Week 9 blog post

On this week , we finished lesson 4.5 -4.7 and we reviewed midterm.

On these lessons, we learned how to change the equation to completing the square from y= ax^2+ bx +c, a > 0, we learned to find x-intercept by to factor. And we learned to graph a quadratic function from its equation in general form, also we learned to use a quadratic function to model a situation.

Ex) Write each equation in standard form

y =x^2 + 4x +8                                           y= 2x^2 +16x -10

y= (x^2 +4x +4-4)+8                                y= 2(x^2 +8x+16-16)-10

y=(x+2)^2 -4+8                                         y= 2(x+4)^2-32-10

y= (x+2)^2 +4                                            y= 2(x+4)^2-42

Ex)  Identify the x-intercepts of the graph of each quadratic function.

a) y= (x-12)(x+11)                                          b) y= (x+5)(x-17)

x-intercept : 12, -11                                    x- intercept : -5, 17

c) y=(2x+2)(x-6)                                            d) y= x(x-9)

x-intercept : -1, 6                                            x-intercept : 0,9

Ex) Determine the intercepts, the equation of the axis of symmetry, and the coordinates of the vertex of the graph of each quadratic function.

a) y= $latex 2x^2 +16x+24

y= 2(x^2 +8x+12)

y= $latex 2(x+6)(x+2)

x-intercept : -6, -2

y-intercept : 24

y= 2(x^2 + 8x+16-16)+12

y= $latex 2(x+4)^2-32+12

y= $latex 2(x+4)^2-20

Vertex : (-4, -20)

axis of symmetry : x=-4

Ex) A rectangular area is divided into 2 rectangles with 200 m of fencing used for the perimeter and the divider. What are the dimensions of the total maximum area.

3W+2L=200

2W/3 + L = 100

L = 100 – 2W/3

 

W * L = Maximum

 

W(100-2W/3)=Maximum

100W-2W^2/3 = Maximum

-2/3($latex W^2 – 200/3+40000/36 -40000/36)=Maximum

-2/3(W-200/6)^2+ 80000/108= Maximum

-2/3(W-200/6)^2+20000/27= Maximum

W = 100/3m

L = 50m

 

 

 

 

 

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