Tag Archives: burtonmath11

Week 12 – Solving Absolute Value Equations Algebraically

A couple of weeks ago we learned how to solve absolute value equations by using algebra.

 

Verifying is how you can check for extraneous roots (Roots that give you a different answer). Make sure you remember it’s in absolute value form, so there’s never any negative values, if there’s ever any negative  you know that there are no solutions. If the left side does not equal the right side it’s an extraneous root.

Week 13 – Graphing Reciprocal Quadratic Functions.

Last week we learned about how to graph reciprocal functions.

The steps to solve.

1.) Graph the parabola

2.) Locate the invariant points

3.) Find the asymptotes

 

Three types of Quadratics

Positive Slope, Positive y-intercept

 

It never touches (0,0)

so no invariant points

 

Positive slope (can be negative), y-intercept=0

The asymptotes are touching the parabola

Positive or Negative Slope, opposite y-intercept

 

 

 

 

Week 9 – Modelling Problems with the Quadratic Equation.

Find two integers with a sum of 36 and the greatest possible product.

Find the variables

I personally like to use the variable “x”, and since there are two integers that means there are two integers. That means the second variable is (36 – x). For example, if the first variable was 4 and using 36-x, would result in 36 – 4, that means the second variable would be 32.

Variables  

1st x= x

2nd x= 36-x

when it says the greatest possible product, that means that we’re looking for the maximum value.

We put the variables in the factored form.

(x) (36 – x)

1st x = 0

2nd x = 36

Now we need to find the vertex

(1st x) + (2nd x) / 2

0 + 36 /2= 18

(18,_)

Then we input the new “x” value

(x) (36-x)

(18) (36-18)

(18) (18)

=324

So the answer is (18,324).

 

Week 8 – Properties of a Quadratic Equation

This week we looked at graphing our quadratic equations. Even looking at a simple equation can help to see the properties of the parabola. A parabola is a shape that forms.

 

y= x^2 + 6, means that the vertex moves up on the y-intercept, another thing we can analyze the equation is the coefficient, if the coefficient was negative then the parabola would be facing down, also you can see that the coefficient is equal to one, that means the parabola will stretch, if the parabola was less than one, the parabola would be compressed.

Week 6 – Perfect Square Trinomials & Solving Quadratic Equations Using Factoring

Perfect squares trinomials 

Perfect squares trinomials have one defining pattern… That’s they are perfect squares.

ex.

x^2 + 6x + 9

These are the simplest of equations because of the coefficient being one for x^2. the middle or 2nd term is the sum of two numbers that are the same like, 3. Because in these case, 3+3= 6 which is the middle term. The end or 3rd term is the product of those same numbers that were the product of the 2nd term.

ex.

(2nd and 3rd term).

6x + 9 

3+3 = 6 – The 2nd term.

3 x 3 = 9 – The 3rd term.

Solving Quadratic Equations Using Fractions. 

You can recognize that the equation is quadratic if the answer is equal to zero.

ex.

x^2 – 36x = 0

(x +6) (x – 6)=0

There are two “x” variables in this equation. So you have to solve for both.

First “X”

 

x + 6 = 0

    -6  = -6

x = -6

 Second “X”

x – 6 = 0

isolate x 

x – 6 = 0

   +6    +6

x = 6

You can solve to verify.

Verifying for  x = -6

-6^2 – 36 = 0

36 – 36 = 0

Verifying  for x = 6

6^2 – 36 = 0

36 – 36= 0

Week 5 – Factoring Expressions

Last year in grade 10, we learned how to factor simple expression, but this week we added another step

ex

1.) Replace the complicated set of varibles with a simple varible (a)

2.) Factor the expression

3.) Once properly factoring substitue the varibles back in

=(x^2 – x) ^2 + 5 (x^2 – x) ^2 – 24

=a^2 + 5 a – 24

=(a – 3) (a + 8)

=(x^2 – x – 3) (x^2 – x + 8)

=(x + 3) (x – 1) (x + 2) (x + 4)

 

 

Week 4 – Addition, Subtraction, Multiplication and Division of Radicals

In Grade 9, we looked at the concept of grouping like terms.

Grade 9; 5x + 6y – 3x + 9y + x

x = 3x

y= 15y

Now in Grade 11, we now group radicals.

Grade 11; 4\sqrt[3]{9}, 6\sqrt[3]{9}, 15\sqrt[3]{9}

= 25\sqrt[3]{9}.

Rule

Radicand and index must be the same.

Subtraction

\sqrt{160} – \sqrt{40}

= \sqrt{16\cdot10} – \sqrt{4\cdot10}

= 4\sqrt{10} – 2\sqrt{10}

= 2\sqrt{10}

 

Division and Multiplication Rule

The denominator CANNOT be a radical

 

Rationalizing the Denominator

\sqrt{5} /    x    \sqrt{6} =  \sqrt{30} /   ——->       \sqrt{30} /

\sqrt{6}       x    \sqrt{6} =  \sqrt{36}      ——->        6

 

Rationalizing the denominator is a way of cancelling out the radical so the denominator is not a radical.

1.) Recognize that the denominator is a radical

2.) Multiply both the numerator and the denominator (what you do the bottom you got to do the top)

3.) Simplify to find the new values

4.) See if there are any perfect squares (if the index is for example 3, then look for perfect cubes)

5.)Simplify the new values

6.) MAKE SURE A RADICAL IS NOT IN THE DENOMINATOR PLACE.

Week 3 – Absolute Value, Roots

This week we learned some new concepts and reviewed concepts from grade 10.

2.1 Absolute Value – The principal square root of the square root of a number.

ex. \mid{-7}\mid      absolute value = 7

the absolute value is the amount that the given number is different from 0.

Note – The radicand can only be positive

\mid{7-16}\mid

7 – 16 = -23

absolute value = 23

-23 & 23, are both 23 digits away from 0.

2.2 Roots and Radicals

Perfect Squares 4,9,16,25,36,49,64,81,100

Perfect Cubes 8,27,64,125

Simplify

\sqrt [3]{40}

=\sqrt[3]{8\cdot5}

=2\sqrt[3]{5}

 

 

 

 

Week 2 – Geometric Sequences

This week in math we saw some questions that displayed percentages.

For example. 

Billy a wanted to estimate the value of his car (New) after several years of use. The car was worth $34 000 new and depreciates in value by 10% each year. What will be the value of his car at the end of 1 year of use and 5 years of use?  

Analysing the Question.

Value of New Car = 34 000

t_{1} = 34 000 x  100% – 10% (Starting value x years of depreciation)

t_{1} = 34 000 x .90 = 21 420
de·pre·ci·ate
dəˈprēSHēˌāt/
verb
  1. diminish in value over a period of time.

5th year 

Formula = t_{5} = ar^{n-1}

Take t_{1} r ^{n-1}

34 000 (.90) ^ 5-1 = 20 075.66