Math 11 – Page 2 – Manroop's Blog

Week 8 – Analyzing Quadratic Functions

manroopt2015 on April 8, 2018

This week in Pre-Calculus 11 we began the Analyzing Quadratic Functions unit. This week we learned about the different properties of quadratic functions. To graph a quadratic function it is very important to have the ability to identify the properties of a quadratic function by looking at the equation.

What is General Form? General form is $ax^2+bx+c$ where c represents the y-intercept. In general form “a” represents the direction the parabola will open (if “a” is positive the parabola will open up and if “a” is negative the parabola will open down). “A” also represents the scale of the parabola and whether the parabola will be compressed or stretched.

The Scale of a Parabola: The scale of a parabola can be determined by the coefficient in front of the $x^2$ . If the coefficient is 1 the scale of the parabola will be 1-3-5 which means it is congruent to the parent function ( $y=x^2$ ). The 1-3-5 pattern means that when you start at the vertex you go one to the right and rise one and then you go one to the right and rise 3… If the coefficient of $x^2$ is not 1 then you just have to multiply 1-3-5 by the coefficient to figure out the scale.

Vertex: The vertex is the highest or lowest point of the parabola.

Axis of Symmetry: The axis of symmetry intersects the parabola at the vertex. The axis of symmetry splits the parabola exactly in half.

Minimum and Maximum Point: The minimum and maximum point of the parabola is also determined by the coefficient on $x^2$ . The minimum and maximum point is whether the vertex of the parabola is at the highest or lowest point it could possibly be at. If the vertex is at it’s minimum point that means it opens up and the coefficient on $x^2$ is positive. If the vertex is at it’s maximum point that means it opens down and the coefficient on $x^2$ is negative.

Domain: The domain of a quadratic function is all the possible values of x, the domain is alway $x\in\Re$ .

Range: The range is all the possible y values, this is dependent on the vertex.

Ex. $y=x+4x-3$

y intercept= -3

Congruent to $y=x^2$

Scale: 1-3-5

Minimum Point (opens up)

Domain: $x\in\Re$

Ex. $y=-7x^2+5x-8$

y-intercept= -8

Not congruent to $y=x^2$

Scale: -7- -21 – -35

Maximum Point (opens down)

Domain: $x\in\Re$

What is Standard Form? Standard Form is $y=a(x-p)^2+q$ . Standard form is also known as vertex form because p and q represent the vertex’s coordinates. P also tells us the horizontal translation and q also represents the vertical translation.

Ex. $y=4(x-3)^2+1$

Vertex: (3,1)

Minimum Point (opens up)

Axis of Symmetry: 3

Domain: $x\in\Re$

Range: $y\geq1$

Not Congruent to $y=x^2$

Scale: 4-12-20

Horizontal Translation: 3 to the right

Vertical Translation: 1 up

How to Convert General Form to Standard Form: You can easily go from general form to standard form by completing the square.

Ex. $y=x^2-10x+3$

$\frac{-10}{1}\times\frac{1}{2}$

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

$-5^2=25$

$y=x^2-10x+25-25+3$

$y=(x-5)^2-22$

Week 7 – Pre Calc 11

manroopt2015 on March 19, 2018April 14, 2018

This week in Pre-Calculus 11 we finished the solving quadratic equations unit. This week we learned about the discriminant of a quadratic equation. The discriminant can be very helpful when solving a quadratic equation because it helps determine how many solutions the equation has.

What is the Discriminant? The discriminant is from the quadratic formula. The discriminant can be calculated by using $b^2-4ac$ . Depending on whether the discriminant of a quadratic equation is positive, negative or zero you can determine how many solutions the equation may have. If the discriminant is a positive number the equation will have two solutions, if the discriminant is negative there will be no solutions, and if the discriminant is equal to zero then the equation will have one solution.

Ex. $4x^2+3x-15=0$

a= 4

b= 3

c= -15

$b^2-4ac$

$3^2-4(4)(-15)$

$9+256$

265

The example above will have 2 solutions because the discriminant is positive.

Ex. $x^2+5x+7=0$

a= 1

b= 5

c= 7

$b^2-4ac$

$5^2-4(1)(7)$

$25-28$

-3

The example above will have 0 solutions because the discriminant is negative. The discriminant is the radicand of the quadratic formula and because there cannot be negatives in the radicand there will be no solution.

Ex. $9x^2+6x+1=0$

a= 9

b= 6

c= 1

$b^2-4ac$

$6^2-4(9)(1)$

$36-36$

The example above will have one solution because the discriminant is equal to zero. Before solving for the discriminant you can simplify the equation by taking out a common factor from all of the terms.

$2x^2+6x-8$

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

a= 1

b= 3

c= -4

$b^2-4ac$

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

$9+16$

The discriminant above is positive which means that there will be two solutions to the equation.

Week 6 – Pre Calc 11

manroopt2015 on March 11, 2018March 17, 2018

This week in Pre-Calculus 11 we continued the quadratics unit. This week we learned about different ways to solve quadratic equations and how to factor perfect square trinomials.

What is a Perfect Square Trinomial? A perfect square trinomial is when the first term and the third term are perfect squares. A perfect square trinomial cannot have a third term that is negative.

Ex. $4x^2+20x+25$

$(2x+5)^2$

The example above is a perfect square trinomial because both the first and third term are perfect squares. To factor the perfect square trinomial I found the square root of the first term and the square root of the third term.

Ex. $x^2-9x-36$

$(x-12)(x+3)$

Although the first and third term are perfect squares, the expression is NOT considered a perfect square trinomial because the last term is negative. The expression is not a perfect square trinomial but it can still be factored.

What is Completing the Square? Completing the square is when the second or third term in a perfect square trinomial is missing. When the second term is missing you have to take the third term find the square root of it and then multiply the square root by 2.

Ex. $x^2$ +___+100

$\sqrt100=10$

$10*2=20$

$x^2+20x+100$

$(x+10)^2$

When the third term of a perfect square trinomial is missing you can take the second term divide it by 2 and then square it to find the third term.

Ex. $x^2+7x$ + ___

$\frac{7}{1}\times\frac{1}{2}=\frac{7}{2}$

$(\frac{7}{2})^2=\frac{49}{4}$

$x^2+7x+\frac{49}{4}$

$(x+\frac{7}{2})^2$

What is Solving a Quadratic Equation? Solving a quadratic equation is when you have to find a value of x. When solving a quadratic formula there will always be an equals sign at the end of the equation. After factoring the equation you have to isolate the variable in order to find the value of x.

Ex. $x^2-6x-27=0$

$(x-9)(x+3)=0$

$x-9=0$

$x=9$

$x+3=0$

$x=-3$

$x=9,-3$

Solving a Quadratic Equation By Completing the Square: Another way to solve quadratic equations is by completing the square. Although this strategy can be time consuming it’s easy and it works.

Ex. $x^2+6x+23=0$

$3+3=6$

$3^2=9$

$x^2+6x+9-9+23=0$

$(x+3)^2-9+23=0$

$(x+3)^2+14=0$

$(x+3)^2=-14$

$\sqrt{(x+3)^2}=\pm\sqrt{14}$

$x+3=\pm\sqrt{14}$

$x=-3\pm\sqrt{14}$

In the example above in front of $\sqrt{14}$ is the plus or minus sign ( $\pm)$ . The $\pm$ is there because the square root of 14 can be -7 or 7. I did not calculate the value of $\sqrt{14}$ because by leaving it in the square root makes it an exact value.

The Quadratic Formula: The quadratic formula is another method to solve a quadratic equation. The formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . By plugging in the values of a, b and c, you can simply calculate the value of x.

$ax^2+bx+c=0$

Ex. $4x^2+5x=8$

$4x^2+5x-8=0$

a= 4 b= 5 c= -8

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-5\pm\sqrt{5^2-4(4)(-8)}}{2(4)}$

$x=\frac{-5\pm\sqrt{25+128}}{8}$

$x=\frac{-5\pm\sqrt{153}}{8}$

$x=\frac{-5\pm\sqrt{9*17}}{8}$

$x=\frac{-5\pm3\sqrt{17}}{8}$

Week 5 – Solving Quadratic Equations

manroopt2015 on March 4, 2018March 10, 2018

In the fifth week of Pre-Calculus 11 we began the Quadratics unit. This week we reviewed factoring polynomials and learned a few new strategies about factoring.

What is Factoring? Factoring is taking an expression and breaking it up into pieces so that you can multiply those pieces together to get the original number. For example, the number 9 can be factored to $9*1$ or $3*3$ . To factor polynomials it’s helpful to think about the acronym CDPEU.

C- Is there anything COMMON within the terms?

Ex. $7x^3+14x$

$7x(x^2+2)$

In the expression above, the greatest common factor is $7x$ because $7x$ can factor into both of the terms.

D- Is there a DIFFERENCE of squares? A difference of squares requires the expression to be a binomial, the terms have to be subtracting (difference) and both of the terms have to be perfect squares.

Ex. $100x^2-4$

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

P- Does the expression have the right PATTERN? To factor a trinomial the expression has to have the correct pattern by having $x^2$ , x, and a number.

Ex. $3+y-x$

The expression above cannot be factored because it does not have the pattern $x^2$ , x, and a number. It also does not have any common terms.

Ex. $x^2+7x+12$

$1*12=12$ $1+12=13$

$2*6=12$ $2+6=8$

$3*4=12$ $3+4=7$

$(x+4)(x+3)$

The trinomial above can be simplified because it has the pattern we are looking for. To find out what numbers I needed to use I listed all of the factors of 12. After listing the factors of 12 I figured out which one of the factors adds up to the middle term (7).

E– Is the expression EASY to factor? To determine the complexity of an expression you have to check if the $x^2$ has a coefficient. If the $x^2$ does not have a coefficient then it is an easy expression to factor.

Ex. $x^2-12x+20$

$-1*-20=20$ $-1+-20=-21$

$-2*-10=20$ $-2+-10=-12$

$-4*-5=20$ $-4+-5=-9$

$(x-10)(x-2)$

U- Is the expression UGLY to factor? A trinomial that’s difficult to factor is when the $x^2$ has a coefficient.

Ex. $2x^2-9x+4$

$(2x-1)(x-4)$

To factor the expression above I multiplied the first and the last term together. After multiplying the two terms together I got $8x^2$ , then I began to list all the factors of $8x^2$ . After figuring out the factors of $8x^2$ I chose the two factors that add up to -9x. Lastly, I placed the two factors in the chart above and found the GCF of the terms that were horizontal and vertical from each other.

Week 4 – Pre Calc 11

manroopt2015 on February 24, 2018February 24, 2018

In the fourth week of Pre-Calculus 11 we continued the absolute value and radicals unit. This week we learned about addition and subtraction of radicals.

Addition and Subtraction of Radicals: To add and subtract radicals, the radicands have to be of the same value. Not only do the radicands have to be the same, but the index also has to be the same. The coefficients in front of the radical are added or subtracted, not the radicand.

Ex. $\sqrt{23}+\sqrt[4]{23}$

The example above cannot be added together because the index’s are not the same. Although, the radicals have the same radicand, the index of the radicals are different.

Ex. $2\sqrt{21}+\sqrt{21}$

$3\sqrt{21}$

The expression above could be added together because the radicand and the index are the same. I added the coefficients of the radical together and not the value inside the radical.

Ex. $\sqrt{20}+8\sqrt{18}-2\sqrt{50}+\sqrt{45}$

$\sqrt{4\times5}+8\sqrt{9\times2}-2\sqrt{25\times2}+\sqrt{9\times5}$

$2\sqrt{5}+24\sqrt{2}-10\sqrt{2}+3\sqrt{5}$

$5\sqrt{5}+14\sqrt{2}$

In the example above, before adding/subtracting the radicals I simplified everything inside the radical. By simplifying the radical first, I was able to tell whether I could add/subtract the radicals together.

How to Multiply Radicals: To multiply radicals you have to multiply the coefficients by the coefficients and multiply the radicand by the radicand.

Ex. $3\sqrt{5}\times7\sqrt{2}$

$21\sqrt{10}$

It’s best to simplify the radical before multiplying, if possible. By simplifying the radical beforehand, it’ll be easier to multiply the radicals together because the numbers will be smaller.

Ex. $2\sqrt{5}(\sqrt{75}-7\sqrt{8})$

$2\sqrt{5}(\sqrt{25\times3}-7\sqrt{4\times2})$

$2\sqrt{5}(5\sqrt{3}-14\sqrt{2})$

$10\sqrt{15}-28\sqrt{10}$

In the example above, I simplified the radicals to make the numbers more manageable to multiply. After simplifying the radicals I began to distribute inside the brackets.

How to Divide Radicals: When dividing a question that involves radicals you can only divide radicals by radicals and coefficients by coefficients.

Ex. $\frac{\sqrt{2}}{4}$

The example above cannot be simplified any further. Although 4 and 2 have a common factor of 2, they cannot be simplified because they are not in the same form.

Ex. $\frac{6\sqrt{10}}{9\sqrt{5}}$

$\frac{2\sqrt{2}}{3\sqrt{1}}$

$\frac{2\sqrt{2}}{3}$

The most important rule for dividing radicals is that a radical cannot be left in the denominator of a fraction. If the denominator does contain a radical you have to rationalize the denominator.

Rationalize the Denominator: To rationalize the denominator with a radical you have to multiply the whole fraction by the denominator.

Ex. $\frac{\sqrt{35}}{\sqrt{10}}$

$\frac{\sqrt{7}}{\sqrt{2}}$

$\frac{\sqrt{7}}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}$

$\frac{\sqrt{14}}{\sqrt{4}}$

$\frac{\sqrt{14}}{2}$

When the denominator is a binomial containing a radical you have to multiply by the conjugate. To find the conjugate of a binomial you have to flip the sign in the middle of the binomial. The conjugate of $6+\sqrt{3}$ is $6-\sqrt{3}$ .

Ex. $\frac{\sqrt{5}}{3-2\sqrt{7}}$

$\frac{\sqrt{5}}{3-2\sqrt{7}}\times\frac{3+2\sqrt{7}}{3+2\sqrt{7}}$

$\frac{3\sqrt{5}+2\sqrt{35}}{9+6\sqrt{7}-6\sqrt{7}-4\sqrt{49}}$

$\frac{3\sqrt{5}+2\sqrt{35}}{9-28}$

$\frac{3\sqrt{5}+2\sqrt{35}}{-19}$

Week 3 – Absolute Value & Radicals

manroopt2015 on February 17, 2018February 24, 2018

This week in Pre-Calc 11 we started the Absolute Value and Radicals unit. We learned the basics about absolute value and reviewed simplifying radicals.

What is the Absolute Value of a Real Number? The absolute value of a real number is the principle square root of the square of a number.

Sign: $\mid\mid$

The absolute value of a number is it’s distance away from 0. The absolute value of a number is always positive.

Ex. 1 $\mid-10\mid=10$

Ex. 2 $\mid10\mid=10$

In the examples above I showed the absolute value of a positive and negative number. -10 and 10 have the same absolute value because they are the same distance away from 0; both numbers are 10 spaces away from 0.

How do you simplify an expression? The $\mid\mid$ act like brackets. You simplify everything inside the absolute value sign and then move to the outside.

Ex. $9\mid3+6(-6)\mid$

$9\mid3-36\mid$

$9\mid-33\mid$

$9(33)$

297

In the example above, I solved everything inside the absolute value symbol. Then, I figured out the absolute value -33 which is 33. Lastly, I multiplied the absolute value by the number outside of the absolute value symbol.

How to Simplify Radicals: To simplify radicals you have to break the radicand down by using perfect squares.

Ex. $\sqrt18$

$\sqrt9*2$

$\sqrt9\times\sqrt2$

$3\sqrt2$

In the example I broke down the square root of 18 by using the number 9. 9 can divide into 18 and is a perfect square. After I figured out that $9*2=18$ I removed 9 from inside the radical by finding the square root of 9 and moving it to the outside.

Week 2 – Pre-Calc 11

manroopt2015 on February 10, 2018

This week in Pre-Calc 11 we continued to learn about different series and sequences. This week we learned about Geometric Sequences and Series.

What is a Geometric Sequence? A Geometric Sequence is a group of numbers that is multiplied by a constant ratio.

Ex. 2 ,10 ,50 ,250…

The common ratio for this sequence is $\frac{5}{1}$ or you can represent the common ratio by using the number 5. The common ratio is represented by the letter $r$ . A Geometric Series can NEVER have a dividing pattern. The common ratio of a geometric sequence CANNOT equal to 0 or 1.

Ex. 12, 6, 3, 1.5…

For this pattern $r=\frac{1}{2}$ although at first you might think that the pattern is dividing by 2 each time. In a geometric sequence the numbers have to be MULTIPLIED by a constant ratio each time, so this pattern can be represented as $r=\frac{1}{2}$ because multiplying by $\frac{1}{2}$ is the same as dividing by 2 each time.

How do you find r? The formula to find r is $r=\frac{t_n}{t_(n-1)}$

Ex. $\frac{1}{3},\frac{-1}{6},\frac{1}{12}$

$t_n=\frac{1}{12}$

$t_(n-1)=\frac{-1}{6}$

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

$r=\frac{1}{12}\div\frac{-1}{6}$

$r=\frac{1}{12}\times\frac{6}{-1}$

$r=\frac{6}{-12}=\frac{-1}{2}$

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

How do you find $t_n$ ? The formula for $t_n$ is $t_n=a*r^{n-1}$ . Instead of the first term being represented as $t_1$ , geometric sequences represent the first term as “a”. Similar to arithmetic sequences, geometric sequences also use “n” to represent the number of terms.

Ex. Find $t_{11}$ for the following geometric sequence: 5, 10, 20, 40…

$r=2$

$a=5$

$t_n=a*r^{n-1}$

$t_{11}=5*2^{10}$

$t_{11}=5*1024$

$t_{11}=5120$

What is a Geometric Series? A Geometric Series is similar to a geometric sequence except you replace the commas with a plus sign to find the sum of a certain number of terms.

Ex. $2+4+8+16$