Math 11 – Manroop's Blog

Week 18 – Final Review

manroopt2015 on June 18, 2018June 18, 2018

This week in Pre-Calculus 11 we are reviewing for the final. The top 5 things I learned in Pre-Calculus 11 are calculating the discriminant, CDPEU, solving equations, finding the vertex of a parabola, and special triangles.

Calculating the Discriminant: The discriminant has been helpful throughout Pre-Calculus 11 because with the discriminant you can find out how many solutions/roots/x-intercepts an equation has or whether it is an extraneous solution. The discriminant is derived from the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . The discriminant is the part that is under the square root sign $b^2-4ac$ . If the discriminant is negative that means that there are no solutions, it does not have a x-intercept and/or it is an extraneous solution because there cannot be a square root of a negative number. If the discriminant is equal to zero that means that the equation has one solution. If the discriminant is positive that means there are two solutions.

Ex. $x^2+3x+30$

$b^2-4ac$

$3^2-4(1)(30)$

$9-120$

$-111$

CDPEU: CDPEU was introduced in the quadratics unit. The acronym helps you remember the steps of factoring. The letter C represents “Is there anything in common?”.

Ex. $2x^2+8x-10$

$2(x^2+4-5)$

$2(x+5)(x-1)$

In the example above all of the terms shared a common multiple of 2, so I factored it out. After factoring the 2 out I continued on and factored.

The letter D represents “Is there a difference of squares”. A difference of squares only occurs in binomials. A difference of squares requires two square roots where one is negative.

Ex. $x^2-49$

$(x-7)(x+7)$

The example above is a difference of squares. Both the first term and the second term are perfect squares which means I could factor the expression easily. You have to make sure it is a DIFFERENCE of squares and not a SUM of squares, otherwise you will not be able to use this strategy.

The P represents the word pattern. The pattern that we look for when factoring is $a^2+bx+c$ . If the expression or equation does not have this pattern it is linear. If it is a quadratic then you have to categorize it using the next two letters.

$9x-3$

$3(x-1)$

The example above is not a quadratic expression but a linear expression because there is no $x^2$ . Even though this is a linear expression you can still factor it.

The letter E represents the word easy. This is when there is not a coefficient in front of the $x^2$ and it is easily factorable.

Ex. $x^2+x-20$

$(x+5)(x-4)$

The example above, is easy to solve because there are two numbers that multiply to 20 and add to 1.

The letter U represents the world ugly. An ugly quadratic equation is when there is a coefficient in front of the $x^2$ . This causes the quadratic expression to be hard to factor.

Solving Inequality Equations: To solve inequality equations algebraically we used the method of substitution. After finding each of the values of x and y it is important to always verify the solution by plugging it back in.

Ex. $y=x^2+6x+2$ and $y=2x-1$

$y=2x-1$

STEP 1: $2x-1=x^2+2x+4$

$0=x^2+6x-2x+2+1$

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

$0=(x+3)(x+1)$

$x+3=0$

$x=-3$

$x+1=0$

$x=-1$

STEP 2: $y=2x-1$ $x=-3$

$y=2(-3)-1$

$y=-6-1$

$y=-7$

$(-3,-7)$

STEP 3: $y=2x-1$ $x=-1$

$y=2(-1)-1$

$y=-2-1$

$y=-3$

$(-1,-3)$

STEP 4: $y=2x-1$ $(-1,-3)$

$-3=2(-1)-1$

$-3=-2-1$

$-3=-3$

STEP 5: $y=2x-1$ $(-3,-7)$

$-7=2(-3)-1$

$-7=-6-1$

$-7=-7$

In the example above, the first step was to isolate one of the variables from one of the equations but that was already done. The next step was to plug in the value we found in the previous step into the other equation. After that I brought all of the terms onto one side of the equation and then factored the equation to find out the $x$ values. Then I plugged in the x values into one of the equations to find the $y$ value. To make sure that both of the points that I found were solutions I plugged those values back into one of the original equations and made sure both of the side were equal to each other.

Finding the Vertex: Finding the vertex of a parabola is one of the most important things I’ve learned this year because we have used parabolas throughout the course. To find the vertex of a parabola you have to complete the square. When you complete the square you divide the middle term by two and then square it.

$y=x^2+8x-23$

$\frac{8}{2}^2=\frac{64}{4}=16$

$y=x^2+4x+16-16-23$

$y=(x+4)^2-39$

Vertex: $(-4,-39)$

In the example above I used completing the square to find the vertex. There was not a coefficient in front of the $x^2$ which made the equation easier to factor.

Special Triangles: One of the most important things I learned was the special triangles. Special triangles prevent the use of calculators because they always have the same patterns. One of the special triangles has a $1-\sqrt{3}-2$ pattern which means that the angles are 90-60-30. Another one of the special triangles is the $1-1-\sqrt{2}$ which has the angles of 45-45-90.

Ex. What is the exact cosine ratio for $225$ degres?

$225-180=45$

$\cos45=\frac{1}{\sqrt{2}}$

Week 17 – Trigonometry

manroopt2015 on June 11, 2018

This week in Pre-Calculus 11 we started the trigonometry unit. This week we learned about the sine law and the cosine law.

What is the Sine Law? The sine law is typically used when you are not dealing with right triangles. Although the sine law could be used with right triangles it involves more steps. When solving a triangle that doesn’t have a right angle you cannot use SOH CAH TOA, you have to use the sine law or cosine law. The formula for finding an angle is $\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{sin(C)}{c}$ . The formula for finding a side length is the reciprocal $\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$ .

Ex. In a triangle $\angle(B)=34$ , side $c$ is 14.0cm and $\angle(C)=65$ . What is the side length of $b$ to the nearest tenth?

$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$ .

$\frac{a}{\sin(A)}=\frac{b}{\sin(34)}=\frac{14}{\sin(65)}$

$\frac{b}{\sin(34)}=\frac{14}{\sin(65)}$

$b=\frac{14\times\sin(34)}{\sin65}$

$b=\frac{7.82}{\sin65}$

$b=8.638$

$b=8.6cm$

In the example above, the first step I did was plug in the known values into the formula. I knew this triangle was not a right triangle because it does not have a $\angle90$ . The second step was to isolate the variable and find it’s value.

Cosine Law: The Cosine Law is used when the Sine Law cannot be used. The cosine law can calculate a missing side length or angle. The formula for finding a missing side length is $a^2=b^2+c^2-2bc\cos(A)$ . The formula for finding an angle is $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$ .

Ex. $\triangle{MKT}$ , side m is 12 cm long, side k is 7 cm and side t is 13 cm long. What is $\angle{K}$ to the nearest degree?

$\cos{K}=\frac{m^2+t^2-k^2}{2(mt})$

$\cos{K}=\frac{12^2+13^2-7^2}{2(12)(13)}$

$\cos{K}=\frac{144+169-49}{312}$

$\cos{K}=\frac{264}{312}$

$\cos{K}=0.846$

$K=\cos^{-1}{0.846}$

$K=32.2$

$K=32$ degrees

In the example above I used the cosine law to solve for $\angle{K}$ . The first step was to fill in the values. After that, I solved for $\angle{K}$ by calculating one side of the equation and then using the inverse function of cosine to solve for $\angle{K}$

Week 16 – Pre Calc 11

manroopt2015 on June 3, 2018

This week in Pre-Calculus 11 we finished the Rational Expressions unit. This week we learned about the application of rational equations in real life situations.

Word Problem Basics: Word problems always contain valuable information for solving the problem, so it is always useful to read the question slowly and carefully. To understand the problem better you can create a chart or organize the information in a format you like. Being able to identify key words like greater than (+), less than (-), the sum (x+x), the difference (x-x), the quotient ( $\frac{x}{y}$ ), the product of ( $x*y=$ ), etc. These words are helpful to remember when solving a word problem. When giving the answer to a word problem always answer with a sentence and with the units.

Ex. Sally can do all the house’s chores in 55 minutes. When Sally and Jake work together, they can do all the chores in 32 minutes. How long does it take Jake to do the chores?

What We Know:

From the clues in the word problem we know the time it takes Sally to do all the chores and how long it takes Jake and Sally together to do all the chores. Because they complete all of the chores in that amount of time we know that they complete 100% of the chores, which can be represented as $\frac{1}{1}$ .

The Equation: $\frac{32}{55}+\frac{32}{x}=\frac{1}{1}$

The first fraction represents the portion which it takes Sally and Jake to do all the chores over the amount of time Sally takes by herself. The second fraction represents the amount of time Sally and Jake take to do all the chores over the time Jake can do the chores. The time that Jake takes to do the chores is represented by $x$ because we are solving for that variable. The sum of both the rational numbers is $1/1$ because they complete all of the chores.

Solving: $\frac{32}{55}+\frac{32}{x}=\frac{1}{1}$

$(55x)(\frac{32}{55}+\frac{32}{x})=(55x)(\frac{1}{1})$

$32x+1760=55x$

$1760=55x-32x$

$1760=23x$

$\frac{1760}{23}=x$

$76\frac{12}{23}=x$

Final Answer: It takes Jake around 76.5 minutes to do all the chores by himself.

Week 15 – Pre Calc 11

manroopt2015 on May 26, 2018

This week in Pre-Calculus 11 we continued the rational expressions unit. This week we learned how to add and subtract rational expressions and how to solve equations.

How To Add and Subtract Rational Expressions: When you add rational expressions you always have to find a common denominator. To find a common denominator it is best to figure out the lowest common multiple by factoring both of the denominators.

Ex. $\frac{3}{5}+\frac{7}{8}$

$(\frac{3}{5}\times\frac{8}{8})+(\frac{7}{8}\times\frac{5}{5})$

$\frac{24}{40}+\frac{35}{40}$

$\frac{59}{40}$ or $1\frac{19}{40}$

In the example above, I had to add two fractions together. The first step was to find a common denominator, the lowest common multiple between 5 and 8 is 40. Then I added the numerators together. After adding the fractions together you must check if the fractions simplify any further.

Ex. $\frac{13}{9}-\frac{4}{3}$

$(\frac{13}{9}\times\frac{2}{2})-(\frac{4}{3}\times\frac{6}{6})$

$\frac{26}{18}-\frac{24}{18}$

$\frac{2}{18}$

$\frac{1}{9}$

In the example above I subtracted two fractions together. The steps were the exact same as when I added. The first step was to find a common denominator, the lowest common multiple between 9 and 3 is 18. After changing the denominators I subtracted the numerators. Then I simplified the fraction.

How To Add and Subtract Rational Expressions with Variables: Similarly, when dealing with variables while adding and subtracting you still need to find a common denominator.

Ex. $\frac{12}{5x}+\frac{7x}{2}$

$(\frac{12}{5x}\times\frac{2}{2})+(\frac{7x}{2}\times\frac{5x}{5x})$

$\frac{24}{10x}+\frac{35x^2}{10x}$

$\frac{35x^2+24}{10x}$

$x\neq0$

In the example above, I found a common denominator of 10x and then I added the fractions together.

Ex. $\frac{8}{6x+9}+\frac{3}{4x-4}$

$(\frac{8}{6x+9}\times\frac{4(x-1)}{4(x-1))}+(\frac{3}{4x-4}\times\frac{3(2x+3)}{3(2x+3)})$

$\frac{32x-32}{3(2x+3)(4(x-1))}+\frac{18x+27}{3(2x+3)(4(x-1))}$

$\frac{32x-32+18x+27}{3(2x+3)(4(x-1))}$

$\frac{50x-5}{12(2x+3)(x-1)}$

$\frac{5(10x-1)}{12(2x+3)(x-1)}$

$2x+3\neq0$

$2x\neq-3$

$x\neq\frac{-3}{2}$

$x-1\neq0$

$x\neq1$

In the example above, the first thing I did was factor both of the fractions. After, I found a common denominator by multiplying both of the denominators together. Then, I made a big fraction and added the like terms together and simplified. One key step is to remember to find the non-permissible values.

How To Solve Fractional Equations: There are a number of different ways to solve fractional equations that involve variables. If the equation has one fraction on both sides of the equals sign then you can cross multiply the fractions. Because we are solving, you have to isolate the variable and find out it’s value.

Ex. $\frac{x+2}{x-3}=\frac{x-1}{x-2}$

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

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

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

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

$4x-7=0$

$4x=7$

$x=\frac{7}{4}$

$x-3\neq0$
$x\neq3$

$x-2\neq0$

$x\neq2$

In the example above, the first step was to cross multiply. After cross multiplying I collected all the like terms. Then, I isolated x and found x’s value.

Week 14 – Rational Expressions

manroopt2015 on May 20, 2018

This week in Pre-Calculus 11 we started the rational expressions unit. This week we learned about equivalent rational expressions and how to multiply and divide rational expressions.

What is a Rational Expression: A rational expression is a number that can be expressed as a fraction. For example, the number 3 can be represented as $\frac{3}{1}$ , therefore it is a rational number. Irrational numbers are numbers that form an endless amount of decimals and are non-repeating.

Non Permissible Values: When there is a variable in the denominator of a fraction you have to give it a restriction. This is an important step because the denominator of a fraction can never be zero because that would make the fraction undefined.

Ex. $\frac{1}{x-4}$

$x-4\neq0$

$x\neq4$

In the example above, because there was a variable in the denominator I had to create a restriction. The first step is to make the denominator equal to zero. Then, I isolated the variable to figure out a value that would make the fraction undefined.

What are Equivalent Rational Expressions? Equivalent rational expressions are fractions where the numerator and denominator have something in common, so that they can simplify.

Ex. $\frac{10}{25}$

$\frac{10}{25}=\frac{5\times2}{5\times5}$

$\frac{10}{25}=\frac{2}{5}$

In the example above, I simplified a fraction by taking something out that both the numerator and denominator had in common. I took something in common out by factoring the fraction first, then cancelling out equal values. Similarly, in fractions that involve variables you have to factor the numerator and denominator first and then take out common terms.

Ex. $\frac{3x-12}{x^2+x-20}$

$\frac{3(x-4)}{(x+5)(x-4)}$

$\frac{3}{x+5}$

$x-4\neq0$

$x\neq4$

$x+5\neq0$

$x\neq-5$

In the example, the first thing I did was factor the numerator and denominator. Then, I took out the factors that they had in common. After, I figured out the non permissible values for the original expression and the factored expression.

How To Multiply and Divide Rational Expressions: To multiply and divide rational expressions you just have to multiply straight across. But it is important to factor first because it will be easier to simply the expression later.

Ex. $\frac{2x+2}{3x}\times\frac{x^2-2x}{4x+4}$

$\frac{2(x+1)}{3x}\times\frac{x(x-2)}{4(x+1)}$

$\frac{2(x+1)(x)(x-2)}{3x(4)(x+1)}$

$\frac{2x(x-2)}{3x(4)}$

$\frac{2x(x-2)}{12x}$

$\frac{x-2}{6}$

$3x\neq0$

$x\neq0$

$x-2\neq0$

$x\neq2$

The first thing I did was factor both of the rational expressions and then I took out the factors that were the same. After that, I simplified the fraction because 2 and 12 share a common factor of 2. Then, I found the non permissible values for the original and factored rational expression.

How To Divide Rational Expressions: To divide rational expressions you do the exact same as multiplying, by factoring and then taking out common factors. But when dividing fractions you have to remember to multiply by the reciprocal of the second fraction. You also have to find the non permissible values for the variable.

Ex. $\frac{x+5}{x-4}\div\frac{x^2-25}{3x-12}$

$\frac{x+5}{(x-4)}\div\frac{(x-5)(x+5)}{3(x-4)}$

$\frac{x+5}{(x-4)}\times\frac{3(x-4)}{(x-5)(x+5)}$

$\frac{(x+5)(3)(x-4)}{(x-4)(x-5)(x+5)}$

$\frac{3}{x-5}$

$x-4\neq0$

$x\neq4$

$x+5\neq0$

$x\neq-5$

$x-5\neq0$

$x\neq5$

For dividing the rational expression, the first step I did was to factor both of the fractions. Then, I reciprocated the second fraction and multiplied them together. After, I took out the common factors and everything that was left over become the new rational expression.

Week 13 – Absolute Value And Reciprocal Functions

manroopt2015 on May 14, 2018

This week in Pre-Calculus 11 we started the Absolute Value and Reciprocal Functions unit. This week we learned how to graph an absolute value linear function/quadratic function and we learned about reciprocal functions.

What is an Absolute Value Function? An absolute value function is always in the top half of the graph because the y value has to be positive. Linear absolute value functions make a v-shape and quadratic absolute value functions make a w shape.

Critical Point: The critical point or the point of inflection is where the line or parabola changes direction. The critical point is always the x-intercept.

How To Graph an Absolute Value Function: When graphing an absolute value function it is always useful to graph the original function first. For linear functions, after graphing the original function, you have to look where the line hits the x-axis and then you flip the line by making the slope the opposite.

Ex. $y=\mid2x+9\mid$

Original Function: $y=2x+9$

Absolute Value Function: $y=\mid2x+9\mid$

y-intercept: (0,9)

x-intercept: (-4.5,0)

Domain: $x\varepsilon\mathbb{R}$

Range: $y\geq0$

Piecewise Notation: $f(x)=2x+9,x\geq-4.5$

$f(x)=-(2x+9),x<-4.5$

In the example above, I graphed a linear absolute value function. In the example, as soon as the line hit the x-axis it changed directions. The slope of the original function is $\frac{2}{1}$ but once the line hit the x-axis the slope became $\frac{-2}{1}$ .

Ex. $y=\mid$ $x^2-x-6\mid$

Original Function: $y=x^2-x-6$

Absolute Value Function: $y=\mid$ $x^2-x-6\mid$

y-intercept: (0,6)

x-intercept: (-2,0) and (3,0)

Domain: $x\varepsilon\mathbb{R}$

Range: $y\geq0$

Piecewise Notation: $f(x):x^2-x-6,x\leq-2$ or $x\geq3$

$f(x):-(x^2-x-6),-2<x<3$

In the example above, I graphed a quadratic absolute value function. The first step was to graph the original function. After graphing the original function, I saw where the parabola hit the x-axis and then flipped everything below the x-axis. The original vertex was (0.5,-6.25) but with the absolute value function it turned into (0.5, 6.25) because the y value had to become positive.

What is a Reciprocal Function? The reciprocal function of a quadratic or linear function is one over the original function.

Ex. $y=x+2$

Reciprocal Function: $y=\frac{1}{x+2}$

When a reciprocal function is graphed it creates two curves also known as the hyperbolas. The reciprocal of 1 is $\frac{1}{1}$ which equals 1, so it stays the same. Similarly, the reciprocal of -1 is $\frac{-1}{1}$ which equals -1. The points of 1 and -1 on the y-axis are the invariant points (they do not change) these are used to draw the asymptotes. The asymptote is used to separate the hyperbolas. This year, one of the asymptotes is always going to be the x-axis.

Ex. $f(x)=3x+4$

$f(x)^{-1}=\frac{1}{3x+4}$

In the example above, the first thing I did was graph the original function. After graphing the original function I found where the line meets at (_, 1) and (_ , -1) to draw my asymptotes. Then I drew my hyperbolas accordingly.

Week 12 – Pre Calc 11

manroopt2015 on May 7, 2018

This week in Pre-Calculus 11 we finished the solving quadratic inequalities unit. This week we learned how to solve linear-quadratic systems and quadratic-quadratic systems by using algebra.

How To Solve Systems Algebraically: To solve systems algebraically you have to use a method called substitution. Substitution is a concept that we learned in the systems unit from Math 10.

Substitution: Substitution is when you isolate a variable of an equation and then plug it into the other equation. After that you use the value that you solved for and plug it in into one of the equations to figure out the second variables value.

Linear-Quadratic Systems: Linear-Quadratic Systems can have 0,1, and 2 possible points of intersection. You can figure the points of intersection by graphing or using substitution algebraically.

Ex. $y=-2x^2+8$ / $3x-y=-3$

STEP 1: $3x-y=-3$

$3x+3=y$

STEP 2: $3x+3=-2x^2+8$

$2x^2-8+3x+3$

$2x^2+3x-5$

$(2x+5)(x-1)$

STEP 3: $2x+5=0$

$2x=-5$

$x=\frac{-5}{2}$

$x-1=0$

$x=1$

STEP 4: ( $\frac{-5}{2}$ ,y)

$3x+3=y$

$3(\frac{-5}{2})+3=y$

$\frac{-15}{2}+3=y$

$\frac{-15}{2}+\frac{6}{2}=y$

$\frac{-9}{2}=y$

$(\frac{-5}{2}.\frac{-9}{2})$

STEP 5: ( $1$ , y)

$3x+3=y$

$3(1)+3=y$

$3+3=y$

$6=y$

(1,6)

In the example above I solved the linear-quadratic system by breaking up the process in steps. In Step 1, I isolated $y$ because it was the easiest variable to isolate. By isolating $y$ I was able to figure out it’s value. In Step 2, I plugged in the value we found in step 1 into the other equation. After plugging in the value I found for $y$ I factored the equation to find out the x-intercepts (the values for x). In Step 3, because we factored the equation in the previous step I used those factors to isolate x and find the roots of the equation (the values for x). In Step 4, I took the first value I found for x and plugged it into one of the equations to find the value of y, that gave me a point where the linear-quadratic system intersects. In Step 5, I did the same thing as step 4 except I used the other value of x we found and plugged it into one of the equations to find the other point of intersection.

Quadratic-Quadratic Systems: Quadratic-Quadratic Systems can have 0,1,2, and an infinite amount of points of intersection. In order to figure out the number of points of intersection you can graph the parabolas or solve it algebraically by using substitution.

Ex. $y-4=x^2$ / $y=-x^2+12$

STEP 1: $y-4=x^2$

$y=x^2+4$

STEP 2: $y=-x^2+12$

$x^2+4=-x^2+12$

$x^2+x^2+4-12=0$

$2x^2-8=0$

$2(x^2-4)=0$

$2(x-2)(x+2)=0$

STEP 3: $x-2=0$

$x=2$

$x+2=0$

$x=-2$

STEP 4: $(2,y)$

$y=x^2+4$

$y=(2)^2+4$

$y=4+4$
$y=8$

(2,8)

STEP 5: $(-2,y)$

$y=x^2+4$

$y=(-2)^+4$

$y=4+4$

$y=8$

(-2,8)

In the example above, I did the exact same steps as I did in the linear-quadratic system example. The first step was to isolate a variable, then I plugged that value into the other equation, after that I factored the equation and found the values for x, and finally I plugged those values for x into one of the equations to figure out the values for y.

Week 11 – Solving Quadratic Inequalities

manroopt2015 on April 29, 2018April 29, 2018

This week in Pre-Calculus 11 we started the Solving Quadratic Inequalities unit. This week we learned about how to solve inequalities with one variable, graphing linear inequalities, and graphing quadratic inequalities.

How to Solve an Inequality with One Variable: To solve an inequality you have to isolate the variable in order to figure out the possible values of the variable, then you should test a number accordingly. When solving an inequality there are some rules that you need to remember. In order for the inequality to be true when you are diving by a negative number or multiplying by a negative number you MUST flip the inequality symbol.

NOT FLIPPING THE SIGN

Ex. $\frac{1}{-2}x>2$

$x>(2)(-2)$

$x>-4$

TEST: $x=0$

$\frac{1}{-2}\times0>2$

$0>2$

FLIPPING THE SIGN

Ex. $\frac{1}{-2}x>2$

$x<2\times-2$

$x\prec-4$

TEST: $x=-5$

$\frac{1}{-2}\times\frac{-5}{1}>2$

$\frac{-5}{-2}>2$

$2.5>2$

In the example above I showed how the inequality is not true unless you flip the sign when multiplying and dividing by a negative number. When I didn’t flip the sign, the inequality for $x$ did not work which meant that it was false. The graph in the example represents the inequality. The shaded area represents all of the possible values for x in order to make the inequality true. The shaded area and the unshaded area is separated by a broken line this represents the < sign.

How to Solve a Quadratic Inequality: To solve a quadratic inequality you have to factor the inequality and find the values of x. After finding the possible values of x you can test points between the values of x using a number line.

Ex. $x^2+12x+20>0$

$(x+10)(x+2)$

$x+10=0$ $x+2=0$

x=-10 x=-2

$-10>x>-2$

In the example above, I used a number line to solve the quadratic inequality. The first step I did was to find the zero’s of the quadratic function by factoring. Then I placed the zero’s on a number line and tested numbers between the zero’s. After testing the values using the original inequality I found out what numbers satisfied the inequality.

Week 10 – Pre Calc 11

manroopt2015 on April 21, 2018April 21, 2018

This week in Pre-Calculus 11 we prepared for the midterm. Although we didn’t learn anything new, we did review many things from the previous units that I forgot about.

Sequences and Series

Sequences and Series was the very first unit of the year. One of the concepts that I found confusing was dealing with percentages in word problems.

Ex. Billy makes $1240 a month. His boss recently told him that he will receive a 3.02% raise at the end of every month. How much will he make at the end of the first month? How much in total will he make at the end of the ninth month?

First Month:

a= $1240

$r=1.00+0.0302=1.0302$

$1240\times1.0302$

$1277.448$

$1277.45

Ninth Month:

a= $1277.45

$r=1.00+0.0302=1.0302$

n= 9

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{1277.45(1-1.0302^9}{1-1.0302})$

$S_n=\frac{1277.45(1-1.307055142)}{-.0302})$

$S_n=\frac{1277.45(-.307055142)}{-.0302})$

$S_n={-392.2475909}\div{-.0302}$

$S_n=12988.33082$

$12,988.33

In the example, the first thing I did was figure out what the common ratio was. I knew that the common ratio was above 1.0 because in the example it says that Billy will receive a 3.02% raise in his cheque, which means that his pay is increasing. Then, I found the amount he made at the end of the first month by multiplying $1240\times1.0302$ because that would be the “a” value in the second part of the question.

Infinite Sum: An infinite sum can be calculated when a geometric series converges because the series decreases a sum can be calculated. The sum can only be calculated if $-1< r< 1$ . This means that r is a fractional value and cannot be an improper fraction. The formula to calculate the infinite sum is $S_{\infty}=\frac{1}{1-r}$

Ex. 3, 9, 27, 81…

$r=\frac{t_n}{t_{n-1}}$

$r=\frac{9}{3}$

$r=3$

INFINITE SUM CANNOT BE CALCULATED

The infinite sum for the example above cannot be calculated because the common ratio is 3.

Ex. 10, -5, $\frac{5}{2}$ …

$r=\frac{t_n}{t_{n-1}}$

$r=\frac{-5}{10}$

$r=\frac{-1}{2}$

$S_{\infty}=\frac{a}{1-r}$

$S_{\infty}=\frac{10}{1-\frac{-1}{2}}$

$S_{\infty}=\frac{10}{1+\frac{1}{2}}$

$S_{\infty}=\frac{10}{\frac{3}{2}}$

$S_{\infty}=\frac{10}{1}\times\frac{2}{3}$

$S_{\infty}=\frac{20}{3}$

Absolute Value and Radicals

The absolute value and radicals unit was the second unit of the year. In this unit we learned about how the absolute value of a number is really the number of spaces it is away from 0.

Ex. $\mid-7x+x^2-10\mid$ , $x=6$

$\mid-7(6)+6^2-10\mid$

$\mid-42+36-10\mid$

$\mid-16\mid$

In the example above, I was given a value for the variable so I just had to plug in the value to find the absolute value of the expression.

Analyzing Quadratic Functions

This was the most latest unit we finished. In this unit I struggled with some of the word problem questions involving substitution.

Ex. A graph of a quadratic equation passes through (6,2) and (0,9), the axis of symmetry is x= 7. Determine an equation of the function.

STEP 1: $y=a(x-x_1)^2+q$

$y=a(x-7)^2+q$

$2=a(6-7)^2+q$

$2=a(-1)^2+q$

$2=a+q$

STEP 2: $y=a(x-x_1)^2+q$

$y=a(x-7)^2+q$

$9=a(0-7)^2+q$

$9=49a+q$

$9-49a=q$

STEP 3: $2=a+q$

$2=a+9-49a$

$2=-48a+9$

$2-9=-48a$

$-7=-48a$

$\frac{7}{48}=a$

STEP 4: $q=9-49a$

$q=9-49(\frac{7}{48})$

$q=9-\frac{343}{48}$

$q=\frac{89}{48}$

STEP 5: $y=a(x-x_1)^2+q$

$y=\frac{7}{48}(x-7)^2+\frac{89}{48}$

In the first step, I began by using standard form because I had the axis of symmetry. Then, I plugged in the point (6,2) into the x and y, this created an equation with two unknown variables. In step two, I figured out the value of “q” by using the other point given in the question (0, 9). In step three, I plugged in the value I got for “q” into the first equation I developed to find a value for “a” because we found a value for “q” we just needed to isolate “a” to find the numerical value of it. In step 4, I took the value I got for “a” and plugged it into the equation $q=9-49a$ to find the numerical value of q. Then, in step 5 I took all the information we figured out in the 4 steps and made it into an equation.

Week 9 – Pre Calc 11

manroopt2015 on April 15, 2018April 15, 2018

This week in Pre-Calculus 11 we finished the Analyzing Quadratic Functions unit. This week we learned about factored form and modelling problems with quadratic functions. Factored form is very helpful when looking for the x-intercept and it can also help with finding the axis of symmetry.

What is Factored Form? The formula for factored form is $y=a(x-x_1)(x-x_2)$ . This formula is very useful when looking for the roots of a quadratic function. You can easily convert from general form to factored form by factoring and finding the solutions of the function.

Ex. $y=x^2+6x-16$

-1 x 16= -16 -1+16= 15

-2 x 8= -16 -2+8= 6

-4 x 4= -16 -4+4= 0

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

$x+8=0$

x= -8

$x-2=0$

x= 2

x-intercepts= (-8,0) and (2,0)

y-intercept= (0, -16)

Opens up (minimum)

Congruent to $y=x^2$

How to Find the Axis Of Symmetry: To find the axis of symmetry you have to find the average of the x-intercepts by adding them together and then dividing the sum by two. After finding the axis of symmetry you can find the vertex by plugging the x value into the equation.

Ex. $y=2x^2+7x-4$

$y=(2x-1)(x+4)$

$2x-1=0$

$2x=1$

$x=\frac{1}{2}$

$x+4=0$

$x=-4$

x-intercepts= $(-4,0)$ and $(\frac{1}{2},0)$

AOS: $x=\frac{\frac{1}{2}+-4}{2}$

AOS: $x=\frac{\frac{1}{2}+\frac{-8}{2}}{2}$

AOS: $x=\frac{\frac{-7}{2}}{2}$

AOS: $x=\frac{-7}{4}$

$y=(2(\frac{-7}{4}-1)(\frac{-7}{4}+4)$

$y=(\frac{-7}{2}-\frac{2}{2})(\frac{-7}{4}+\frac{16}{4})$

$y=(\frac{-9}{2})(\frac{9}{4}$

$y=\frac{-81}{8}$

Vertex: $(\frac{-7}{4},\frac{-81}{8})$

How to Deal with Word Problems: To deal with word problems involving quadratic functions the first step is to find a relationship between the variables. After finding the relationship between the variables find the x-intercept. Then find the axis of symmetry and plug it into the equation to find the y value. The answer of the word problem is going to involve the vertex of the quadratic function.

Ex. Find two integers with the difference of 11 and the greatest product.

$x-y=11$