*Hey Guys!*Well on Friday's class we had

*Mr. Clark once again to fill in for Mr. K, and no time was wasted. To kick off the class,*

**was assigned and is expected to be done.**

*Exercise 20: Review 2*As soon as that was established, our class

*as a whole*had asked Mr. Clark to look over a few questions on

**(which was previously assigned) that we were having problems with.**

*Exericse 21: Circles on a Coordinate Plane***4.**Find the

*centre*and

*radius*, and

*sketch the graph*for the following circles.

**(c) x²+y2+6y-12=0**

**HINT:**To solve,

**complete the square**

**standard form**

**of the equation of a circle**

x²+y²+6y-12=0

x²-10x+25+y²-4y+4=25+4

(x²-5)2+(y-2)²=29

**centre:**(5,2)

**radius:**√29

**9.**

*Sketch the graph*of

**y=x²-2x+5**. State the

*domain*and the

*range*.

**HINT:**The

**degree is 2**, we have a

**parabola**!

y=x²-2x+5

y=(x²-2x+1)+5-1

y=(x-1)²+4

**HINT:**

**Vertex**is

**(1,4)**and

**Y-Intercept**is

**(0, 5)**

**Domain:**(-, ∞)

**Range:**[4, ∞)

**10.** Write a *quadractic equation* with **roots 2±√3**** HINT:** Find

**S**um

**o**f

**R**oots +

**P**roduct

**o**f

**R**oots

*s.o.r.*= 2+√3 + 2-√3

=4

*p.o.r.*= (2+√3)(2-√3)

=4-3

HINT:

**(√x)(√x)=x**

=1

**eq'n =**x²-4x+1

**11.** Two sidewalks meet at *right angles*. At noon, * Person A is 12 km north of the intersection*,

*.*

**walking south at 2 km/h***. At*

**Person B**is**18 km east of the intersection**,**walking east at 4 km/h****?**

*what**time is the area of AOB a maximum*HINT: Maximum.. think parabola.. think sign of a quadratic equation.. think vertex

Original Area = Base(Height)/2

= 18 km(12 km)/2

= 108 km²

After X hours **A will be 12 km-2x**

After X hours **B will be 18+4x**

New Area = (18+4x)(12-2x)/2

= (216-36x+48x-8x²)/2

= (-8x²+12x+216)/2

= -4x²+6x+108

HINT: lets **complete the square** to find the **vertex**= -4x²+6x+108

= -4[x²-(3/2)x+(9/16)]+108+(9/4)

= -4[x-(3/4)]² + 441/4

**Vertex**(3/4, 441/4)

So that was all with ** Exercise 21: Circles on a Coordinate Plane**.

Students then asked questions about our *next* assignment*Exercise 22:Distances between Points and Lines*

We were all familar with the** line distance formula****distance =√(x _{2}, x_{1})²+(y_{2}-y_{1})²**

Mr. Clark then presented to us the "

*new*" distance formula

Below is the

**points and lines distance formula**

**distance=ax**

_{1}+by_{1}+c/√a²+b²**i.e) Find the distance from Point (2, -5) and the line 3x+4y+9=0**

distance=ax_{1}+by_{1}+c/√a²+b²

=3(-2)+4(-5)+9/√3²+4²

=35/5

=7

**HINT:** Keep the **Pythagurian Triples** in mind when solving these equations**a²+b²=c²**

*3, 4, 5

*6, 8, 10

*5, 12, 13

*12, 16, 20

*8, 15, 17

7, 24, 25

So I guess homework for the weekend is all uncompleted exercises up until..**Exercise 22: Distances between Points and Lines***Have a good weekend and enjoy!*

Scribe for Monday will be ** Lauressa**since you expressed your ready-ness to blog!

P.S. Images will be placed tomorrow if I find the energy, sorry.

Great post but just a comment from an older audience member. YOur use of small font is hard to read. Are you using it for a reason? I had a hard time reading it and I have perfect vision. As an edit I would recommend making the font normal and if you need the small font to explain something further then use it.

ReplyDeleteKeep on Scribing.

Mr. Harbeck

Sargent Park School.