Week 11 – Solving Quadratic Inequalities

April 29, 2018April 30, 2018 keishan2015

This week in Pre-calculus 11 we learned all about inequalities. Something that I learned that really stuck with was how to solve linear inequalities and graph them.

Signs and symbols to know:

> greater than / < less than
≥ greater than or equal to / ≤ less than or equal to
An open dot is used to indicate that the boundary point isn’t apart of the solution / A closed dot is used to indicate that the boundary point is a part of the solution
A dotted line is used to is used when the values along the line aren’t included in the solution [> or <] / A solid line is used when the values along the line are included in the solution [≥ or ≤]

Example 1:

First, you need to make it so that one side is equal to zero.
Next, you factor the inequality and find its zeros. You do this using the zero product law. In this case the zeros would be: -5 and 1.
Next you would write these on a number line and pick three test numbers. You pick one number from each of the three sections of the number line. (In this case you would pick a number less than -5, one in between -5 and 1, and one greater than 1).
after doing that you test the points by putting them into the inequality. This will help you write your solution. Out of the three test points there will be some that prove to be true and others that will not. You take the one(s) that are true and use them to write your solution. (In this case our test point in between -5 and 1 proved to be true).
Write your solution. Since we know that the value of x is in between -5 and 1 we would write it as {-5<x<1}.

Example 2:

*Note: another method used to solve is a sign chart*

First you need to factor the expression. In order to do this you need to make sure that one side is equal to zero. To do this you would just rearrange the inequality and move things from one side to the other.
After factoring you find the zeros of the inequality. (in this case they would be -2 and 6).
Next you would create a sign chart. To do this you create two number lines with the zeros written on them. You would then write one of the binomials beside one of the number lines and the one beside the other and circle the zero that corresponds with that binomial. For the first number line we would write in each section of the line if the solution would be + or -. If we know that putting -2 into the binomial (x+2) = 0, then we know that anything less than it would result in a negative and anything greater than it would result in a positive. For the second line we do the same thing. If we know that when we put 6 into the binomial (x-6) = 0, then we also know that any value less than 6 would result in a negative and anything greater than 6 would result in a positive.
Next you multiply the signs that are across from each other. We know: [(-)(-)= +/ (+)(-)= -/ (+)(+)= +]. You then compare your inequality to the new values and use the sections that prove the inequality to be true to write your solution. (in this case you would compare the inequality to + – +)
Then you write the solution. We know that the solution is -2≤x≤6 because the only time that zero is greater than or equal to x is when x is in the middle section (in between -2 and 6).

Week 10 – Review: Infinite Geometric Series

April 22, 2018April 23, 2018 keishan2015

This week in Pre-calculus 11 we reviewed the first four units in preparation for the midterm. Though it was review, one thing that I had to go back and take a look at was lesson 1.5 [Infinite Geometric Series].

what to know:

There are two types of geometric series, diverging [no sum] / converging [sum].

Diverging (no sum):

when a geometric series diverges that means that the numbers are getting bigger
If a geometric series diverges that means:
r>1
r<-1
If a geometric series diverges then it also means that it has no sum

Converging (sum):

When a geometric series converges that means that the numbers are getting closer together
If a geometric series converges that means:
0<r<1
-1<r<0
If a geometric series converges then that means that you can find its infinite sum: $s_\infty=\frac{a}{1-r}$

Example:

First, we need to find the common ratio to determine if this geometric series converges or diverges.
To find the common ratio, you take one of the terms and divide it by the preceding term.
Once you find the common ratio compare it to the restrictions of both a converging and diverging geometric series. The common ratio in this equation is r = 4, 4>1 and therefore diverges. Since this geometric series diverges we can not find its infinite sum.

Example:

First, we need to find the common ratio to determine if this geometric series converges or diverges.
To find the common ratio, you take one of the terms and divide it by the preceding term.
Once you find the common ratio compare it to the restrictions of both a converging and diverging geometric series. The common ratio in this equation is r= 0.25 or $\frac{1}{4}$ . since 0< $\frac{1}{4}$ <1, this geometric series converges and therefore has an infinite sum.
to calculate the infinite sum of this geometric series we use the formula: $s_\infty=\frac{a}{1-r}$ and fill it is with what we know. Note: a is used to represent $t_1$ .
Once you have solved the equation you should be left with the geometric series infinite sum.

Week 9 – Equivalent Forms

April 15, 2018April 16, 2018 keishan2015

This week in Pre-Calculus 11 i learned a number of different things about quadratic formulas and graphing. One thing that really stuck with me was how to convert from the general form to standard form and so on. Knowing how to convert between the two formulas can help you find more information about a graph.

Example:

Note: When going from the general form to the standard form, you need to use the difference of squares method [creates a set of zero pairs])
First what you would do is check and see if there is a coefficient in front of the $x^2$ , if there is a coefficient in front then you would need to get rid of it to make the question easier. To get rid of the coefficient all you would need to do is divide the first TWO terms by the coefficient. *Do not divide the third term by the coefficient*
You would then put brackets around the first two terms and bring the coefficient of the first term in front.
Next, you create the zero pairs and add them inside of the brackets. To do this you take the term in the second part of the bracket (bx, ie. 4x), divide it by two and then square it. You then write it inside the brackets using +/- to make it a zero pair. *The first three terms inside the brackets should create a perfect square trinomial*
Then next step is to get the 4th term in the brackets outside of the brackets and simplify the perfect square trinomial [write in format of: $(x-p)^2$ ]. All you need to do to get the 4th term out of the brackets is to multiply it by the coefficient that is in front of the brackets and add it to the number at the end of the equation.
To change the format of the perfect square trinomial you take the third term and divide it by two, you then write it into the formula $(x-p)^2$ where the variable, p is. *make sure that the sign in between the x and p matches the one in the middle term of the perfect square trinomial*

Example:

In order for you to change the standard form to the general form all you need to do is expand the equation.
First, to make things easier, you would rewrite the equation so that the squared binomial is side by side and not written in a condensed form. After rewriting it out you multiply the brackets together.
Next you add up all of the like terms, and your equation should be written in the general form.

Week 8 – Analyzing the Standard Form

April 8, 2018April 9, 2018 keishan2015

This week in Pre-calculus 11 I learned a lot about graphing quadratic functions and how to analyze them. Something that stuck with me was how to analyze quadratic functions using the standard form: $y=a{(x-p)}^2+q$ .

Things to know:

General Form: $y=ax^2+bx+c$ , when using the general form, c tells you what the y-intercept is, and a tells if the parabola will open up or down (if a is negative it will open down, if a is positive it will open up) as well as if the parabola will be stretched or compressed.
Standard Form: $y=a{(x-p)}^2+q$ , when using the standard form, a will tell you if the parabola will open up or down, if the parabola will be stretched or compress and the scale. In the equation, p will tell you the horizontal translation and q will tell you the vertical translation. The vertex will always be represented by (p,q).
Scale: The scale of a parabola can be found by looking at the coefficient in front of the $x^2$ . If the coefficient in front is equal to one then the parabola will have a scale of 1-3-5 (to check this: Start at vertex go up 1 over 1, up 3 over one, up 5 over one etc. if it matches then the parabola has a scale of 1-3-5), having a scale of 1-3-5 also means that it is congruent to the parent function $y=x^2$ , it can also have a scale of 2,6,10.
Vertex: The highest or lowest point of the parabola.
Axis of Symmetry: intersects the parabola at the vertex. Splits the parabola perfectly in half.
Minimum & Maximum: used to indicate if the vertex is at the highest or lowest possible point. If the parabola is opened up it will be at its lowest height (minimum), if the parabola is opened down the parabola is at its highest point (maximum).
Domain: All possible values of x, will always be $x\in\Re$
Range: All possible values of y

Example:

Firstly since our equation is written in the proper format ( $y=a{(x-p)}^2+q$ ) we should write down what we know and put it into the formula. We know: a =1, p =2, and q =6 [ $y=1(x-2)^2+6$ ].
To start off it is best to find what the vertex is since it tells you a lot about a quadratic function. To find the quadratic function we use what we know: (p,q) –> (-2,6). If we know that the vertex is (-2,6) then we know that the axis of symmetry is: x= -2 (the axis of symmetry is located where the vertex is). If we then take a look at the value of a we know that the parabola of this equation will open up and have a scale of 1-3-5 (congruent to parent function: $y=x^2$ ). We know that it will open up because the value of a is positive and we know it will have a scale of 1-3-5 because the value of a is 1. If we know that the parabola opens up then we know that it is at its lowest point (minimum) where y =6. Finally based on what we know we can write down the domain and range. Since this is a parabola the domain will always be $x\in\Re$ . By looking at the function we know that the range will be y≥6 because the lowest that the parabola goes on the y axis is 6, and since it is opening up, all possible values for y would be greater than or equal to 6.

Example:

Firstly since our equation is written in the proper format ( $y=a{(x-p)}^2+q$ ) we should write down what we know and put it into the formula. We know: a =-2, p =-3, and q =5 [ $y=-2(x+3)^2+5$ ].
Since we have put what we know into the formula we can write what the vertex is: (p,q) –> (3,5)
If the vertex is (3,5) then we know that the Axis of symmetry is x = 3
Since the value of a is -2 we know that the parabola will open downwards and will be at its highest point (maximum). We also know that it will have a scale of 2-6-10 (because it has a value of 2 and not 1, you double the scale of 1-3-5 –> 2-6-10) and not congruent to the parent function.
We also know that the domain is $x\in\Re$ and the range is y≤5 (highest point on the y axis is 5, and the parabola is facing downwards, there would be no greater value than 5).