Quadratic Functions - Parabolas (1 Viewer)

[Petinga]

Help Plz asap

1. Find the equation of the parabola which passes through the points (-1,0) and (0,-1) and is symmetrical about the Y-axis.

2. Find the equation of the parabola through the points (2,0), (4,0) and (0,8).

 

[Trev]

1)
Since it's symmetrical it would also pass through (1,0), you should recognise this as the parabola y=x²-1; but for working:
y=ax²+bx+c
(-1,0); 0=a-b+c
(0,-1); -1=c
(1,0); 0=a+b+c
Solve these simultaneously so y=x².

2)
Same method as above:
y=ax²+bx+c
(2,0); 0=4a+2b+c
(4,0); 0=16a+4b+c
(0,8); 8=c
Solving simultaneously gives y=x²-6x+8

 

[dave28]

hi, could someone please show & explain the answer to the 2003 HSC question 7 b) ii) velocity given by v = 2 - 4cost for 0 equal or less than t which is equal to or less than 2 pi.
what is max velocity of particle during this period?

 

[word.]

Um. You could easily start a new topic...
v = 2 - 4cos{t}
and -1 <= cos{t} <= 1. by inspection max velocity occurs when cos{t} = -1,
i.e. v = 2 - 4(-1) = 6 m/s