Notesale: Turn your study into money

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Section 1:

Task 1:
• 2 practical examples of parabolas are found in the world; basketball game, shooting
the basketball into the hoop
• Water coming out of a fountain vertically
• Practical use of a water fountain, is for decoration,
and a natural occurrence happens when the water is
shot out from the bottom, reaching its peak and
coming back down to the ground is the form of a
parabola
...





Figure 1
Task 2:
A)
Graph of Parabola’s





















B) Similarities between the parabola’s in the graph above include:
• all three cross the y-axis
...
y = x² - 2 contains ‘2’
• y = x² is a basic parabola, containing no ‘c’ value in the equation
• Translation, y = x² + 3 and y = x² - 2 both have vertical shift
• Dilation is wider y = x² - 2 and narrower for y = x² + 3
• C) When a quadratic equation is in the form y = a (𝓍 – h)² + k
...

•

D)

Graph of Parabola’s


E) Similarities between the parabola’s in the graph above include:
• all three intercept the x-axis only once
• all are concave up and have a minimum turning point
• none are reflected
• dilation is the same
Differences include:
• Translation, y = 𝓍² + 6𝓍 + 9 and y = 𝓍² - 4𝓍 + 4 have a horizontal shift on the x-axis
• y = x² doesn’t intercept the y-axis
F) In each graph the equations where in the form y = a (𝓍 – h)² + k
...
Both the graphs have either a horizontal or vertical shift present within the three
equations
...
This
is because in this form the equation y = a (𝓍 – h)² + k, the turning point is represented by
(h,k), thus the turning point (4,5) ⎯ as seen in the graph below: (PTO)

H) 𝑥 $ − 10𝑥 + 25
𝑥 $ − 10𝑥 + 5$ + 25 − 5$
𝑥 − 5 $ + 25 − 5$
𝑥 − 5 $ + 25 − 25
𝑥 − 5 $ + 0
𝑥 − 5 $

i) 𝑥 $ − 10𝑥 + 32
𝑥 $ − 10𝑥 + 5$ + 32 − 5$
𝑥 − 5 $ + 32 − 25
𝑥 − 5 $ + 7

J) 1
...
𝑥 $ + 8𝑥 + 12
𝑥 $ + 8𝑥 + 4$ + 12 − 4$
𝑥 + 4 $ + 12 − 16
𝑥 + 4 $ − 4
J) 3
...
The first
step is to make the ‘a’ value 1, by factoring out any other number
...
Work out the RHS, and put the perfect square trinomial
$
into brackets as a binomial, and square outside the brackets
...


Task 3:
1
...
Formula:

4𝑥 $ − 16𝑥 − 9 = 0
𝑎 = 4, 𝑏 = −16, 𝑐 = −9
SOLVE
2

𝑥 =
𝑥 =
𝑥
𝑥
𝑥
∴
𝑥
∴







−𝑏± 𝑏 −4𝑎𝑐

2𝑎
16± −162 −4(4)(−9)



2(4)
16± 256+144
=

8
16± 400
=

8
16±20
=

8
16+20
16−20
𝑥=
𝑜𝑟 𝑥 = 8
8
36
−4
= 𝑜𝑟 𝑥 =
8
8
9
1
𝑥 = 𝑜𝑟 𝑥 = −
2
2

3
...
6461 𝑜𝑟 𝑥 = 4
...
6461 − 6 𝑜𝑟 𝑃 = 16 4
...
3376 (𝑐𝑎𝑛𝑡 ℎ𝑎𝑣𝑒 𝑛𝑒𝑔𝑖𝑡𝑖𝑣𝑒 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟) 𝑜𝑟 𝑃 = 66
...
55968
∴ 𝑃 = 67𝑐𝑚

Task 5:
A) Allow ‘T’ to = 0








$
𝐻 = 2 + 20𝑡 − 5𝑡








$
𝐻 = 2 + 20 0 − 5 0








𝐻 = 2𝑚

B) 𝐻 = 2 + 20𝑡 − 5𝑡 $








$
17 = 2 + 20 1 − 5 1







17 = 2 + 20 − 5









17 = 𝑅𝐻𝑆












C) Maximum height and time = turning point of parabola
𝐻 = 2 + 20𝑡 − 5𝑡 $
0 = −5𝑡 $ + 20𝑡 + 2
𝑎 = −5, 𝑏 = 20, 𝑐 = 2
SOLVE
𝑇𝑃 =
𝑇𝑃 =

−𝑏

2𝑎
−20

2(−5)

𝑇𝑃 = 2
Allow ‘T’ to = 2
𝑦 = −5𝑡 $ + 20𝑡 + 2
𝑦 = −5 2 $ + 20(2) + 2
𝑦 = −20 + 40 + 2
𝑦 = 22𝑚
∴TP = (2,22)

D) Sub ‘17’ into 𝐻 = 2 + 20𝑡 − 5𝑡 $
17 = 2 + 20𝑡 − 5𝑡 $
17 − 17 = −5𝑡 $ + 20𝑡 + 2 − 17
0 = −5𝑡 $ + 20𝑡 − 15
a = −5, b = 20, c = −15
SOLVE
2

𝑥 =

𝑥 =
𝑥 =
𝑥 =

−𝑏± 𝑏 −4𝑎𝑐

2𝑎
−20± 202 −4(−5)(−15)
2(−5)
−20± 400−300

−10
−20± 100

−10



𝑥 = 2 ± −1
𝑥 = 2 + −1 𝑜𝑟 = 2 − −1
∴ 𝑥 = 1 𝑜𝑟 ∴ 𝑥 = 3
∴ 𝑥 = 3 as the larger number will be the time value because the ball is on its
way down

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