Monthly Archives: October 2018

Week 8 – Pre Calc 11

This week in Pre Calc 11 we learned about properties of Quadratic Functions. We learned how to determine the vertex of a Quadratic function and if it is minimum or maximum, the line of symmetry, the x and y intercepts, the domain and range and we also learned how to determine if a function is going to be congruent to the parent function: y=x^2. Using all of these properties that we learned, we were then able to analyse Quadratic Functions written in Standard Form or as Mrs. Burton likes to call it Vertex Form, to later be able to graph them very easily. This is an example of a Quadratic Equation written in Vertex Form: y=-3(x-4)^2-6. This Vertex Form is also known as y=a(x-p)^2+q. Knowing what each of the different variables represents in this form can help you analyse the function and allows you to get a really good sense of what it would look like on a graph.

In Vertex Form, y=a(x-p)^2+q:

q, is the vertical translation, which means that it will translate either a certain number of units up or down along the y axis. It also tells us what the y in the vertex is going to be.

p, is the horizontal translation, which means that it will translate either a certain number of units to the right or to the left, moving along the x axis. It also tells us the line of symmetry and what the x in the vertex is going to be. However we must remember that the sign changes when you take the p value out of Vertex Form and put it into your vertex. Example: y=-3(x-4)^2-6 the vertex for this function would be (4,-6).

a, tells us the parabola is going to be congruent to the parent function, if it going to be a reflection or not and if it is going to be a stretch or a compression. If the value of a is is a number other than 1, then the parabola will not be congruent to the parent function. If the sign is negative the parabola will have a maximum vertex, which means that it opens down and that it will be a reflection, however if positive the parabola have a minimum vertex which means that it will open up. If the value for a is a fraction then the parabola will become wider and will be a compression and if the value is a whole number, then the parabola will become skinnier and will be a stretch. Example: y=-3(x-4)^2-6 in this function, the parabola will be a reflection, it will have a maximum vertex which means that it will open down and it is a stretch.

In the example y=-3(x-4)^2-6, that I used throughout this post, we discovered that the vertex is going to be (4,-6), that it is not congruent, that it is a reflection and has a maximum vertex. If I were to graph this function with the information that I have determined, the function would look like this:

Week 7 – Pre Calc 11

One thing that I learned this week in Pre Calc 11 is how to find the discriminant of a quadratic equation. The discriminant does not solve a quadratic equation, however it can help you avoid attempting to solve an equation with no solutions and it can also help determine the nature of the roots. The formula used to find the discriminant is also part of the quadratic fromula, b^2-4ac. There are three possible outcomes when you calculate the discriminant, it will either show you that the equation will have one solution, two solutions or no solutions at all. The discriminant is a really easy and fast way to check to determine whether or not it is worth taking the time to solve an entire equation.

If the discriminant is a number that is greater than zero then it will have two possible solutions and it also means that if you were to graph the equation it would intercept the x axis two times. For example, if the discriminant is 64, than it is a distinct root, with two solutions, it is a rational root, because it is a perfect square and it is also a real root because it is greater than zero.

If the discriminant is a number that is equal to zero then it will have one possible solution and it will only intercept the x axis one time on a graph. For example, if the discriminant is 0, than it is an equal root, with one solution, it is an irrational root and it is also a real root because it is greater than or equal to zero.

If the discriminant is a number that is less than zero then it will have no possible solutions and it will not intercept the x axis at all. For example, if the discriminant is -17, than it is not a real root because it is less than zero (a negative number) and it will have no solutions.

I have included an example below that shows the detailed steps I would take to find the discriminant of a quadratic equation.

Week 6 – Pre Calc 11

This week in Pre Calc 11 we learnt three different methods on how to solve Quadratic Equations. The first method is Factoring, it is the fastest and the easiest method and there are two differnt ways that you can factor a Quatratic Equation. The first one is the Grouping Method which is the one that I prefer or the second one is the Box Method, which is a much more visual way to factor the equations. When factoring using the Grouping Method the first step is to determine which two numbers can multiply to the third term and add to the second term. For example, in the equation x^2+8x+15=0 you have to find two numbers that multiply to +15 and two numbers that add to +8. Those two numbers would be +5 and +3. From there, you can determine what the two factors of the first equation would be: (x+3)(x+5)=0. Once you have found the factors you can then solve for x. In order to solve for x you must first isolate it. x=0-3   x=0-5 which mean that x=-3 and x=-5.

The second method that we learned is Completing the Square. This method can be used to solve any Quadratic Equation especially those that can not be solved by factoring. When Completing the Square of a Quadratic Equation you can first verify that the equation is not factorable, for example: x^2+6x-13=0 is not factorable because there are not two numbers that multiply to -13 and add to +6 which means that you can use the Completing the Square method. Once you have verified that the equation is not factorable, you can then rewrite the equation leaving space to add in your zero pairs: b = zero pairs. Ex: x^2+6x+b-b-13=0. To find the zero pairs, you divide the middle number by two and then you square it. Ex: \frac {6}{2}=3 and 3^2=9. Once you have your completed equation, which includes the zero pairs, x^2+6x+9-9-13=0 you can factor the first three terms and since the last two terms are like terms you can combine and simplify them as well. Ex: (x+3)^2 -22=0. You are now solving to isolate x, which means you would add +22 to both sides of the equation. Ex: (x+3)^2 =22. From there you square root both sides. Ex: x+3=+/-\sqrt 22. Next you subtract 3 from both sides to give you your final answer. Ex: x=-3+/-\sqrt 22.

The third method that we learned is the Quadratic Formula. This method can also be used to solve any Quadratic Equation but it is a little more complex and there are more steps, leaving you more opportunities to make a mistake. To start you have to determine which numbers are a, b and c and then from there you can put the numbers into to formula to easily solve the equation.

I have included an example below that shows the detailed steps I would take to solve a Quadratic Equation using the Quadratic Formula.

Week 5 – Pre Calc 11

This week in Pre Calc 11 we focused a lot on reviewing how to factor polynomial expressions. Mrs. Burton gave us a fun acronym to help us remember the steps to follow when figuring out how to factor the epressions: CDPEU.

Common: 7x+14 = 7(x+2)

Difference of squares (Binomial): x^2-25 = (x-5)(x+5)

Pattern: x^2+x+4

Easy: 1x^2

Ugly: 4x^2

One thing that I learned this week is how to factor Ugly equations. These equations are harder and more complex than the equations that we learned in Grade 10. For example: x^2+1.5x+0.5 may look like a very daunting and “ugly” expression because of the decimals, however if you were to change the expression so that the numbers were fractions instead of decimals than it would make it much easier to solve. Ex. x^2+\frac{3}{2}x+\frac{1}{2}. Once you have changed the decimals into fractions, you must put each term under a common denominator. Ex. \frac{2}{2}x^2+\frac{3}{2}x+\frac{1}{2}. From there you follow the acronym CDPEU and you find what they have in common and take it out of the equation. Ex. \frac{1}{2}(2x^2+3x+1). Once you have removed what they have in common you may than finish factoring the expression using one of the four methods that we learned in class. Ex. \frac{1}{2}(2x+1)(x+1). Onve you have finished factoring the equation, you can always check using the acronym to make sure that it can not be factored any further.

I have included an example below that shows the steps I would take to factor another challenging “ugly” expression that starts using fractions and not deciamls.