1x2-10x+25 = 0
After you are done, you can click this button to see the solution, to check if you got it right.
Actual Graph of y = 1 x2-10 x+25
x = 5 (sometimes solutions may be close approximations of the actual solutions)
Expression factored: 1 x2-10 x+25 = (x-5)(x-5)
Solution ExplainedEquation:
a = 1 b = -10c = 25
Discriminant: b2-4ac = 10 2-4*1*25 = 0
Discriminant (0) is zero. There is only one solution. x = -b/2a
or
x = 10/(2*1 ) = 10/(2) = 5
Practice Problem 4
Solve this on paper, preferably without a calculator:
3x2-30x+78 = 0
After you are done, you can click this button to see the solution, to check if you got it right.
Quick Answer
Actual Graph of y = 3 x2-30 x+78
x = no real solutions (sometimes solutions may be close approximations of the actual solutions)
Expression factored: 3 x2-30 x+78 = Expression cannot be factored
Solution ExplainedEquation:
a = 3 b = -30c = 78
Discriminant: b2-4ac = 30 2-4*3*78 = -36
Discriminant (-36) is less than zero. No solutions are defined.
Practice Problem 3
Solve this on paper, preferably without a calculator:
1x2-4x+3 = 0
After you are done, you can click this button to see the solution, to check if you got it right.
Quick Answer
Actual Graph of y = 1 x2-4 x+3 x = 3, 1 (sometimes solutions may be close approximations of the actual solutions)
Expression factored: 1 x2-4 x+3 = (x-3)(x-1)
Solution ExplainedEquation:
a = 1 b = -4c = 3
Discriminant: b2-4ac = 4 2-4*1*3 = 4
Discriminant (4) is greater than zero. The equation has two solutions.
or
or
x1,2 = (4 ± 2) / 2*1
or
x1 = 6 / 2 = 3
x2 = 2 / 2 = 1
or
x1,2 = 3, 1
Equation factored:
(x-3)(x-1)
Practice Problem 2
Solve this on paper, preferably without a calculator:
2x2-8x+8 = 0
After you are done, you can click this button to see the solution, to check if you got it right.
Actual Graph of y = 2 x2-8 x+8
x = 2 (sometimes solutions may be close approximations of the actual solutions)
Expression factored: 2 x2-8 x+8 = 2 (x-2)(x-2)
Solution ExplainedEquation:
a = 2 b = -8c = 8
Discriminant: b2-4ac = 8 2-4*2*8 = 0
Discriminant (0) is zero. There is only one solution. x = -b/2a
or
x = 8/(2*2 ) = 8/(4) = 2
Practice Problem 1
Solve this on paper, preferably without a calculator:
1x2+3x+0 = 0
After you are done, you can click this button to see the solution, to check if you got it right.
[Don't scroll down until you've done the problem first!]
Quick Answer
Actual Graph of y = 1 x2+3 x+0
See lesson on Graphing x = 0, -3 (sometimes solutions may be close approximations of the actual solutions)
Expression factored: 1 x2+3 x+0 = (x+0)(x+3)
Solution Explained
Equation:
a = 1
b = 3
c = 0
Discriminant: b2-4ac = 3 2+4*1*0 = 9
Discriminant (9) is greater than zero. The equation has two solutions.
or
or
x1,2 = (-3 ± 3) / 2*1
or
x1 = 0 / 2 = 0
x2 = -6 / 2 = -3
or
x1,2 = 0, -3
Equation factored:
(x+0)(x+3)
Introduction
What to Expect:
-More practice questions
-Fun, interactive media
-A great learning atmosphere
What to Do:
-Explore the site by viewing similar posts with tabs
-Visit links of neat mathematical concepts
-Post questions and get answers
-Share your views on math with others
