### Archive of ‘Math 11’ category

Over this semester of Pre-Calculus 11, we covered many units. But the most important thing I learned, was how to work smarter and not harder. There’s always an easier way to do every problem, and a catchy little acronym to go with it. Each unit we covered brought new insight and challenges into my world of math and problems that stretched my brain as far as possible. But I didn’t hate it. Can Divers Pee Easily Underwater? I know the answer to that question. Does Ms. Burton love Slurpee’s as much as we do? Absolutely. Pre-Calculus 11 had it’s many ups and downs, but i learned so many new things that I’ll take with me into the future. The top 5 lessons I learned this year were:

Arithmetic Sequences:

Arithmetic sequences are a group of numbers (i.e. 4, 9, 14, 19, 24…) that increase by a steady amount. In this case 5. Using the equation $t_n=t_1+(n-1)d$ we can find any number within the sequence, with $t_n$ representing the unknown number, $t_1$ represents the first number in the sequence, $d$ represents the difference between numbers, and $n$ being the number we’re trying to find. The plugged in equation would look like $t_n=4+(n-1)5$ which in turn is $t_n=5n-1$.

With this equation you can implement any number of the sequence into the equation to find it’s value.

If we factor a quadratic equation, we can find the roots, as well as the axis of symmetry. With this information, we can draw a quick graph depicting what our parabola would look like.

ex. Graph the equation $y=-2x^2-6x+20$

graphing the equation with only the factored form won’t give us a complete parabola, but it will give us enough information to get a rough idea.

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

$latex y=-2(x+5)(x-2) replace the “y” with 0. 0=-2(x+5)(x-2) rearranging the bracketed “x’s” will give us our roots (x+5) -> x=-5, (x-2) -> x=2 with the x-intercepts, we know that when graphed will be in the format (x,y). But since the x-intercept is where the parabola crosses the x-axis, the y-intercept will equal 0. The x-intercepts will be (-5,0) and (2,0). The axis of symmetry is half way between both roots, so the AOS will be x=-1.5. Using the Discriminant: The discriminant is the area under the square root in the quadratic equation: if you have a quadratic equation (equation equal to zero with 3 distinct parts), you can use the quadratic formula to solve. Depending on the answer, we can figure out whether the equation will have 1,2 or 0 solutions. With a quadratic equation in the format $ax^2+bx+c$ we can substitute the a, b and c values into the discriminant equation to find out how many roots. Parabolas: Graphs of quadratic functions all have the same shape which we call parabola, that can be graphed to give us information such as the roots, y-intercept, axis of symmetry, etc. In order to find this, we have to first graph it using some of these easy tricks. A translation is an image of the original parabola, but moved either horizontally or vertically from the original parent equation $y=x^2$ Depending on the equation, you can determine where the parabola has moved on the number line. If the coefficient of $x^2$ changes, the parabola will either stretch or compress. A coefficient less than 1 will cause the parabola to compress, while a coefficient more than 1 will cause it to stretch. If the coefficient $x^2$ is negative the parabola will open down, while if it’s positive it will open up. In the case that your equation looks similar to $y=(x+3)^2$, with your numbers in brackets, then the “c” will effect whether your parabola translates left or right. If the number is positive, it will translate left, and if it’s negative it will translate right. It does not follow the general idea that negatives go left and positives go right. Sine and Cosine Law: Sine and Cosine Law are used to solve for angles and sides of a triangle that isn’t a right triangle. Sine Law is used when you’re able to use the formula $\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$. In order to use the formula you must have all values of one fraction, and half of 1 other. This way, you can solve for an angle/side you’re missing. It can be any part of the formula, as long as one + half of another is given. Cosine law is used when two sides and one angle are given to you, but you’re trying to find the missing side. The formula used is $a^2=b^2+c^2-2bc(cos A)$ for finding a side, or $cos A=\frac{b^2+c^2-a^2}{2bc}$. This week we learned how to use Sine Law and Cosine Law to find the angles and sides of triangles that aren’t right angles. Sine and Cosine Law are used to solve for angles and sides of a triangle that isn’t a right triangle. Sine Law is used when you’re able to use the formula $\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$. In order to use the formula you must have all values of one fraction, and half of 1 other. This way, you can solve for an angle/side you’re missing. It can be any part of the formula, as long as one + half of another is given. Cosine law is used when two sides and one angle are given to you, but you’re trying to find the missing side. The formula used is $a^2=b^2+c^2-2bc(cos A)$ for finding a side, or $cos A=\frac{b^2+c^2-a^2}{2bc}$. This week we spent some time reviewing trigonometry from grade 10, so that we would be refreshed for our trigonometry unit this year. Important things to remember: 3 main ratios in trig are Sine (sin), Cosine (cos), and Tangent (tan) An abbreviation we use to remember this is SohCahToa SohCahToa is used to find angles of right triangles. For example: H=8, O=4. Find angle x. This diagram labels the hypotenuse, opposite and adjacent, but not all diagrams are labeled. Too figure out which is which is easy! Hypotenuse will always be your longest side, and is usually straight across from your right angle. Adjacent is the side directly connecting your right angle and missing angle. Opposite is the side leftover or directly across from the missing angle. In our abbreviation Soh Cah Toa: “S” stands for Sine, “C” stands for Cosine and “T” stands for Tangent. Other than these, your smaller letters all stand for something: Your “o”pposite, “h”ypotenuse and “a”djacent. Knowing where each side is can help us dictate whether to use Sine, Cosine or Tangent to solve for the angle. In our diagram, we’re given the length of the hypotenuse and the opposite. Knowing that we decide to use “Soh” from our abbreviation because it’s the only one with opposite and hypotenuse. To find the angle we use the “sin” button on our calculator, but the inverse function of it because we’re solving for the angle and not a side. When punched in the equation should look like $sin^-1(\frac{4}{8})$ If you were instead solving for a side, the regular “sin” button would be used instead. This week we learned how to add and subtract rational expressions. These are a bit harder than multiplying and dividing because the denominator has to be the same throughout the expression. Steps to simplify: 1. determine lowest common denominator (LCD) 2. rewrite each fraction as an equivalent fraction with LCD 3. Combine numerators by adding/subtracting 4. reduce if possible For example: $\frac{x-9}{2x}-\frac{3x}{x-4}$ where “x” cannot equal 0 or 4. 1. For this expression, the least common denominator would be $2x(x-4)$ because neither denominator can be simplified any farther or multipled to create a common denominator, so we multiple the two denominators together to create one LCD. 2. Since we’re multiplying the bottom, we also have to do the same to the top, so the equation becomes: $\frac{(x-9)(x-4)}{2x(x-4)}-\frac{(3x)(2x)}{(x-4)(2x)}$. 3. This simplifies to $\frac{x^2-13x+36-6x^2}{2x(x-4)}$ 4. The final simplified expression will be $\frac{-5x^2-13x+36}{2x(x-4)}$ This week we learned how to multiply and divide rational expressions, which are much simpler than adding and subtracting. It’s easiest to explain how to do the question as an example so, e.x. $\frac{x+2}{(x-4)(x+3)}*\frac{3(x-4)}{2x(x+2)}$ When multiplying and dividing, you can cancel out expressions that are in both the numerator and denominator. In this case: $(x+2)$ and $(x-4)$ After doing this, your equation should look like $\frac{3}{2x(x+3)}$ Because you can’t factor the expression anymore, that’s as far as it can be simplified. Another thing to always do with expressions is label your restrictions. In this case, the restrictions are number that can’t be used in the place of variable “x” that would cause the denominator to become 0. For this equation x cannot equal 4, -3, 0, -2 This week we learned how to graph quadratic reciprocal functions, and learned some new vocabulary. quadratic reciprocal functions will look one of three ways when graphed: 1. the parent function only touches the x-axis in one spot 2. the parent function doesn’t touch the x-axis at all 3. the parent function crosses the x-axis in two places When graphing the parabola, you first find the asymptotes, invariant points, and then the location of the hyperbola. Asymptotes are located at where the parabola crosses the x-axis. At the x-intercept(s) you can drop a dotted line vertically, and a dotted line horizontally. Depending on how many x-intercepts you have will dictate how many asymptotes you have. Dotted in red: Invariant points are located where your parabola meets the points on the y-axis 1 and -1. The parabola can have 0-4 invariant points. These dictate where you’ll draw the hyperbola. Circled in blue: Your hyperbola is the graph of the reciprocal of your original equation. For example: $2x^2+4$ becomes $\frac{1}{2x^2+4}$ When you graph the hyperbola, there’s no need to graph each individual point. Instead, you can just draw a curved line going through your invariant point and approaching zero. Colored in purple: This week we learned how to solve systems algebraically with more than one variable. For example: x+6y=4 and 0=x-4y In these types of systems you used steps. 1. Isolate a variable 2. Plug the equation for that variable into the second system 3. solve for that variable 4. plug variable into original equation, solve for the missing variable 5. Check! This may not make complete sense in steps, so I’ll demonstrate using the first example. 1. x+6y=4 can be rearranged to make x=4-6y 2. next you plus your new equation x=4-6y into your second equation: 0=x-4y which turns into 0=(4-6y)-4y. This becomes y=2/5. 3. Next, we plug our value for y into our original first equation x+6y=4, which becomes x+6(2/5)=4 4. Simplifying the equation: x=1.6 5. Check y=2/5 and x=1.6: x+6y=4 -> 1.6+6(2/5)=4 Yes! 0=x-4y -> 0=1.6-4(2/5) Yes! So we know we did it correctly This week we learned how to graph inequalities with two variables. These inequalities are essentially forms of the equation y=mx+b, used to graph lines. The same idea can be applied to graphing the inequalities. However depending on the inequality the line will either be broken or solid. solid line = $\leq$ or $\geq$ broken line = $>$ or $<$ using the same idea as y=mx+b, we can find the slope, and the y-intercept. “m” represents our slope, if it’s a whole number it can be read as $\frac{n}{1}$ with “n” being your rise(y), and 1 being your run(x). For instance: If you were to graph $y>x+2$, your y-intercept would be 2, and your slope would be $\frac{1}{1}$ It would look like this: The area shaded, is figured out by testing x and y coordinates in your equation, until the equation is made true through the coordinates chosen. $\sqrt{3x}+7=x+1$ When faced with a problem where you have to solve for “x” but aren’t able to get rid of the $\sqrt{3x}$ by squaring both sides right away, you first have to isolate both “x” variables on one side of the equation. Now we can get rid of the square root by squaring both sides. $(\sqrt{3x})^2=(x-6)^2$ $3x=x^2-12x+36$ On the right side of the equation, $x-6$ turns into $x^2-12x+36$ because we have to expand. You aren’t able to just do the math. Now, since our equation is quadratic we can make our equation equal to zero, and factor it to find our roots. $0=x^2-15x+36$ Lucky for us this equation factors nicely:$latex 0=(x-12)(x-3)

So, x=12 and/or x=3

To check we plug one of our newly found “x’s” into our equation. If one doesn’t work, there’s still a chance that the other one will. ALWAYS check both!

$\sqrt{3x}+7=x+1$

$\sqrt{3*3}+7=3+1$

$\sqrt{9}+7=4$

$3+7=4$

$10=4$

This one DOES NOT work!!! However, we’ll check x=12 to check if that one does work.

$\sqrt{3x}+7=x+1$

$\sqrt{3*12}+7=12+1$

$\sqrt{36}+7=13$

$6+7=13$

$13=13$

So x=12 is a root, but x=3 is not. Therefore, the equation only has one root.

This week we learned about graphing quadratic equations in factored form. In factored form, we can find the roots, as well as the axis of symmetry. With this information, we can draw a quick graph depicting what our parabola would look like.

ex. Graph the equation $y=-2x^2-6x+20$

graphing the equation with only the factored form won’t give us a complete parabola, but it will give us enough information to get a rough idea.

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

\$latex y=-2(x+5)(x-2)

replace the “y” with 0.

0=-2(x+5)(x-2)

rearranging the bracketed “x’s” will give us our roots

(x+5) -> x=-5, (x-2) -> x=2

with the x-intercepts, we know that when graphed will be in the format (x,y). But since the x-intercept is where the parabola crosses the x-axis, the y-intercept will equal 0. The x-intercepts will be (-5,0) and (2,0). The axis of symmetry is half way between both roots, so the AOS will be x=-1.5.