SIT190 Assignment 2

SIT190 Assignment 2 Trimester 2, 2018 Due
Date: Friday 24th August, 5.00 pm
Complete the following problems, showing all your working.
Marks are allocated to your steps, not just the final answer.
1. Factorise and solve the following quadratic
equations: (i) −4X2 + 8X = −12
4x2-8x−12 =0
4(x2+2x−3)
4(x+1)(x-3)
(ii) 144 – 36x2 = 0
= 36x2 – 144 = 0
-36 • (x2 - 4) =0
Factoring: x2 – 4
(x + 2)(x - 2)
= -36(x + 2)(x - 2)
2. Use the Quadratic Formula to
solve: (i) −3x2 – 5x + 2 = 0
The quadratic formula =
x=−b ± √b2−4 ac
2 a
A=-3, b=-5 and c=2
5± √ −52−4 x−3 x 2
2 x−3
5± √ 49
−6
5± 7
−6
x=−2∨1
3
(ii) 2x 2 + 10x + 8 = 0
A=2, b=10 and c=8
−10 ± √102−4 x 2 x 8
2 x 2

−10 ± √ 36
4
−10 ±6
4
x=4∨−1
3. For y = −x 2 + 2x + 3,
(i) find the y-intercept
y intercept occurs when x = 0
replacing x with 0; y = −0 2 + 2*0 + 3, y intercept = (0,3)
(ii) find the x-intercepts
x intercept occurs when y = 0
replacing y = 0
0= −x 2 + 2x + 3
Solving the quadratic equation
x2−2x−3=0
Add 3 to both sides:
x2−2x=3
x2−2x+1=4
x2−2x+1=4
x−1=2x−1=2 or x−1=−2x−1=−2.
x=3x=3 or x=−1x=−1.
Therefore x intercept = (3,0) and (-1,0)
(iii) determine whether the curve has a highest or lowest point
f(x) = −x 2 + 2x + 3
f’(x)= −2x+ 2
f”(x) = -2
the value of f’(x) and f”(x) is an negative meaning the point is a highest
point of maximum
(iv) find the coordinates of the highest or lowest point
f = −x 2 + 2x + 3
= −2x+ 2
X = 1
Replacing x=1 in y = −1 2 + 2*1 + 3, we get y = 6,
The point is (1,6)
(v) sketch the curve.
4. A rocket is fired into the air from the top of a hill 100 metres above
sea level. The trajectory taken by the rocket is given by the
formula:h = −1
3 t2+ 20
3 t+ 100
where ℎ is the height of the rocket above sea level and t

is the time (in seconds) after it was fired. The rocket eventually falls
into the sea.
(a) How high is the rocket after:
(i) 3 seconds
h = −1
3 t2+ 20
3 t+ 100 replacing t=3
h = −1
3 32+ 20
3 3+ 100 = -3+20+100 = 117 therefore h=117
(ii) 9 seconds
h = −1
3 t2+ 20
3 t+ 100 replacing t=9
h = −1
3 92+ 20
3 9+ 100 = -27+60+100 = 133 therefore h=133
(b)
(i)How high does the rocket go before it begins to fall?
F’(h) = =−2
3 t+ 20
3 = 0
−2
3 t+ 20
3 = 0
20
3 = 2
3t
T =10 seconds h at this point = −1
3 (10)2+ 20
3 (10)+ 100 = 133.33
(ii) After how many seconds does the rocket fall into the sea?
Solving h = −1
3 t2+ 20
3 t+ 100
x=−b ± √b2−4 ac
2 a
A= −1
3 , b= 20
3 c,= 100
−20
3 ± √(20
3 )
2
−4 x −1
3 x 100
2 x −1
3
−6.67 ± √ 80 0
3
−0.67
−6.67 ±16.33
−0.67
t=34.33 seconds