This semester of Pre-Calc 11 has really been a challenge. There had been so many new concepts learned and many of what I learned in Math 10 finally made sense and applied to the new concepts.

Here are some things that I just learned or finally understood

 

Domain and Range

In math 10, domain and range was never a big topic. For linear graphs, the domain and range were generally all real numbers since they never really had any restrictions of where they could go up or down. It was never important to know for tests and it didn’t help with understanding the concept being taught. This year, domain and range was greatly used for graphing and explained the qualities of a graph.

We graphing quadratics, absolute values, and reciprocals, the domain and range explained what values were available and helped explain the graph.

In a quadratic, the range helped explain that the vertex is the maximum or minimum of the graph. This is because the range shows the end of all y values in the graph.

In an absolute value, the domain and range help express where the original was and how all the terms are positive. The range is always that y is positive. The domain is two parts, one for the original line that includes the critical point and for the reflection of the negative terms.

The domain and range for reciprocals showed the values that were not included. This helped create an understanding that there is a boundary, the asymptotes, that the reciprocal will get close but never touch.

 

 

 

Factoring 

Last year, I struggled with factoring. This year, since factoring is greatly important for finding x-intercepts of a quadratic, I learned and practiced to make factoring come easier to me.

I learned to organize what I know first. Since the middle term in the sum of two factors of the last term, writing beside factors of the last term helps organize and show every possibility to help you find the right answer.

I also learned quicker ways to factor quadratics with an a term.

Using a square with 4 boxes, we can put the first term in one end and the last term in the other and then find their product. We then find a factor of that product that has the sum of the middle term and input those two terms in the remaining boxes. We can then go down each row horizontally and vertically to find like terms. This gives us two factors. We have then factored the expression.

 

 

 

Discriminant 

The discriminant was a very useful thing I learned this year when working with quadratics. The discriminant is b²-4ac that refers to the quadratic standard form ax²+bx+c and is a part of the quadratic formula 

 

The discriminant was helpful because it helped identify where the quadratic would be located roughly on the graph and if it had x-intercepts or not.

If the discriminant had a positive answer, b²-4ac > 0, the equation had 2 answers or 2 x-intercepts.

If the discriminant had 0 as the answer, b²-4ac = 0, the equation had exactly one x-intercept that was the vertex and that touched the x-axis.

If the discriminant had a negative answer, b²-4ac < 0, the equation had no x-intercepts so was either above or below the x-axis.

The discriminant can help give a visual. If you need to find the solution using y = 9 and do not have a graph and the discriminant shows that there are no x-intercepts, it is down facing with a vertex of (-3, -5), then we are sure that 9 will never cross the quadratic so there are no solutions.

 

 

Restrictions and Values of Variables 

Something new I learned in this class that I found interesting was that in some cases, variables can have certain values or restrictions. It was something that I never thought would be a thing. A value is when there is a variable under a root symbol.

For example, √x

Since the index is a even term, 2, everything under the root symbol has to be 0 or higher. x ≥ 0

√(x+5)

x + 5 ≥ 0

x ≥ -5

√x²

Since the exponent is even, x can be any number because the negative with cancel out.

xεR

When a variable is a denominator, restrictions have to be shown after it is fully factored for numbers that cannot replace the variable. We need to find what x equals that allows the factor to equal 0 and that number is the restriction. This is because 0 cannot be in the denominator. So, we are looking for the 0s of the factored term.

if x-4 is in the denominator, it is fully factored.

x-4 = 0

x = -4

x ≠ -4

 

x² is in the denominator

x ≠ 0

This was a new and interesting concept because it created the idea that variables are not always equal to just one number.

 

Creating Triangles 

During the trigonometry unit, there were many questions that asked for triangles to be created with given angles and length and for the rest of the triangle to be answered. Something interesting I learned is that, when creating a triangle without a given image, there can be more than one answer that your calculator gives. This is because all angles in a triangle equal 180 degrees, so when only one angle is given and you find the second angle, you must assume it can be a quadrant one reference angle or a quadrant 2 reference angle. If the next number you found can be subtracted from 180 and does not have a sum with the other angle higher than 180 than there are two triangles.

For example

After using the sin law, I was able to find the angle for C. Since I had to create this triangle, I had to make sure if the triangle could be found in quadrant 2. For this, we use the angle as a reference angle, so subtract it from 180 to find the angle. If the new angle and the angle is given has a sum less than 180, that means that it is to scale with the triangle and there are two options for angle C.

When looking at a triangle that is not a right angle triangle, we can use the sine law to find either the side or the angle.

The capital letters (A, B, C) refers to the angles

The lower case (a, b, c) refers to the length across the same letter angle.

When looking for a side, we make sure that sin(A), sin(B), sin(C) is as the denominator with the lengths (lower case letters) on the top.

When looking for an angle, we use the reciprocal (the inverse)

Since there are there are three equal signs, we fill out the law and only use two of the terms, keeping the equal sign in between.

So we are looking for R

We fill in the law with the sin on the top since we are looking for an angle

sin(R)⁄41 = sin(39)⁄28 = sin(P)⁄p

We can eliminate sin(P)⁄p since it doesn’t give us enough information to solve anything

We must use sin(39)⁄28 since it has all the information

sin(R)⁄41 = sin(39)⁄28

From here, we solve for R

If we multiply both sides by 41, we can cancel out one of the denominators

sin(R) = 41sin(39)⁄28

Then we can take the sin to the other side which becomes sin inversed

R = sin^–1 [41sin(39)⁄28]

Plugging this into our calculators, we get that R = 67°

Now, we need to make sure this makes sense

All triangles angle summed together equal 180°

180 – 67 – 39 = 74

This makes sense because the biggest angle should be across the biggest side.

 

This week we learned how to apply rational equations to real life scenarios.

In order to do this, there are some tips for reading and solving these word problems.

1. Read problem and understand what is being asked
2. Introduce a variable to represent an unknown number
3. Write an equation using given info
4. Solve
5. Write answer in sentence. Check that the solution makes sense

What this means is that when presented with a word problem, it is good to read it a couple of times in order to completely understand and not miss anything. It would be a good idea to highlight any keywords like added, times, then, etc.

The variable is what is being asked for, it is the answer to the question. Using all that you know, you can then write an equation and solve it. Then answer in a sentence. If this sentence makes sense to the question asked, it should be the answer.

An example of using rational expressions to answer a question with proportions:

Koolaid is added to water to make a solution that is a drink. How much Koolaid must be added to 10L of water to make a solution that contains 70% Koolaid?

70% becomes ₇⁄10

₇⁄10 = x⁄10+x

because it says added to 10, it becomes x over 10 + x

Then we can cross multiply, so 7(10 + x) = 10x

70 + 7x = 10x

Then solve for x

70 = 3x

23.3 = x

23L Koolaid should be added to water in order for it to be 70% Koolaid

 

To add and subtract rational expressions, it is easiest to start with a common denominator. This can be found by factoring.

With this example

There needs to be a common denominator, so we can find what they both need or have in common. Since this is a binomial denominator, we can leave them in the factored forms.

We can then put all the numerator terms together to create one big fraction

Once the expression is simplified, we can go back to find the non-permissible values.

To do this, we can look at the original expression and find the number that, if replaced x, would result in a 0. If x = 2 and 2 is subtracted, that is 0, which is not allowed to be the numerator. If x is -2 and 2 is added, it becomes a 0. The non-permissible values for x are -2, 2

 

 

 

When multiplying and diving rational expressions, we can take the rational expressions and factor them, then find the restrictions, remove any common factors, and multiply through.

It is important to factor first then find the restrictions. Restrictions are there to find terms for the variables that could lead to having a 0 in the denominator. That is why restrictions are found by = the factor to 0 and finding the answer for the variable.

Next thing to note is that everything is always fully factored. This helps cancel out factors. After this, just multiply through. Factors can be left in factored form.

This week, we learned how to graph absolute values. Since absolute values mean the positive of the number, all numbers under 0 were reflected upwards.

Originally, this graph crossed -2 and went further down, since it is the absolute value, all the negative y values reflect to the positive.

So, we can say y = -2x -4 if x≤-2 and y = -(-2x -4) (the other possibility for an absolute value) if x>-2

the original line has an equal to because it includes the critical point of the middle point, whereas the reflected line does not include that number.

There is a certain part of the parabola that is reflected upwards. We can say that y = (x-4)² -5 if x≤1.8 and x≥6.2 and y=-((x-4)² -5) is 1.8<x<6.2

 

This week, we learned how to take a linear equation and quadratic equation and graph them to find a solution and taking 2 quadratics to find the solution.

When presented with two equations, y=-x+2 and y = (x – 4)² – 4, we can graph them to see where they intersect. The intersections will be their solutions.

Their solutions are (2,0) and (5,-3)

If we are given the equation y = (x – 4)² – 4 and y = (x – 6)² – 4, we can graph them to find their solutions.

They have 1 solution that is (5,-3)

Two equations can have 0 solutions if they never touch, 1 solution if they only meet in one spot, 2 solutions if they meet in two spots. They can have infinite solutions if they are the same equation.

Graphing linear inequalities is building off of graphing linear equations. They are graphed the same way, the number before x represents the slope(rise over run) and the second term shows the y-intercept.

The only difference is that when graphed, there is going to be a section shaded. This section represents all the numbers that satisfy the equation. If the line is dotted, those numbers are not included and will be represented by <>. If the line is solid, those numbers will also satisfy the equation and the inequality sign will have an equal on it ≤≥.

To find what the inequality is if the graph is already given.

y _ -2x + 2

You create the equation using y = mx + b or slope y-intercept form.

Now, we can take a number in the shaded area and make a true statement.

(0,0) is a coordinate in the shaded area.

0 _ -2(0) +2

0 _ 2

Since 2 is greater than 0 and the line is dotted, we know we need < to make the statement true

0 < 2

We can go back and add < to the blank space of our equation

y  < -2x + 2

 

If we are given the equation to graph, we follow similar steps.

3x + 6 ≤ 2y

Divide by 2 to bring the equation in slope y-intercept form

3⁄2x+3 ≤ y

Once graphed, you can choose any coordinate in one of the sides, (5,0)

3⁄2(5)+3 ≤ 0

15⁄2 + 6⁄2 ≤ 0

21⁄2 ≤ 0

So when graphing, we know to show a solid line because of the inequality symbol and to shade in the side that includes the coordinate that satisfies the equation.

While reviewing, I remembered the importance of restrictions, something that I had to relearn and remember for the coming up midterm.

A restriction makes sure only certain numbers can replace x in a radical so that the radical still works.

For this radical, ∛×, because the radical index is 3 and not a factor of 2, the restriction will always be x∈R x is the element of the real numbers. This is because no matter if the number is positive or negative, it is to the power of 3, it will solve to be positive.

For numbers in square roots and roots with the index like 4, 6, 8, 10, etc, the numbers must be greater than 0 so that they can solve.

√x, x≥0

if there are more numbers under the radical, we solve the inequality.

√(x+5)

x+5 ≥ 0

-5

x ≥ -5

That is how restrictions are found. It is important that restrictions are always shown and matched up with the answers of x to make sure x truly follows its restriction.

 

There are three ways to solve and write the quadratic equation.

General Form

The general form is y =ax² + bx + c

This form tells us the y-intercept, which is c. It tells us the scale, which is a, that is what number changes the 1, 3, 5 scale. This equation does not help us graph it, it tells us it is quadratic by the x², but we can only graph it if it is in factored or standard form.

y =x² + 6x + 9 is an example of a quadratic formula in general form.

Factored Form

Factored form is the first form that is able to be used to solve quadratic equations and graph them. When an equation is formed to factored, it tells you the x-intercepts. It also tells you the scale.

y =x² + 6x + 9 can be factored to y = (x+3)(x+3), which means the only x-intercept is -3, found by taking the factors and solving them to 0 (x+3 = 0, x + -3). This also tells us this is up facing and has the scale 1, 3, 5. Since this quadratic crosses x only once, we can assume it starts there and goes up.

y =x² + 7x + 10 can be factored to y = (x +2)(x+5), so the x-intercepts are -2 and -5 ( x+2 = 0, x+5 =0). This doesn’t help us too much to fully graph because we know where the x-intercepts are, it is up facing, and has the general scale, but we do not know where it starts.

Standard or Vertex Form

This is the most helpful form for graphing, y = (x – p)² + q, (p,q) is the vertex (x,y).

If we have an equation in general form, we can change it to standard form by finding the square.

y =x² + 6x + 8

Find the square means taking the middle term, dividing it by two and then taking it to the power of 2 and then adding that as a 1 (a positive form and negative form). This helps create the factor.

y =x² + 6x + 8

6/2 = 3

3² = 9

A 1 is like 1/1 just like 3/3 is also just 1.

We can then add the 9 back in as a 1, so +9 and -9.

y =x² + 6x +9 -9 + 8

We can then create x² + 6x +9 as a factor (x+3)²

y =(x+3)² -9 + 8

y =(x+3)² -1

We have now a new form that can help us factor. This form tells us the vertex which is (p,q). We also must remember that since the equation has a negative p, the term in its place has been multiplied by a negative. So, when looking for the vertex, it would be (-3, -1). We then find this point of the graph and start from there going up as indicated since there is no negative before the bracket affecting the scale, and going up by 1, 3, 5.

Quadratic Formula

The last way to solve a quadratic equation that cannot be factored is using the quadratic formula. 

y =x² + 3x – 4

First, we list a = 1, b = 3, c = -4

Then it is just filling in the equation.

x = -(3)±√(3)²−4(1)(-4)/2(1)

x = -3±√9+16/2

x = -3±√25/2

x = -3+5/2 0r x = -3-5/2

x = 1 0r x = -4