Derive a quadratic equation using the following points: (1, -4), (2, -5), (3, -10)

y = ax^2+bx+c 
(1) (1, -4) -> -4=a+b+c , 
(2) (2, -5) -> -5= 4a+2b+c, 
(3) (3, -10) -> -10=9a+3b+c 

(2)-(1): 
-5+4=4a-a+2b-b+c-c 
-1=3a+b (or 3a+b=-1) 
(3)-(1): 
-6=8a+2b (or 4a+b=-3) 

taking the difference of these 2 we get 

4a-3a+b-b=-3+1 (or a=-2) 

and thus b=5 
Now from (1):
-4=-2+5+c (so c=-7) 



y=-2x^2+5x+-7 

:) 

x+(x+y)i=3+8i

Real: x=3 
Imaginary: x+y=8, 3+y=8, y=5 

:)

Find the values of x and y, where x and y are real numbers (3x-4y)+(5/3)yi=-6+5i

3x-4y)+(5/3)yi=-6+5i 
Re:
3x-4y=-6 

Im:
5/3 y=5
y=3 

3x-4(3)=-6 
3x=12-6
3x=6
x=2 
:)

find real numbers x and y if: (x+i)(3-iy)= 1+13i

(x+i)(3-iy)= 1+13i 
(3x+y)+(3-xy)i=1+13i 

(1) 3x+y=1, 
(2) 3-xy=13 
(2) 3-xy=13, xy=-10, y=-10/x 

(1) 3x+y=1, 3x+(-10/x)=1, 3x^2-x-10=0, x=2, x=-5/3 

then 
(x,y)=(2,-5) 
or 

(x,y)=(-5/3, 6)

interval does increase y=x^2+4x

vertex is on the y-axis, opens upwards, so it must increase when x>0, i.e.
(0,oo) 
:)

solve: 8x^2-3=2x

8x^2-3=2x 
8x^2-2x-3=0 
8x^2+4x-6x-3=0 
(8x^2+4x)-(6x+3)=0 
4x(2x+1)-3(2x+1)=0 
(4x-3)(2x+1)=0 

x=3/4 or x=-1/2

a=sin(theta), express sec2(theta) as a function of a.

 
 
therefore we can write