# Week 13 Blog Post

This week we learned chapter 8: Absolute Value and Reciprocal Functions

1: Write each absolute value function in piecewise notation

a) Y=|2x-7|

there are two possibilities: 2x-7>=0   or                   -(2x-7)<0

2x>=7 divided by 2         x<7divided by 2

the answer is : y={2x-7>=0, x >=3.5

{-(2x-7)<0,x<3.5

# Week 12 blog post

Things I learned this week:

There is one thing that I always will  misunderstanding it,  is that: solid line represents the sign is >= or<=, however, the broken line means the sign is < or >. This is one of the most important part while drawing the equation. And also, after drawing the line, we should check the sign again to make sure the shaowded part is correct and fits the sign.

# Week 10 blog post

For the quadratic function y=-3($x-2^2$) + 5

ask for y-intercept, for y-intercept, means x=0, in this quadratic function made x-2=0, so y-intercept = -3*$-2^2$ + 5=-7

I’ve always messed the vertex and the intercepts, I thought y in the vertex means y-intercept so I made a lot of mistakes.

# Week 9 Blog post Analyzing Quadratic Functions of the Form Y=a$latex x^2$+bx+c

Factored Form: y=a(x-x1)(x-x2)

y=-2$x^2$ -6x+20

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

y=-2(x1+5)(x2-2)

x1=-5  x2=2(x-intercepts)

so the axis of symmetry is (-5+2) devided by 2 equals to -1.5

# Week 8 blog post 4.1 Properties of a Quadratic Function

Ms Burton, I learned this chapter on Youtube by myself. And I took some notes.

The vertex of a parabola is its highest or lowest point.(_,_) The vertex may be a maximum or minimum point.

If the graph opens up, x should be positive, the vertex amy be the minimum point, the domain can be any  real numbers, and the range must be y>= (greater than the vertex)

If the graph opens down,x should be negative, the vertex amy be the maximum point, the domain still can be any  real numbers, and the range must be y<= (lease than the vertex)

# Week 7 Developing and Applying the Quadratic Formula

When we solve this kind of questons, we need to follow the steps and check carefully after his work. We need to calcuate again and compare our work to his work, that helps find the mistakes.

(PS: Ms Burton, hope you enjoy your wonderful spring break 🙂 )

# Week 6 Developing and Applying the Quadratic Formula

$x^2$ – 6x + 4=0

For me, quadratic formula is the easiest problem to solve because the only way to solve this is to remember the formula, and put each pattern in the foumula. Firstly, we need to find a, b and c for this formula in this question. A = 1, b= -6, c=4. Secondaly,  is to find the formula: .   Finally, place each pattern in  the formua.

# Week 5 Factoring Polynomial Expressions

$x^2$ -11x +30

=(x-5) (x-6)

Before solving this type of question, we should remind ourselves  of the ‘CDPEU’ chart, then find the right direction for solving this kind of question. According to the chart, this question is a trinomial one. The first part($x^2$) and the last part(30) always provides informtaion. For example, if we want to find the factor of this one, we need to look at the first part and the last part. The last part is 30.  30=3 × 10=2×15=5×6, the middle part is -11x. Due to -11, we can eliminate 3&10, 2&15 becasue regardless of adding or substracting,  3&10 or 2&15 still can not get the answer of 11. There is only answer: 5&6 becasue 5+6=11, but there is one  thing very important is that the number is -11(it’s negative). The last part is a postive 30, so the final answer is -5&-6.

# Week 4 Multiplying and Dividing Radical Expressions

Expand and simplify: (P127)

a) ($\sqrt{3}$ +8) ($2\sqrt[1]{3}$ -1) – $\sqrt{3}$ ($7\sqrt[1]{3}$ -4)

=6-$\sqrt{3}$$16\sqrt[1]{3}$ -8-(21-$4\sqrt[1]{3}$)

=-2 + $15\sqrt[1]{3}$ -21 + $4\sqrt[1]{3}$

=-23+ $19\sqrt[1]{3}$

Firstly, when we solve this type of questions, we need to find the like terms becasue when adding/ substraction, the coefficients are combined when the radicands are the same. For example, in this question, like terms are $\sqrt{3}$$2\sqrt[1]{3}$ and $\sqrt{3}$($7\sqrt[1]{3}$ becasue you can find they all have $\sqrt{3}$, and that means we need to think of a way to combine them together to get a simplified answer.  Secondly, do the calculation carefully and be aware of the like terms. Final step is to check you answer is already simplified  or not.