Week 13 – Graphing Reciprocals

Posted on May 12, 2018 by shannonf2015

This week we have learned how to graph reciprocals for both linear functions and quadratic functions.

For graphing reciprocal functions, we have to determine the asymptotes.

The x-axis is a horizontal asymptote and a vertical asymptote is when x has a certain value, a vertical line that the graph approaches but never reaches.

For all reciprocal functions, y cannot be 0 since $\frac{1}{f(x)}$ can never be 0.

When we take the reciprocal of a number, the only numbers that stay the same are -1 and 1. Therefore, when we graph, we find the points that line up to where y=1 and y=-1. These are the invariant points.

Let’s look at an example of a linear function.

$\frac{1}{-3x+9}$

First step is to graph the line.

Next is to locate the invariant points.

Next is to find the asymptotes. Horizontal asymptote is always y=0.

Once you have the line graphed, now you must draw a curved line that goes only through the variant points while approaching the asymptotes but not actually touching it.

let’s look at another example but with quadratic functions.

Example:

y = $2x^2-4x-6$ and y = $\frac{1}{2x^2-4x-6}$

First is to graph the parabola. Start by factoring to find the x-intercepts.

Now that you have the x-intercepts, you can add them together and divide it by two to find the axis of symmetry and plug that and a point in standard form to find q for the vertex.

When you draw out the parabola, circle the invariant points and draw in the asymptotes.

Draw a curved line through the invariant points.

If the parabola has one x-intercept, then it will be divided into 4 zones and will have two invariant points:

If the parabola has no x-intercepts, then it will divided in two zones and it won’t have any invariant points or a vertical asymptote:

Week 12 – Absolute Value Functions

Posted on May 5, 2018 by shannonf2015

This week, I have learned how to graph the absolute value of a linear function, the absolute value of a quadratic function and how to write it in piecewise notation.

Example of a linear function:

y = I3x-5I

Let’s start by graphing this.

We know the y-intercept is -5, and the slope is 3. Start by plotting the given information on the graph for y = 3x-5

Now, to be able to place the absolute value, we must remember that absolute values cannot be negative. So the line that goes past the x-axis will instead change to positive values which will make the line bounce upwards after touching the x-axis.

To write this in piecewise notation, we can’t really tell what the x-axis is here, so we can use the equation to determine the x-intercept.

Let’s look at an example of a quadratic function:

Let’s start by graphing this. We know the vertex is (-1, 1) and it’s congruent to $y=x^2$ which follows the pattern 1,3,5,7,9…

Once you have it graphed without the absolute value symbol, you now can start graphing the equation when it’s in the absolute symbol. To do this, all the negative values must switch to positive values. Therefore, the outsides of the parabola will flip up.

Now that you have this, we can use this graph to determine the intercepts and domain and range. The x-intercept is -2 and 0 and the y-intercept is 0. The domain is always the element of all real numbers (xER) and the range is y ≥ 0 as the values of y must be +.

To write this in piecewise notation, we must first look at the x-intercepts. We can see that the outer part changes but the inner part does not whether it’s in the absolute value symbol or not. So y = $-(x+1)^2+1$ is if -2 ≥ x ≥ 0 (The inner part of the graph; we are including the + values and 0).

For $y=-(-(x+1)^2+1)$ if x < -2 and x > 0 (The outer parts of the parabola; all the negative values).

So the piecewise notation is:

Week 11 – Graphing Inequalities and Systems of Equations

Posted on April 28, 2018 by shannonf2015

For the past week, I have learned how to write solutions for quadratic inequalities and also how to graph them.

Let’s start with an example:

Example 1: Given an equation

$2x^2-8x-10<0$

First step is take out what is common then factor to be able to find the x-intercepts.

Common: 2

2( $x^2-4x-5$ )<0

2. Factor

2(x+1)(x-5) < 0

x-intercepts: -1 and 5

Now that you have the x-intercepts, we can draw these on a number line.

By looking at the equation, we can tell the parabola will be opening up as the coefficient next to $x^2$ is positive. So there is a +, then a – , then a + as the parabola will dip past the x-axis since it has two x-intercepts.

Looking at the inequality symbol, it’s using <. This means they are asking for the numbers that are less than 0, so negative numbers.

The negative numbers are in the middle so the solution would look like this:

-1 < x < 5

The circles are not coloured because it is not including “or equal to”. This equation is just using less than.

Now let’s look at another example where we are given a graph and we must write an inequality.

Let’s go over some important things:

≤ and ≥ : has a solid line

< and >: has a dash line

Example:

First step is to find the y intercept, which is -1. Next is to find the slope. We can tell by the points that it’s going up by 4 and moves to the right once. So the slope is 4. This linear graph is tilted up so it’s positive.

The equation so far looks like this:

y ☐ $4x^2-1$

To find the inequality symbol, we must test a number from the shaded section that is not a point on the line and determine which symbol to choose which will result in a correct answer.

Let’s use (5,0)

Plug in 5 in x and 0 in y:

0 ☐ $4(5)^2-1$

0 ☐ 4(25)-1

0 ☐ 100-1

0 ☐ 99

To make this true, we must use ≤ because 99 is greater than 0 and the line is solid.

So the equation will be:

y ≤ $4x^2-1$

Week 10 – Radicals, Arithmetic Series and Infinite Geometric Series

Posted on April 21, 2018 by shannonf2015

During review, I have recalled some concepts I have forgotten.

One of them is when a value or variable has an exponent that is a fraction. When it’s a fraction, the denominator represents the root. We can remember this as Flower Power.

Example:

$64^\frac{1}{3}$ —> $\sqrt[3]{64}$ —> 4

Another simple method I recall is adding an arithmetic series that is finite. The first step is taking the first term and the last term in the series and adding them together. Next, count how many terms there are in total and divide the amount by two. Multiply the terms divided by two with the sum of both first and last terms.

Example:

10 + 5 + 0 -5 -10 -15

10 + (-15) = -5
Number of terms in total: 6
$\frac{6}{2}$ = 3
(-5)(3)
Sum = -15

Another concept I did not talk about last time was infinite geometric series. We can tell if they converge or diverge by looking at the value of the common ratio.

Diverges: r > 1; no sum

Converges: -1 < r < 1

Formula:

Image result for infinite geometric series formula

Example:

2 + 3 + 4.5 + 6.75 +…

r = $\frac{3}{2}$

The common ratio is greater than 1, so it is infinite (diverges); no sum.

-0.5 – 0.05 – 0.005 – 0.0005 – …

r = $\frac{-0.05}{-0.5}$ = 0.1

The common ratio is greater than -1 and smaller than 1 so it is finite (converges); has a sum.

Let’s look at an example of how to use the formula with a geometric series.

8 + 2 + 0.5 + 0.125 + …

Information given:

a = 8
r = $\frac{1}{4}$

Plug in the information given in the formula and solve.

Remember to flip the fraction (reciprocals) when doing division.

Week 9 – Quadratic Functions

Posted on April 14, 2018 by shannonf2015

Throughout the week, I have learned how to convert a quadratic function in general form to standard form and factored form to be able to find different information about how the graph looks like.

Let’s start with an example:

$y=2x^2-20x+32$

Right away, we can tell that the y-intercept is 32 because it is in general form.

We can also tell the direction of opening as it is positive, which means it opens up and has a minimum value.

Start by removing the greatest common factor which is 2.

$y=2(x^2-10x+16)$

Next, factor the three term in the bracket to find the x-intercepts (the roots)

$y=2(x-2)(x-8)$

x intercepts: 2, 8

Now we need to find the vertex, to do that we need to change the general form to standard form by completing the squares. We cannot change from factored form to standard form.

$y=2x^2-20x+32$

First take out what is common between the first two terms, leaving the 32. Leave two spaces for a + value and – value between middle term and last term to create zero pairs.

y = 2( $x^2$ -10x+___-___) + 32

Take middle term and divide it by 2 and square it.

$-\frac{10}{2}=5$ —> $(-5)^2=25$
$y=2(x^2-10x+ 25-25) +32$

Multiply the coefficient (2) with the negative number of the zero pair (-25). Then do the calculations with the last term.

$y=2(x^2-10x+25)+2(-25)+32$
$y=2(x^2-10x+25)-50+32$
$y=2(x^2-10x+25)-18$
$y=2(x-5)^2-18$

Standard form: $y=a(x-p)^2+q$

Vertex: (p,q)

The vertex is (5, -18); the value of p changes sign when written for the vertex.

Now that we know the vertex, we can determine the axis of symmetry, domain and range.

AOS = x=5

Domain: xER

Range: y ≥ -18

This equation is also congruent to $y=2x^2$ .

Another example is when we are given some information about the quadratic function and we must write an equation.

Example: The graph of a quadratic function passes through (1, -7) and the zeros of function are -6 and -1. Write an equation in general form.

Information given:

Point: (1, -7)

X-intercepts:

X1 = -6
X2 = -1

By looking at the information, we cannot use the standard form because we are not given the vertex.

We therefore have to use factored form as we have x1 and x2 as well as a point.

Factored form: $y=a(x-x1)(x-x2)$

Replace

y = -7
x = 1
x1: -6
x2 = -1

$-7=a(1+6)(1+1)$

When x1 is replaced with -6 and when x2 is replaced with -1, the sign changes to positive because there are two negatives.

Next is to isolate a.

$-7=a(7)(2)$
$-7=14a$
$\frac{-7}{14}=\frac{14}{14}$
$-\frac{1}{2}=a$

Now that you have a, put it in the equation. Remember, they want it in general form.

$y=-\frac{1}{2}(x+6)(x+1)$
$y=-\frac{1}{2}(x^2+7x+6)$
$y=-\frac{1}{2}x^2-\frac{7}{2}-3$

Week 8 – Quadratic Functions and Graphs

Posted on April 7, 2018 by shannonf2015

I have learned many things about graphing such as what information we can take out to determine how a graph looks like of a quadratic function. I have also learned the most important information to determine the equation of a quadratic function and how to draw it as a graph.

Parabola: graph of every quadratic function which is a curve

General Form: $y=ax^2+bx+c$

Standard Form: $y=a{(x-p)}^2+q$

Vertex : highest or lowest point; (p,q)

Minimum Point: when the graph opens up (The coefficient next to $x^2$ is positive)

Maximum Point: when the graph opens down (The coefficient next to $x^2$ is negative)

Axis of Symmetry: intersects the parabola at the vertex

Domain : All possible values for x (always the element of real numbers)

Range: All possible values for y

X intercepts: zeros of function (values of x when the function is 0, when y=0)

Pattern of parent function, $y=x^2$ : 1,3,5,7,9…

Throughout the week we have different types of transformations:

Parent Function : $y=x^2$

$y=x^2+q$ : depending on the value of q, the image of the graph $y=x^2$ moves a vertical translation (moves however many times up or down) while keeping the same points, size and just sliding the vertex. This quadratic equation is therefore congruent to the parent function, $y=x^2$ , because it does not change, it just moves up or down.

$y={(x-p)}^2$ : depending on the value of p, the image of the graph $y=x^2$ moves a horizontal translation (however many times to the right or left).

${(x-3)}^2$ —– when the sign is negative, the vertex of the graph moves to the right
${(x+3)}^2$ —– when the sign is positive, the vertex of the graph moves to the left

This is also congruent to the parent function.

$y={ax}^2$ : the graph will stretch vertically when a > 1.

When the value of a is between 0 and 1 (a fraction), the graph will compress vertically : 0 < a < 1

This transformation is not congruent to the parent function.

When these transformations are all combined, the equation becomes standard form.

Let’s look at some examples of using the standard form to determine an equation of a graph.

1) Vertex: look at where the highest or lowest point is and write down the coordinates.

(4,1)

2) Plug in the p and q values from the vertex

$y=a{(x-p)}^2+q$

$y=a{(x-4)}^2+1$

3) Find a point on the graph and use the coordinates to replace x and y.

I’ll use the point : (3, 4)

$4=a{(3-4)}^2+1$

4) Isolate a

$4=a{(3-4)}^2+1$
$4=a(-1)^2+1$
$3=a(-1)^2$
3 = a

The equation is : $y=3(x-4)^2+1$

Let’s look at how we can determine the domain and range and the intercepts by the quadratic function:

Example: $y=3{(x-2)}^2+1$

It is always important to first find the vertex as it is the essential piece of information to be able to determine the form of the graph.

$y=3{(x-p)}^2+q$

Vertex: remember the vertex is (p,q). In our example, p is -2 which becomes +2 because when you place -2 in the standard form, it will become positive with both negatives:

$y=3{(x-(-2))}^2+q$
$y=3{(x+2)}^2+q$

q is the y coordinate of the vertex (not the y-intercept like in general form) which will be 1.

Our vertex is (2, 1)

2. Domain: x∈R – Domain is always the element of real numbers.

3. Range: we should think about the direction of opening to be able to tell the range. Since the coefficient is +3, the graph opens up. This also means that the vertex is a minimum point with y-coordinate 1.

Therefore, y ≥ 1

4. Direction of opening: Opens up

5. Equation of the axis of symmetry: This is the line that cuts through the vertex.

AOS: x = 2

6. The intercepts:

To find the y-intercept, x=0. Replace x with zero in the equation and solve.

$y=3{(0-2)}^2+1$

y = 13

To find the x-intercept, y=0. Replace y with zero in the equation and solve.

$0=3{(x-2)}^2+1$
$-1=3{(x-2)}^2$
$-\frac{1}{3}={(x-2)}^2$

This equation has no solution, so there are no x-intercepts.

Week 7 – Interpreting the Discriminant

Posted on March 18, 2018 by shannonf2015

Last week, we have learned how to determine the number of solutions of a quadratic equation without solving the equation. To be able to do that, we look at the discriminant!

The discriminant is the radicand of the quadratic formula: $b^2-4ac$

This is how to determine the number of solutions by the discriminant.

Two real roots when $b^2-4ac$ > 0

If the discriminant is positive (so greater than 0), it has two solutions. (It touches the x-axis twice)

Exactly one real root when $b^2-4ac$ = 0

If the discriminant is zero, that means there is exactly one solution (Just touches the x-axis once)

No real roots when $b^2-4ac$ < 0

If the discriminant is a negative number (so smaller than 0), it has no solutions (Does not touch the x-axis)

Now that we understand how to use the discriminant to determine the number of solutions, let’s apply these to some equations.

First example: Calculate the value of the discriminant and determine how many solutions.

$4x^2+2x+3=0$

Remember the quadratic equation is : $ax^2+bx+c=0$

Therefore: a = 4, b = 2, c = 3

Plug in the numbers to determine the discriminant.

$b^2-4ac$

$2^2-4(4)(3)$

4-16(3)

4-48

-44

Since the discriminant is negative, it has no solutions; no real roots.

Now let’s look at another example where we have to create an equation with a given number of roots.

Determine the values of y of the equation which has two real roots.

$5x^2+3x+y=0$

To have two real roots, the value of its discriminant must be greater than 0.

So we use: $b^2-4ac$ > 0

Substitute: a = 5, b = 3, and c = y

$3^2-4(5)(y)$ > 0

Simplify.

9 – 20y > 0

Move the 9 to the other side to isolate the variable.

-20y > -9

Divide both sides by -20.

y > $\frac{9}{20}$

Since we divided both sides by a negative number, we must switch the inequality symbol.

y < $\frac{9}{20}$

Week 6 – Solving Quadratic Equations

Posted on March 10, 2018 by shannonf2015

This week, we have learned how to solve quadratic equations by completing the square and applying the quadratic formula when the equation is not factorable.

A quadratic equation includes at least one squared variable and has the following form: ax² + bx + c = 0

a, b and c are constants or numerical coefficients and x is the unknown variable.

Let’s start with the method of completing the squares. This method is quite a long process, it depends on the equation. The best would be to see if it is factorable, if it is not or factoring is too difficult for the equation, you can use the completing the squares method or using the quadratic formula.

Completing the Squares

Lets start with an example:

${x^2}+{4x}=2$

To be able to solve this equation, it must equal to zero as it is a quadratic equation. In this example, we have to move the 2 to the other side so that it becomes zero:

${x^2}+{4x}-2=0$

It becomes -2 because when you move it to the other side of the equation, you switch the signs.

Now that we have it written in the proper equation for us to solve, first check if it is factorable. As there are no numbers that both multiply to equal to two and add to equal 4 (only option is $2\times{1}$ ). So we must use either the completing the squares or formula. In this example, I will be using the completing the squares method.

First, we make two places that are both + and – to get zero pairs in between the inner term and the constant:

Then we find what is the missing constant. To do that, we divide the inner term in two and square it. The answer will be 4. Then we factor for those three terms.

Now that you got ${(x + 2)}^2$ , write the other numbers that are next to it: -4 -2 which equals to -6.

Isolate the variable by moving -6 to the other side of the equation. Then square both sides to get x + 2 = $\sqrt{6}$ . Make sure to include both + and – signs next to the radical because it has two options.

Then move the 2 to the other side of the equation to isolate x.

x = -2 ± $\sqrt{6}$

Now let’s see how we can apply the quadratic formula.

${2x}^2-2x-1=0$

This is the same form as:

${ax}^2+bx+c=0$

a=2, b=-2, c=-1

Formula:

Substitute the variables with the numbers given.

Once you substitute it, simplify it while making sure to keep the + and – sign next to the radical.

Week 5 – Factoring Polynomials

Posted on March 3, 2018March 4, 2018 by shannonf2015

I haven’t really learned much starting this week, but I have learned quite interesting new techniques on factoring polynomials.

Let’s start off with this acronym I have learned to help with the process.

CDPEU – Can Divers Pee Easily Underwater?

Common

Difference of Squares

Pattern

Easy

Ugly

Let’s breakdown what each term means before getting into the factoring process.

Common means find what is common between the terms, the GCF (Greatest common factor).
Difference of Squares means if there are two binomials and they are subtracting.
Pattern means if they have three terms following the pattern: $x^2$ $x$ #
Easy means the easy ones to factor out (straightforward) : $x^2$
Ugly ones are the polynomials that not so straightforward and need more work. We’ll see what method we can use later on. : $a{x^2}$

Now let’s apply these steps!

To start, refer to the list above we just talked about.

The first step is find what is common. To do so, write all the possible factors of both coefficients and see what variables both terms have in common.

After we found all possible factors of 49 and 14, we can see that the common number is 7. Both terms also have at least one x variable. Therefore, you would factor out 7x :

Write out the rest of the values in the bracket next to 7x, making sure when you multiply the values in the bracket with 7x, it will give you the same original expression.

Do we stop here? Well let’s check the list.

Difference of squares is the next one. To tell if it’s a difference of squares, it must have two terms and also be able to form conjugates (subtraction of two terms).

In our example, (7x – 2) does not have an $x^2$ , so we cannot factor them out any further. Therefore, it is not a difference of squares.

Next step is pattern. Well this doesn’t work because it doesn’t have three terms, therefore, we cannot simplify it any further.

It is therefore : 7x(7x – 2)

Now let’s look at an example with three terms.

First step is looking for something in common. As we can see, there’s nothing in common so let’s go to the next step.

Second, is it a difference of squares? No! There is three terms.

So, thirdly, let’s check if there’s the pattern : $x^2$ $x$ #

Yes!

The first thing to do is draw two brackets and start with putting in x to get $x^2$ . Then factor out all the possibilities for 24.

To decide which two numbers we must use for 24, the two numbers that are multiplying must equal to 10 (the inner term) when added together. So the pair to choose would be 6×4. The sign depends on the inner term. It’s positive and so is 24 so therefore we use ++.

This is the final expression factored.

(x + 4)(x + 6)

To check if it’s correct, we can foil back in the values, meaning distribute all the numbers back by multiplying.

So far, these are both easy. But let’s get into the uglier ones.

There are no common factors here, and there are no differences of squares. There is the pattern though with the three terms, but this one seems a little more complicated than the other one we just did.

A method I learned for these ugly trinomials is using the box.

You split a box into four sections, writing in first the value with $x^2$ which in this example is $25{x^2}$ on the top left corner and the constant on the bottom right corner.

We multiply both $25{x^2}$ and 4 which equals to $100{x^2}$ .

Write out all the possible factors of 100 and find which pair adds together to equal to the inner term.

When all values are placed in the box, find the greatest common factor in all the values that are next to each other, not diagonally from each other.

Final thing that I have learned which for me personally is such a life saver, is being able to replace values with variables.

Here is what I mean.

If we are given this example:

We can simply replace (3y + 1) with any variable to make the expression more simple.

Now that we have the expression 2ya – 4a, we need to find the greatest common factor, which is 2a.

Then we have to replace the variable back with 3y + 1.

The final expression factored is: 2(3y + 1)(y – 2).

Week 4 – Simplifying Radical Expressions

Posted on February 24, 2018 by shannonf2015

I learned how to expand and simplify a radical expression.

( $3\sqrt{5}$ + $7\sqrt{8}$ )( $\sqrt{10}$ – $6\sqrt{5}$ )

There are multiply ways of starting to simplify this expression, but the first thing that I do is see if I can simplify the radicals given.

Looking at the expression, I can see that I can simplify $7\sqrt{8}$ , as 8 has a square root, which is 4.

$7\sqrt{8}$ —> $7\sqrt{{4}\times{2}}$ —> $7(2)\sqrt{2}$ —> $14\sqrt{2}$

The next step is to distribute the values in the first bracket and multiply with all the values in the second bracket.

This process is called FOILING.

F – first terms

O – outer terms

I – Inner terms

L – left over terms

Rule: We can simply multiply the radicals together, unlike adding and subtracting where they have to have the same index and radicand.

So in my example, it doesn’t simplify by a lot as they don’t have any like terms.

The answer would therefore be : -90 + $15\sqrt{2}$ + $28\sqrt{5}$ – $84\sqrt{10}$

Since my example isn’t the best to show how to simplify when it comes to like terms, let’s look at this example.

I did this equation in two ways which does give you the same answer. The first way, I simplified the radicals first, then added the like terms together. Note: They have to have the same radicand and same index in order to add them or subtract them together.

The second method, I just added the like terms together first as $4\sqrt{x^3}$ had the same radicand and index as $-7\sqrt{x^3}$ and when you simplify $2\sqrt{x^2}$ , it becomes 2x which can be added with 3x. So when you add the coefficients together keeping the same radicand and index, you are combining like terms.

Then, I simplified the radicals, ${-3}\sqrt{x^3}$ can be simplified to ${-3x}\sqrt{x}$ .

Since we have a variable in our expression, we must define x. Our index here is 2 (square root), therefore, x must be greater than or equal to 0 because we cannot have two numbers that are the same that will multiply to give a negative number: x ≥ 0

Last thing that I have learned was simplifying fractions with a radical as the denominator.

Since we cannot leave radicals in the denominator, we must rationalize the denominator in order to make it a real number.

Rationalizing the denominator means multiplying both the nominator and denominator by the denominator. I included the coefficient but you don’t have to. It would probably be best to just multiply $\sqrt{5}$ with the denominator and nominator, but both give the same results.

After I multiplied them, I noticed I could simplify even more. Then finally, I simplified the coefficients as both can be divide by 10.

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PRE-CALCULUS 11

Week 13 – Graphing Reciprocals

Week 12 – Absolute Value Functions

Week 11 – Graphing Inequalities and Systems of Equations

Week 10 – Radicals, Arithmetic Series and Infinite Geometric Series

Week 9 – Quadratic Functions

Week 8 – Quadratic Functions and Graphs

Week 7 – Interpreting the Discriminant

Week 6 – Solving Quadratic Equations

Week 5 – Factoring Polynomials

Week 4 – Simplifying Radical Expressions