Mathematics Education Standards

Introduction to mathematics education standards:

The mathematics education include a different branches of unit conversions, algebra, subtraction, measurement, number sense, multiplication, functions, adding and subtraction of decimals, fractions & mixed numbers, division, algebra, geometry, median problems, algebra function, probability and statistics number using words decimals. This mathematics education supports all type of standards up to higher standards.


Example problems for Mathematics education standards:


Example 1:

Solve the quadratic equation `x^2 +5x + 6 =0`

Solution:

`X^2 +5x +6 =0`

`X^2 +2x +3x + 6 =0`

` x(x +2) +3 (x +2) = 0`

`(x +2)(x + 3) =0`

`x + 2 = 0 `                   `x` ` + 3 =0`

`X = -2 `                      `X =-3`

Example 2:

Solve the quadratic equation `x^2 +4x + 4 =0`

Solution:

` X^2 +4x +4 =0`

`X^2 +2x +2x + 4 =0`

` x(x +2) + 2(x +2) = 0`

`(x +2)(x + 2) =0`

`x + 2 = 0 `              ` x + 2 =0`

`X = -2 `                      `X =-2`

Example of polynomial exponent problems- Mathematics education standards:

Addition of polynomial exponent:

Two or more polynomials, adding the terms,

Suitable example adding polynomial exponent,

Example1:

` (2x^2+3x^3)+(x^2+7x^3)`

`=2x^2+x^2+3x^3+7x^3`

The variable and exponent must be same then we add the polynomial exponent,

`=3x^2+10x^3`

So the result is `=3x^2+10x^3`

Subtraction of polynomial exponent

Example2:

`(3x^2+3x^3)-(x^2+7x^3)`

`=3x^2-x^2+3x^3-7x^3`

The variable and exponent must be same then we subtract the polynomial exponent,

`=2x^2-4x^3 `

So the result is` =2x^2-4x^3 `

Adding polynomials- Mathematics education standards:

Example 1: Find the sum of `6x^2 + 7x + 16 and 1x - 3x^2 -4.`

Solution: By means properties of real numbers, we realize

`(6x^2 + 7x + 16) + (-3x^2 + 1x - 4) = 3x^2 + 7x + 1x + 16 - 4`

`= 3x^2 + 7x + 1x + 16 - 4`

`= 3x^2 + 8x + 12`

So the final result is `= 3x^2 + 8x + 12`

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Examples for finite difference problem- Mathematics education standards:


Example 1: calculate the values of Δ`y` and dy if `y = f(x) = x^3 + x^2 - 2x + 1`

Where x changes (i) from `1 to 1.05` and (ii) from` 1 to 1.01`

Solution:

(i) We have `f(1) = 1^3 + 1^2 - 2(1) + 1 = 1`

`f(1.05) = (1.05)^3 + (1.05)^2 - 2(1.05) + 1 = 1.15.`

and Δ`y = f(1.05)- f(1) = 0.15.`

in general `dy = f ^ 1(x) dx = (3x^2 + 2x - 2)dx`

When `x` ` = 1` , `dx = ` Δ`x =1 and dy = [(3(1)^2+2(1)-2] 1= 3`

(ii) `f(1.01) = (1.01)^3 - (1.01)^2 - 2(1.01) + 1 = -1.01`

∴ Δ`y = f(1.01) - f(1) = 1.99`

Quadratic Equations and Graphs

Definition of quadratic equation : An equation in which one or more of the terms is squared but raised to no higher power, having the general form ax2 + bx + c = 0, where a, b, and c are constants.

Quadratic Equations and Graphs:

The general form of a quadratic equation is

ax2 + bx + c =0

Where a, b and c are real numbers and a`!=` 0

It is called so because it has a second degree term.

The values of the variable involves in a quadratic equation are called its solution.

The following method are usually used to solve a quadratic equation.

completing square method
using the quadratic formula
Factorization method


In general, the shape of the graph of a quadratic equations a parabola.

The steps we follow to sketch the graph of the quadratic equation are:

1. Check if a > 0 or a < 0

If a > 0, then the parabola is u-shaped ie it opens upwards

If a < 0, then the parabola is  n-shaped ie it opens downwards

2. Find Vertex

The x-coordinate of the vertex is –b/2a

Substitute the value of x in the function to have the y coordinate.

Thus we have the (x, y) coordinates of the vertex.

3. Find y-intercept

The coordinates of the y-intercept can be found by substituting x = 0.

4. Find x-intercept

The coordinates of the x-intercept can be found by substituting y = 0 and solving the quadratic equation

5. Finding other coordinates

Substitute any possible value for x to get the corresponding value of y or vice versa

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Quadratic Equations and Graphs-example

Graph of the function y =  2x2 - 8x + 6

Here a= 2 b=-8 c=6

1. Since a > 0, then the parabola is u-shaped i.e. it opens upwards

2. The x-coordinate of the vertex is –b/2a = 8/(2*2) = 2

Substituting the value of x in the function y = 2x2 - 8x + 6 to have the y coordinate.

We get y = 2*22 - 8*2 + 6 = -2

So the vertex is (2,-2)

3. The coordinates of the y-intercept can be found by substituting x = 0.

We get y = 2*02 - 8*0+ 6 =6

So the coordinates of the y-intercept is (0,6)

4. The coordinates of the x-intercept can be found by substituting y = 0 and solving   the quadratic equation

We get

2x2 - 8x + 6=0

2(x2 - 4x + 3)=0

2(x - 1)(x - 3) = 0

So x = 1, or x = 3.

So the coordinates of the x-intercept is (1,0)  and (3,0)

5. Finding some other coordinates by substituting any possible value for x

We can now plot the graph.

Quadratic Equations and Graphs-graphs