Math 11 – Aliessah's Blog

Category: Math 11 (Page 1 of 2)

Week 17 – PreCalc 11

By aliessahv2016

On June 9, 2019

This week in PreCalc 11, we started to learn about Trigonometry. We learned many concepts regarding trigonometry, but the one thing that stood out to me was the Sine Law. The reason why it stood out to me was because it cuts down the work we had to do last year to solve the same things that the Sine Law does. Last year if we were introduced to triangles that weren’t typical 90 degrees angles, we had to make a line from the top point perpendicular to the bottom line to solve each angle and solve for each sides measurements. Fortunately, the Sine Law helps us do that with less amount of work needed to be done.

The Sine Law is a/sin A = b/sin B = c/sin C, where the lower case letters are the measurements of the side lengths and and the upper case letters are the lower case letters corresponding angle. By corresponding angle i mean that if angle B describing one point of the triangle, then side b would be the side of the triangle that is directly opposite of angle B. Because the Sine Law works with the angles of triangles and the measurements of the triangles side then it’s useful in finding a missing angle or side.

Example:

Work:

Step 1: draw triangle

When given a problem that describes the angles and the side lengths of a particular triangle, the first step is to always draw it. Drawing the triangle gives us a better understanding of what we have and what we’re looking for with little room for error. In this example, we we’re given two side lengths, KN and MK , and angle M. I drew the triangle, labelled the points, and then labelled the characteristic given. That’s when I found out what I was looking for, angle N, angle K, and side MN.

Step 2: find the equation to use first

The easiest way to do this is to label the sides first. In this example, opposite of angle K would be side k, opposite angle M would be side m, and opposite angle N would be side n. Once we have labelled those we should write out the full equation and plug in everything we know. The equation comes with three different ones, but we can only use two at a time. So the important thing is to use two equations that we can solve for one variable meaning all other variables we should already have. In this example, we have Sin M, m, and n, meaning we could find Sin N.

Step 3: solve the equation

Once we have the first equation we are going to use to solve for something, then all we have to do is solve for it. To solve for it, I used cross-multiplication and solved from there. When i moved Sin to the other side, I inverted it.

Step 4: find the last angle

Once we found angle M, we can easily find the last unknown angle with little work. We know that the 3 angles of one triangle have to add up to 180 degrees. So we know that one angle is 70 degrees and another angle is 75, so that means if we take 180 and subtract 70 and subtract another 75, it will give us the final angle. In this case, the final angle is 35 degrees. That means we have angle M as 70 degrees, angle N as 75 degrees, and angle K as 35 degrees.

Step 5: solve for the last side length

The only thing we have left to solve is the final side length. We can use the Sine Law to find the last side length. Since we have all variables of the equations except for k then we can use any two of the equations to find it. The final step is to solve for it. In this example the final side length was 8.6 cm.

Now we know all the triangles angles and side lengths all thanks to the Sine Law!

Week 16 – PreCalc 11

By aliessahv2016

On June 2, 2019

In Math 11

This week in PreCalc 11 we learned how to solve rational equations. Rational equations are equations containing at least one fraction whose numerator or denominator is a variable.

There are two ways to solve rational equations, one of them is multiplying every term by each of the denominators or cross multiplication. Cross multiplication only works when there are two fractions and one is on each side of the equal sign whereas multiplying by the denominator is a strategy that will work with every type of rational equation.

Example:

Explanation:

Step 1: Factor

Factor any polynomials that aren’t already factored. To factor, find two numbers that when multiplied together equal the last term and when added together equal the middle term.

Step 2: Multiply by the Denominator

Multiplying by the denominator is just doing the opposite operation to take the fraction away to make the equation easier to solve. When you have a fraction it means that its the numerator divided by the denominator, so to get rid of the denominator, we should multiply the term by the denominator to get rid of it. If we multiply the term by it, then the same term that is on the denominator will also be apart of the numerator making them cancel out. But what we do to one term, we do to every term, so we have to multiply each term in the equation by the denominator. Typically there is more than one denominator, so we follow the same procedure for the rest of the fractions.

Step 3: Non-Permissible Values

Non permissible values are numbers in which x cannot equal because it will make the denominators zero which can never happen. So stating non permissible values is important. In this case, the denominators consists of (b – 2), (b +2), and (b – 3) so b cannot equal 2, -2, and 3 because it will make the denominators equal zero.

Step 4: Solve

The final step is to solve. Multiply each term by the denominators, cancel out similar factors that are on the numerator and denominator and solve for x.

Week 15 – PreCalc 11

By aliessahv2016

On May 26, 2019

In Math 11

This week in PreCalc 11 we learned how to multiply and divide rational expressions.

Rational Expressions are fractions whose numerator and denominator are both polynomials and this week we learned how to multiply two rational expressions together. Typically when we multiply two regular fractions together, we multiply the top numbers which turns equals the numerator and multiply the bottom numbers which equals the denominator. Then at the end we simplify the remaining fraction. The same idea goes for multiplying two or more rational expressions together.

Example:

Explanation:

Step 1: completely factor

Factor all the polynomials present in the rational expressions. All factors should be completely factored.

Step 2: name non permissible values

Typically with regular factors, the denominator can never equal zero. The same concept goes for rational expressions, the denominator can’t equal zero. That is why if there is ever a variable on the denominator it can’t equal any numbers that would make the denominator zero. In this case, if x equals 1, -1, 9, -9, 0 then the denominator would equal zero so that’s why we must put that x cannot equal these numbers.

Step 3: simplify

To simplify, cross out any factors that are the same on the top and the bottom.

Final Answer:

The final answer will equal all the numbers that remain on top multiplied together to equal the numerator, and the same goes for the denominator.

Week 14 – PreCalc 11

By aliessahv2016

On May 19, 2019

In Math 11, Uncategorized

This week in PreCalc 11 we learned about rational expressions. Rational Expressions are fractions whose numerator and denominator are both polynomials. They are usually given in unfactored form, and the point is to simplify them, just like a regular fraction, and determine which numbers x cannot equal.

Example:

Explanation:

Step 1: Factor Numerator

In this example, the factoring is simple. All that needs to be done is find two numbers that add to 11 and the product of those two numbers equal 30.

Step 2: Factor Denominator

In this example, the denominator is a difference of squares so to factor is two figure out the conjugates of the square roots of both terms of the polynomial.

Step 3: find what x can’t equal

For fractions, in general, the denominator can never equal zero. Same rule applies for rational expressions, the denominator cannot equal zero. In this example, if x was 5 or -5 then the denominator will equal zero which can never happen. So because of that, we must write that x cannot equal 5 or -5.

Step 4: simplify

Once the fraction is completely factored then cross out factors that are the same on the top and the bottom. In this example, (x +5) is on the top and the bottom so we can cross those out. We can only cross out complete factors that are the same, not single terms. For example, if there was a 3 on the top and 3 on the bottom then we couldn’t cross those out. We could cross them out if it was (x + 3).

Step 5: final answer

After simplifying the fraction, then that leaves you with the final answer.

Week 13 – PreCalc 11

By aliessahv2016

On May 12, 2019

In Math 11

This week in PreCalc 11 we learned how to graph absolute value functions.

Absolute Value Functions is a function that contains an algebraic expression within absolute value symbols. When we learned about Absolute Values before, we learned that whatever number is between the absolute value symbols must come out a positive. For example, | -3 | = 3 . Absolute Value Functions are similar in the sense that the y-values for Absolute Value Functions must be equal to or greater than zero.

Example:

y = | -5x + 10 |

Step 1: Graph the Parent Function

The first step to graphing this Absolute Value Function is to graph the parent function. The parent function is the same function but without the absolute value symbols. In this case, the parent function is y = -5x + 10.

Step 2: Move the Negative Y-Values

As I mentioned before, y-values in the parent function can’t be less than zero which is where the absolute value function comes in. As I mentioned earlier, | -3 | equals 3 because they’re both the same distance from zero on the number line. This relates back to absolute value functions because the negative y-values for the parent function, when put in the absolute value symbols, is a positive. So all we have to do is change the negative y-values into a positive and then graph it again.

That is how you graph Absolute Value Functions

Week 12 – PreCalc 11

By aliessahv2016

On May 6, 2019

In Math 11

This week in PreCalc 11 we learned how to solve systems of equations algebraically. A system of equations is a collection of two or more equations with a same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. The systems can either be both linear, be both quadratic, or be one linear and one quadratic. The reason for finding the unknowns (typically x and y) is because it’s the point that, when both the equations are graphed, they intersect. If the $x = 10$ and $y = 3$ then the point the equations intersect when graphed is $(10, 3)$ But sometimes we can’t graph both the equations to find the solution(s). So to figure it out, we have to solve in algebraically by using a method called substitution.

Example:

$x + y = 5$ $y = (x + 1)^2$

Step 1: Isolate One Variable

The first step to solving a system is to make one of the two equation equal either one of the two variables. In this example, one of the equations is equal to y. To make things easier you can make the simpler equation equal one of the variables, too.

Step 2: Substitute

Now that we have one equation equal y we can place that equation into the other equation. At any place of the equation that I see y I’m going to place the expression that y equals

$x + y = 5$ $y = (x + 1)^2$

$x + (x + 1)^2 = 5$

Step 3: Solve for X

$x + (x + 1)(x +1) = 5$

$x + x^2 + 2x + 1 = 5$

$x^2 + 3x + 1 = 5$

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

$(x + 4)(x - 1) = 0$

$x = -4$ or $x = 1$

Step 4: Substitute

Now we have two x- coordinates that could lead to the solution. There is possible to have two solutions, but even if we have 2 x-coordinates we could only end up with 1 solution. With the 2 x-coordinates that we have, we put them back into the other equation and solve for y

$x + y = 5$

$-4 + y = 5$

$y = 9$

$x + y = 5$

$1 + y = 5$

$y = 4$

Step 5: Verify

To make sure $(-4,9)$ and $(1, 4)$ are real solutions, we should plug both values into both equations.

$x + y = 5$

$-4 + 9 = 5$

$y = (x + 1)^2$

$9 = (-4 + 1)^2$

$x + y = 5$

$1 + 4 = 5$

$y = (x + 1)^2$

$4 = (1 + 1)^2$

Since both solutions make both equations true then that means they are real solutions and that is where the two equations intersect on a graph.

Week 11 – PreCalc 11

By aliessahv2016

On April 28, 2019

In Math 11, Uncategorized

This week in PreCalc 11, we learned about graphing quadratic inequalities with two variables. Inequalities state whether two values are equal, not equal, greater than or less than the other. The point of the inequality is to make it true, If the inequality states that $7 > 6$ then it’s true because 7 is greater than 6, but if it states $7 < 6$ , then it’s not true because 7 isn’t less than 6.

Inequalities can be expressed by quadratic equations. For example $y > x^2 - 2x -8$ . We are supposed to graph this quadratic equation in order to find possible solutions that make the statement true.

Step 1: Graph

To start this process, we have to graph the quadratic. We can start graphing this quadratic by one of two ways. One way is to convert this inequality from general form to vertex form. We can do this by the completing the square method. This form tells us where the parabola is located and what the stretch pattern is so we know the pattern to go up by. Another way we can graph this is to factor it. To factor we have to find two numbers that when multiplied equal -8 and when added equal -2:

$y > x^2 - 2x - 8$

$y > (x - 4)(x + 2)$

In this case the numbers were -4 and 2.

Now we know what the zeros of the equation is so that means the opposites are the x-intercepts. The opposite of -4 and 2 are 4 and -2 which means that those are the x-intercepts of the graph. Since we know where the x-intercepts are, we can find out the vertex. We can do this by adding the x intercepts together and dividing them by 2.

$(4)+ (-2) = 2$

$2/2 = 1$

Now we know that 1 is the x value of the vertex. To find the y value, we can put the x value into the original equation to solve for y.

$y > x^2 - 2x - 8$

$y > (1)^2 - 2(1) - 8$

$y > 1 - 2 -8$

$y > -9$

Now we have the y-value which means we know the vertex $(1 , -9)$

We also know that the stretch value, the value that determines the pattern the parabola follows, is 1 because there is no number attached to the $x^2$ in the inequality. That means the parabola follows the pattern of up 1 over 1, up 3 over 1, up 5 over 1, and so on.

The graph should look like the following:

Step 2: Determine the Solution

Now that we know what the parabola looks like, we have to figure out which coordinates on the graph make the inequality statement true. The points that make the statement true are either going to be all the ones inside the parabola or all the ones outside the parabola. To figure this out we can start by picking one point of the graph and dropping the x value and the y value in the original equation. If, when the inequality is solved, the statement is true, then all the points, where its located (i.e. outside the parabola) then all the points outside the parabola are true, so we would shade that area. If it was located inside the parabola then we would shade the inside of the parabola.

$(0,0)$

$y > x^2 - 2x -8$

$(0) > (0)^2 - 2(0) - 8$

$0 > -8$

When the coordinate $(0,0)$ is placed in the original inequality, a true statement came out. Zero is greater than negative eight, so that means, since $(0,0)$ , is inside the parabola, then all of inside the parabola will be shaded.

All the coordinates inside the parabola make the original inequality true.

Week 10 – PreCalc 11

By aliessahv2016

On April 22, 2019

In Math 11

Factoring has been an important part in PreCalc 11 so far and will probably be in the future as well. Although we have had a lot of practice with them, there are some trinomials that are not as easy to factor as we are used to.

Example:

$(x-6)^2 + 7 (x-6) + 10$

Step 1: Create a Variable in the Expression

The polynomials that we are used to factoring are the ones that have the pattern that follow $ax^2 + bx + c$ . Even though we don’t have that pattern in this equation we can make it so that is does. We can do this by substituting something common. Something that’s common in this expression is the $(x-6)$ . We can make $(x-6)$ equal any variable. In this case let’s make $(x-6) = x$ . So wherever we see $(x-6)$ we switch it for x

$x^2 + 7x + 10$

Now the expression follows the pattern that we are familiar with we can factor as usual.

Step 2: Factor

To factor we have to find a number that when added together equal seven and when multiplied equals 10. In this case the numbers are 5 and 2.

$(x+2)(x+5)$

Step 3: Substitute

We can’t forget about the term that we substituted. Since we made $x = (x-6)$ , everywhere we see x in the factored expression we have to switch it for $(x-6)$

$latex ((x-6) +2)((x-6)+5)

Step 4: Simplify

The only thing left to do is to simplify the expression.

$(x-4)(x-1)$

Week 9 – PreCalc 11

By aliessahv2016

On April 14, 2019

In Math 11, Uncategorized

This week in PreCalc 11, we reviewed and practiced how to read a graph and make an equation in the different forms.

The three forms that we worked with this unit was:

standard form or vertex form- $y = a (x-p)^2 + q$ ,where (p, q) is the vertex of the parabola, and a is the stretch value. It’s the easiest equation to graph because we are given a starting point and the pattern that it goes up by.

general form- $y = ax^2 + bx + c$ . A, B, and C are three real numbers. Once these are given, the values for x and y that make the statement true express a set of (x, y) points which form a parabola when graphed. It’s the most useless equation to graph because we are only given the y-intercept and can’t do much else unless it’s converted to one of the other forms.

factored form– $y = a(x-x1)(x-x2)$ . x1 and x2 are the opposite of the x-intercepts of a graph. If the x intercepts are +3 and +5, then x1 would be -3 and x2 would be -5, and the a value is the stretch value. This equation is useful only when given the x-intercepts and is used mostly when trying to search for the a value as long as there is another point given along the parabola to substitute y and x for to find a.

It’s simple to convert the equations between each other because it uses all the skills that we have learned and used earlier in the semester.

Example:

Step 1: Turning It Into Vertex Form

First we have to look at the vertex and see how far it has moved from (0,0). In this picture the vertex has moved 3 units to the left and 2 units down. The 3 units left is our p-value and the 2 units down is our q-value. If we put it into the vertex equation equation, we are left with:

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

$y = a (x-(-2)^2 + (-2)$

$y = a (x+2)^2 - 2$

Now we have the a-value left to find. If the a-value is one then the pattern the parabola would go up is up 1 over 1, up 3 over 1, up 5 over 1, up 7 over one, etc…Typically the amount the first jump goes up by is usually the a-value. If the first jump is up 2 and over one, then the pattern is probably 2. That means the original pattern of up 1 over 1, up 3 over 1 would be multiplied by 2 so the 2 pattern is up 2 over one, up 6 over one, up 10 over one, etc… In this case the pattern is 2 because the first jump is over up 2 over one so that’s the a-value. That means the complete vertex equation is:

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

Step 2: Convert Vertex Form into General Form

To convert from vertex form to general form is quite simple since it’s mostly expanding and simplifying.

Now that we have $y = 2 (x+2)^2 - 2$ , you just expand and simplify!

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

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

$y = 2 (x^2 + 4x +4) -2$

$y = (2x^2+ 8x +6)$

$y = 2x^2 + 8x +6$

Now we have general form of the equation!

Step 3: Turn the Equation into Factored Form

There are 2 ways that we can turn the graph into a factored form equation

First Way:

The first way is to factor the general form:

Step 1 is to factor out the 2 from the equation

$y = 2x^2 + 8x +6$

$y = 2 (x^2 + 4x + 3)$

Step 2 is to find two numbers that when multiplied equal 3 and when added equal 4. In this case, it’s 3 and 1.

$y = 2 (x + 3)(x + 1)$

Step 3 is to find a number that when added to or subtracted from +3 and +1 will equal zero. In this case, the answer is -3 and -1. That means we have found the x-intercepts. We also found out the stretch value.

Second Way:

Second way to get it into factored form is to look at the graph and find the x-intercepts. The x-intercept is where the parabola crosses the x axis. There may 2, or one, or even none. In this case, there are 2 and the x-intercepts are -3 and -1.

Since we are given the x-intercepts we can place them directly into the equation

$y = a (x - x1)(x - x2)$

$y = a (x- (-3)(x - (-1)$

$y = a (x + 3)(x + 1)$

We also know the a-value which is 2 so we can put the a value in. But if we didn’t have the a-value then we could put in another point and substitute x and y for the coordinates, but we know a so we don’t need to do that.

$y = 2 (x + 3)(x + 1)$

Now we have the factored form, and we know how to convert between the rest of equations!

Week 8 – PreCalc 11

By aliessahv2016

On April 6, 2019

In Math 11

This week in PreCalc 11 we learned the properties of quadratic functions and how to analyze the quadratic equation. We learned that when are given the quadratic equation in vertex form then we can graph the equation.

Last unit we learned the quadratic equation $ax^2 + bx + c = 0$ , this unit we learned $y = a(x-p)^2 + q$ . This is combination of the equations $y = (x-p)^2, y = x^2$ , and $y = x^2 + q$ . These equations are all very important in graphing the curve the equations make.

Example 1:

$y = x^2$ (parent function: creates the most basic parabola)

$y = x^2 + 3$ (addition to the parent function)

Explanation 1:

What this equation tells us is based on the +3 at the end of it. This +3 is the vertical translation which means its moving the vertex from (0,0), as it would be in the equation $y = x^2$ because the +3 is replaced with a O, to (0, 3). It moves the y-intercept, in other words, it moves the vertex up or down. If the equation was $y = x^2 - 3$ then it would move the vertex down to (0, -3).

*vertex is either the highest point or lowest point of the parabola

*parabola is the curve the equation makes

*y-intercept is the point where the parabola crosses the y axis

Example 2:

$y = (x - p)^2$ $y = (x-3)^2$

Explanation 2:

The part that tells us about the graph is the -3. This -3 is the horizontal translation which means it moves the vertex from (0,0) to (3,0). It moves the x-intercept, in other words, it moves the vertex left or right. the reason that it is a positive three, but the equation has a negative 3 is because when you place a +3 in the equation $y = (x-(+3)^2$ , the negative overtakes the positive. If it was a -3 then two negatives make a positive so it would be $y = (x+3)^2$

*x-intercept is where the parabola crosses the x axis

Example 3:

$y = ax^2$

$y = 2x^2$ (tall and skinny)

$y = 1/2x^2$ (wide)

Explanation 3:

The 2, in this case, is telling us about the stretch reflection which means that it tells us how wide the parabola will be or how skinny it will be. If the a value is greater than 1 then it’s tall and skinny. If a is less than one and greater than 0 then it’s wider.

That’s why when we are given an equation in vertex form ( $y = a(x-p)^2 +q$ ) then we can easily graph it. It’s called vertex form of the equation because you are given the vertex (the opposite sign of the p value is the x value and the q is the y-value). From the equation we are also given the stretch value which tells us how wide or skinny the parabola is going to be and also tells us the pattern we need to follow to get the correct function. The parent function, $y = ax^2$ gives us the vertex (O,O), and it tells us to follow the pattern up 1, over 1, up 3 over 1, up 5 over one, and it keeps going up by 2. But when there is a stretch value that isn’t one then that pattern changes. If the a value is 2, for example, then you multiply the basic pattern by 2 so the pattern is up 2 over 1, up 6 over 1, up 10 over one, and so on. If the a value is, for example, 1/2 then you multiply the basic pattern by one half so it would be so that you go up 1/2 over 1 and so on.

Example:

So if the equation is: $y = 3(x-6)^2 +4$

That’s how a quadratic equation in the vertex form is graphed.

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Category: Math 11 (Page 1 of 2)

Week 17 – PreCalc 11

Week 16 – PreCalc 11

Week 15 – PreCalc 11

Week 14 – PreCalc 11

Week 13 – PreCalc 11

Week 12 – PreCalc 11

Week 11 – PreCalc 11

Week 10 – PreCalc 11

Week 9 – PreCalc 11

Week 8 – PreCalc 11

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