SIT190 Assignment 2

SIT190 Assignment 2 with problems on quadratic equations, the Quadratic Formula, intercepts, highest or lowest point, curve sketching, solving for x, trigonometric functions, and finding average change in volume.

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

5. Simplify, and express in terms of positive indices:
(i)
(2 x )−3 y5
( 4 y)−2 x3
(2¿¿−3) x−3 y5
4−2 y−2 x3 ¿
42 y2❑ y5
23 x3 x3
16 y7
8 x6
2 y7
x6
(ii)
¿
¿
3−2 x−2 ( 33 x−3 y−6
x6 )
3−2 x−2 ( 33
x3 x6 y6 )
33 −2
x3 +2 x6 y6
3
x11 y6
iii)
3 x−2
5 x−3 +7 x2
3 x3
5 x2 +7 x2
3 x3
12 x2
x
4
iv)
( x
−1
2 )2 /3 x x2 /3
x−3/ 4
( x
−1
2 x 2 /3
)❑ x x2 /3
x−3 /4

( x
−1
3 )❑ x x2 /3
x−3/ 4
x
3
4❑ x x2 /3
x1 /3
x
3
4 + 2
3
x1/ 3
x
17
12
x1 /3
x
17
12 −1
3
x
13
12
v)
81 x3 x 1
2 −3 x5 x 1
2
25 x3 x 1
2
81 x
3
2 −3 x
5
2
25 x
3
2
6.
(a) Simplify (by writing as a single exponential):
(i) e−2xe3+x
e4x+1
e−2 x+(3+ x)
e4 x+1
e−x+3
e4 x+1
e− x+3−(4 x+1)
e−5 x+4
(ii) (e3−2x)−2e5+2x

2
(e3−2x)−2e5+2x
e-6−-4x)e5+2x
e-6+4x+ 5+2x
e-1+6x
(b) Expand and simplify (if possible):
(i) (𝑒3𝑥 − 𝑒−2𝑥)2
(𝑒3𝑥*2− 𝑒−2𝑥*2)
(𝑒6𝑥 − 𝑒−4𝑥)
=𝑒6𝑥--4x
=𝑒10𝑥
(ii) (𝑒3𝑥 + 𝑒−2𝑥)(𝑒3𝑥 − 𝑒−2𝑥)
(𝑒3𝑥+-2x)(𝑒3𝑥--2x)
(𝑒x)(𝑒5𝑥)
𝑒x+6x
𝑒6x
7. (a) Simplify (by expressing as a single natural logarithm):
(i) −3ln 3 + 2 ln 9 − 2ln 81
−3ln 3 + 2 ln 32 − 2ln 34
−3ln 3 + 2*2 ln 3 − 2*4ln 3
(−3+4-8)ln 3
-7ln3
(ii) 5ln(5𝑥3) − 3ln(𝑥5) − ln(25𝑥2) − 3 ln 5
3*5ln(5𝑥) − 3*5ln(𝑥) − 2ln(25𝑥) − 3 ln 5
3ln(𝑥) +5ln(5)− 15ln(𝑥) − 2ln(𝑥)+ 2ln(5)− 3 ln 5
3ln(𝑥) − 15ln(𝑥) − 2ln(𝑥)+5ln(5)+ 2ln(5)− 3 ln 5
− 14ln(𝑥)+4ln(5)
(b) Simplify (without using a calculator):
(i) 5 log 1
128
5 log 2 (1/128)
5 log2-7
-35log 2
(ii) −2𝑒ln(3𝑥−2)
8. Solve for 𝑥:
(i) 3𝑦 = 9 ln (2𝑥−6)
18

(iii) 𝜃
𝑥
80°
40
(iv)
26 𝑥
𝜃
Y = 3ln(2x-6/18)
(ii) −4𝑒1−3𝑥 + 5 = 6𝑦 + 7
−4𝑒1−3𝑥 = 6y+2
9. Use a calculator to evaluate (answer correct to 3 decimal places):
(i) cos 225° = -0.707
(ii) sin 2.9 R
10.Find the unknown length 𝑥 (correct to 1 decimal place) and angle 𝜃 for
each of the following triangles.
(i) (ii)
35𝑜
25
θ°
𝑥
Cos 35= x/25
X = sine
35*25 = 10.7
angle = 180-
35-90 = 55
degrees
𝜃
𝑥
20
50°
Angle = 180-90-50= 4o degree
Length x = sine 50 = 20/x
Length x = 20/sin 50
X = 76.23
Angle = 180-90-80= 10 degree
Length x = tan 80 = x/4
Length x = tan 80*4
X = 36.01
X 2 =26 2 - 10 2
X 2 = 676-100
X 2 = 576
X=root 576
X= 24
Angle = cos w = 10/26
Acrchcos 0.38461538
Angle = 67.38°
.
11. Express in degrees:
(i) 3.3𝑅

(ii) −7𝜃𝑅
4
12. Express in radians in terms of 𝜃:
(i) 195°
1 degree =0.0175 radian
195° = 195°*0.0175 = 3.4125 pi
ii) 750°
1 degree =0.0175 radian
750° = 750°*0.0175 = 53.625 pi
(ii) 0°

13.If cos 𝜃 = −4 and 𝜃 is in the 3rd quadrant, find the exact values of:
5
(iii) sin 𝜃 = 3/5
(iv) tan 𝜃 = 3/-4
14.The following table shows the volume of petrol (in megalitres) in a petrol
station tank over a six hour period commencing at 9 am.
𝒕 (hours) 9am 10am 11am 12noon 1pm 2pm 3pm
𝒗 (m’litres) 15.3 9.9 8.6 7.5 6.8 5.3
(i) Find the average change in volume (m’litres per hour) of petrol in
the tank from 10am to 1pm.
15.3-7.5 = 7.8/4 = 1.95 litres/ hr
(ii) Find the average change in volume over the 6 hours from 9am to 3
pm.
15.3-5.3 = 10/6 = 1.667
(iii) How much petrol is in the tank at 11am if it is known that the average
change in volume from 9am to 11am is −3.2 m’litres/hour.
Total volume change 3.2*2 = 6.4 litres.
Petrol in the tank at 11 am = 15.3-6.4 = 8.9
.
15.Find 𝑑𝑦
𝑑𝑥
I) 3𝑥3+9𝑥2-x
II) -3/4𝑥-4+5/2𝑥-6-𝑥-2
III) 4𝑥-1/5+4𝑥-7/3
IV)1/6 𝑥1/6
V) 1/8𝑥-7/8-1/8𝑥-3/2