Grade 11 – Page 2 – Madisson's Blog

Week 14 – Adding and subtracting rational expressions and equations

madissonr2016 on December 8, 2018

This week in pre-calc, we learned how to add and subtract ration equations and expressions in 4 easy steps.

Determine the lowest common denominator (LCD). The lowest common denominator is the number 2 denominators have in common when they’re multiplied. So if 3 and 4 were in the denominator, the lowest common denominator is 12. Its like lowest common multiple. The first number they have in common when multiplied.
Rewrite each equation with the LCD in the denominator. When you find the lowest common denominator, you multiple each fraction by the LCD until you get the same denominator for all the fractions. You also need to multiply the numerator as well so you can add or subtract all your numbers over one common denominator.
Solve. After rewriting the question, you can simply the whole equation by collecting all your like terms and adding or subtracting them together all under your LCD.
Reduce if possible. You can reduce your equation only if it can be reduced on the top and bottom by the same number.

Week 13 – Graphing Reciprocals of linear functions

madissonr2016 on December 2, 2018December 3, 2018

This week in pre-calc, we learned how to the reciprocals of linear functions. A reciprocal is when you have 2 hyperbolas that are reflections of each other. So if you divide the graph in a slanted direction, you get a reflection of the 2 hyperbolas, one in the positive quadrants and one in the negative quadrants.

The equation for graphing the reciprocal of a linear equation is 1/x+1 (you just take your linear equation and put it over 1.

The first step when graphing the reciprocal of a linear function is to draw out the original linear function. Once you have your line, the second step is to go to 1 and -1 on your y-axis. When you get to 1 and -1, you go across on the x-axis and find the points where the linear equation lines up with 1 and -1.

After you find where they intersect, you can draw a line that is equal distance between the point at -1 and 1 to get your vertical asymptote. This is acting as a magnetic field or a boundary that you cannot cross. It’s acting as your ‘0’ value. Your horizontal asymptote will be ‘0’ overtime for what we’re learning. This means that when you’re graphing your reciprocals, you will not be able to cross over the horizontal and vertical asymptotes. You always get closer and closer to reaching ‘0’ but you will never reach ‘0’ when graphing reciprocals.

Week 12 – Solving systems of equations algebraically

madissonr2016 on November 25, 2018November 26, 2018

This week in pre-calc, we learned about substitution and elimination. Using substitution or elimination can help us find the points of intersection(s) for lines or parabola’s if they intersect. Doing this gives us the co-ordinates of where they intersect and how many there are.

For 2 lines, there can only be 0, 1 or infinite. Parabola’s can have 0, 1, 2 or infinite. Parabola’s and lines can have 0, 1 or 2 solutions.

Using substitution is taking one of your two equations and getting one variable all by itself so you can input the solution for that variable and replace it with any other variable that has the same value.

Ex. x – 2y = -10

3x – y = 0

this becomes 3(2y-10) – y = 0

Then you want to gather like terms

6 y – 30 – y = 0

5y = 30

then you divide the ‘y’ value by it’s co-efficient on both sides to get the value of ‘y’

y=6

Now that you know your x-value, you can input this value into the first equation.

x – 2y = -10

x – 2(6) = 10

x – 12 = 10

x = 2

Now you know your co-ordinates for this system is (2,6)

Ex. x – 2y = -10

3x – y = 0

With elimination, you want to take your 2 equations, and make one variable equal to the other.

x – 2y = -10

6x -2y = 0

Now you can subtract the equations to make 1 equation.

-5x = -10

Once again, you just divide the ‘x’ term with its co-efficient on both sides and you get your ‘x’ value.

x= 2

Now you can input your ‘x’ value into the on of the equations and then you can simplify for ‘y’

6x -2y = 0

6(2) – 2y = 0

12 – 2y = 0

-2y = -12

y = 6

Co-ordinates = (2,6)

Both ways work for finding the point(s) of intersections it just depends on which method you prefer to use.

Week 11 – Graphing linear inequalities in two variables

madissonr2016 on November 18, 2018November 19, 2018

This week in pre-calc we learned how to graph linear equalities.

The first step when graphing, is the find your y-intercept. If your inequality doesn’t have a ‘c’ term in general form or a ‘b’ in the equation y=mx +b, then your y-intercept is going to be at 0.

After you point that down, you can look at you ‘mx’ and figure out your slope. If the number is positive, your slope will rise up and to the right on your graph. For the opposite slope, it will run down towards the left. If your slope is negative, your slope will rise up and to the left on your graph and the opposite slope will run down towards the right.

Once you graph this all down, you can take a look at your inequality sign. ≥ or in the opposite direction means that you have a solid line when you are connecting your points. This also means that all the points on the line are a solution to the inequality. < or > means that your line is broken and that the points on the line are not a solution. A solution for an inequality is to make the statement true.

A way to test to see which side of the line is the solution, you take a test point (0,0) is the easiest to test. If these 2 points make the statement true then you shade in the side of the solution. If it’s not true, you know to shade in the opposite side.

Week 10 – Midterm units

madissonr2016 on November 13, 2018

This week in pre-calc, we had our math midterm. When I was reviewing over the units, I noticed I forgot how to solve a quadratic equation by factoring. There was a question on the midterm about this exact question and I had no idea how to solve it.

When factoring a quadratic equation to solve, you need to get all your values on the same side and have one side being equal to 0. This is the how you can start to factor your quadratic equation.

When you have the equation :

√2x² + 1x +1 = 2x

The first step you need to do in order to be able to move the ‘2x’ value over is to square both sides to get rid of that square root sign in front of the ‘2x²’.After squaring, your equation would be :

2x² + 1x + 1 = 4x²

Then you can move your ‘4x²’ over to the right side making the equation :

-2x² + 1 + 1 = 0

From here, you cannot factor a negative first term so you need to take out a negative from all the terms :

– (2x² -1 -1) = 0

From here you can factor out your quadratic equation and find your solutions for ‘x’.

Week 9 – Solving a quadratic equation using a graph

madissonr2016 on November 5, 2018

This week in pre-calc, we learned how to take a quadratic equation, and be able to take the key points we need in order to graph the equation. There are 3 different equations that all give us different key points we need to graph. Vertex form ( y = a $(x-p)^2$ +q ), general form ( y = a $x^2$ +bx + c ) and factored form ( y = a(x- $y_1$ .)(x- $y_2$ )

The vertex form gives you the vertex of the graph (the lowest or highest point of the graph). To find the vertex, all you do is take ‘p’ as your x value and because the ‘q’ is a constant term, that’s your y value. Vertex = (p,q)

The general form tells you the y-intercept and the direction the parabola is opening up. The y-intercept is always the constant number, and it’s always ‘c’. If there is no constant term (c) then the y-intercept is always 0. In the first term ‘ax’, the ‘a’ tells you the direction it’s opening towards. If the value is positive, it will be opening upwards. If the value is negative, it’ll open down. If it opens up, the vertex is the lowest point, and if it opens down, the vertex is the highest point. The value of ‘a’ can also tell you if the parabola is stretched or compressed. Stretched is any number greater than 1 and compressed is any number smaller than 1.

In factored form, you can find the x-intercepts. If you have y = 2(x – 3)(x + 2), because the numbers in the brackets can’t go lower, you can’t simplify anymore. You take your (x-3) and you find the value of ‘x’ that will make that term equal to 0. So 3-3 = 0, x-intercept is +3. You can just take the number (-3) and take its opposite sign to find the x-intercept too. Opposite of +2 is -2, x-intercept is -2.

Week 8 – Solving a quadratic equation graphically

madissonr2016 on October 28, 2018October 29, 2018

This week in pre-calc, we learned how to find the key points in a quadratic equation so we can use them to graph a quadratic equation.

The first step in solving a quadratic equation (ax + bx + c = 0) for graphing is when you see your equation, you can tell right away if the parabola is opening up or opening down. If the co-efficient on the first term (ax) is a positive, it will always open up and if it is negative, it’ll open down.

You can figure out wether the vertex is at its minimum point or maximum point. If its positive, the vertex will be at its minimum and if its negative, the vertex will be at its maximum point.

The next thing you’ll be able to tell from the quadratic equation is the y-intercept. You can tell by finding the constant term. If there is no constant term, the y-intercept is 0. The y-intercept is always crossed when x is always 0.

One way to find the x-intercepts is to factor the equation. When you do this you’ll have something like (x+a)(x-a) = 0

When you have this, the value of both x’s needs to be equal to 0 so in the first one, x would be equal to -a and for the second, x is equal to +a. So the value of x is always the opposite of the given number. When the x-intercept is crossed, y will always be at 0.

Week 7 – Interpreting the discriminant

madissonr2016 on October 21, 2018October 22, 2018

This week in pre-calc, we learned how to find out how many solutions a quadratic equation can have without solving the equation.

We use the formula $b^2$ – 4ac from the original quadratic equation. Using this formula will make it easier to find out how many roots there is for the equation. For there to be any roots, the radicand needs to be equal to or above 0. If it’s above 0, there are real roots, distinct roots, for perfect squares it’s called a rational roots discriminant and if it’s a non perfect square, it’s called an irrational roots discriminant. If the radicand is equal to 0, there is only 1 root, an equal root. Any number below 0 (in the negative), there is no real roots.

When you have your quadratic equation, you label your ‘a’ value, your ‘b’ value and your ‘c’ value and then you input the numbers into the formula.

Week 6 – Solving Quadratic Equations

madissonr2016 on October 14, 2018October 15, 2018

This week in pre-calc, we learned how to solve a quadratic equation by factoring. The equation $ax^2$ + bx + c = 0. A, B, & C are all constant terms and A cannot be equal to 0. When factoring a quadratic equation, you use the zero product property. This means you take your quadratic equation and you start by factoring. Once you’re done factoring, you’ll have 2 terms that should be equal to 0.

Ex. $x^2$ -5x + 6 = 0

(x – 2)(x-3) = 0

Once you factor your equation, you take each term and find out the value of x in order for the term to be equal to 0.

Ex. (x-2) = a (x-3) = b

a=0

b=0

a = (x-2) = 0, x = +2

b = (x-3)=0, x = +3

Basically you’re taking the opposite number to find x in order to make the term equal to 0.

(x – 2)(x-3) = 0

(2-2)(3-3) = 0

(0)(0)=0

Week 5 – Factoring Polynomial Expressions

madissonr2016 on October 8, 2018

This week in pre-calc 11, we learned how to factor polynomial expressions. There are 5 steps you can use to factor a polynomial, CDPEU. CDPEU stands for Common, Difference of squares, Pattern, Easy pattern and Ugly pattern.

Common means finding any like terms in the expression and combining them together

Difference of squares is used in binomials only (2 term expression) and there must be a — sign because it’s a difference. Difference of squares only works if everything is a perfect square. When factoring a difference of squares, they will always be conjugates so there is no middle term

Ex.64 $x^2$

64 $x^2$ = 8x x 8x

25 = 5×5

(8x+5)(8x-5) = 64 $x^2$

Pattern, Easy pattern and Ugly patterns are used for trinomials only (expression with 3 terms)

Pattern is figuring out if the expression is an easy pattern or an ugly pattern.

Easy pattern $x^2$ +6x+8 it’s simple because there is no coefficient in the first term and you find 2 numbers that add to the middle term and multiply to the term.

Ex. $x^2$ +6x+8

8 = 1×8, 2×4 2+4 = 6

(x+2)(x+4) = $x^2$ +6x+8

Ugly pattern is $6x^2$ +11x-10 an ugly pattern is ugly because there is a coefficient with the first term. An easy way to factor an ugly pattern is using a box

Some expressions do not factor.