Chapter 3 Test
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University of Guelph *
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1080
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Mathematics
Date
Apr 3, 2024
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docx
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Chapter 3 Test – Graphs of Quadratic Equations
/ 14 Knowledge / 14 Application / 6 Thinking / 4 Communication
KNOWLEDGE
1.
Find the zeros, axis of symmetry, y-intercept and vertex of the function y = -2x(x+6) algebraically. Then use that information to sketch the parabola. [5]
(paper)
Y=-2x^2 -12x
(i used a seperate peice of paper to calculate a table of
values to find abcd)
a.
Zeros x=0 x=-6
b.
Axis of symmetry x=-3
c.
Vertex (-3,18)
d.
y - intercept (0,0)
2.
Write the following equations in standard form [4]
a.
y = (2x – 3)(x +6)
b. y = -3(x - 3)
2
a)
y= (2x-3)(x+6)
y= 2x(x+6)-3(x+6)
y= 2x^2 + 12x -3x -18
y=2x^2 + 9x -18
b)
y=-3(x-3)^2
y= -3(x-3)(x-3)
y= (-3x +9)(x-3)
y=-3x(x-3)+9(x-3)
y= -3x^2 +9x +9x -27
y= -3x^2 + 18x - 27
3.
Evaluate each of the following. Ensure your answer is written with only positive exponents.
a) (-2)
5
b) 8
0
c) −
5
−
2
a) (-2)^5
(-2) (-2) (-2) (-2) (-2)= 4
-32
b)
8^0= 1
c) -5^-2
=(-1/5)^2
=-1/25
4.
State whether the graph of the quadratic relation y
= - 3
x
2
– 4
x
+ 1 opens upward or
downward. Explain you thought process.
To determine if the parabola opens upwards or downwards, we need to look at the equation, more
in particular the a.
In this example, a is -3, which means a is negative = a<0
Therefore, when a is smaller then 0, the parabola opens downward .
APPLICATION
1. A parabola has zeros at (-1, 0) and (-3, 0) and a minimum value of -5. Determine the equation of the parabola in standard form.
axis o.s. = (-1 +(-3))/2
Axis o.s. = -2
If the minimum value is -5 and axis o.s is -2 the vertex of the parobala is (-2,-5)
Y= a(x-r)(x-s)
-5= a(-2 +1)(-2 +3)
-5=a(-1)(1)
-5= -1a
5/1= 1a/1
5= a
Y= 5(x+1)(x+3)
(Converting to standard form)
Y= (5x +5)(x+3)
Y=5x(x+3)+5(x+3)
Y= 5x^2 +15x + 5x+ 15
Y=5x^2 +20x +15
2. An emergency flare was shot into the air from the top of a building. The table gives the height of the flare at different times during its flight.
Time (s)
0
1
3
4
5
6
Height (m)
48
60
60
48
25
0
a)
How tall is the building? 48m
b)
Use the data in the table to create a scatter plot and then draw a curve of good fit that includes BOTH x
-intercepts of the curve (hint: extend line). (paper)
c)
Using your curve…
estimate the maximum height of the
flare: 64m
what is the time when the flare hits the ground:
6
seconds
d)
Determine an equation for the curve you drew. Show
your work.
X= 6
X= -2
Axis o.s. = (6+(-2))/2 = 2
Vertex (determined from c) = (2,64)
Y= a(x-r)(x-s)
64=a(2-6)(2+2)
64=a(-4)(4)
64=a(-16)
64/-16=a(-16)/-16
4=a
Y=4(x-6)(x+2) (factored form, didn’t state to put it in standard)
e)
Use your equation to determine the height of the flare at 4.5 seconds. Show your steps.
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Y=4(x-6)(x+2) Y=4(4.5-6)(4.5+2)
Y=4(-1.5)(6.5)
Y=4(-9.75)
Y=39
The height is 39m
f)
Use your equation to determine the maximum height of the flare. Show your work.
(vertex)^
As stated before, the vertex is (2, 64), however we can still calculate it.
Y=4(x-6)(x+2) Y=4(2-6)(2+2) Y= 4(-4)(4)
Y=4(-16)
Y=64
THINKING
1.
Miss Peluso owns a small bakery. She currently charges $12 for each sandwich. At this price, she sells about 60 sandwiches’ a week. Experience has taught her that a $1 increase in the price of sandwiches means a drop of about four sandwiches per week in sales. At what price should Miss Peluso sell sandwiches
to maximize her revenue?
X= $1 increases Therefore, she should sell her sandwiches at 13.5$ to maximize her revenue.
Y= revenue
12+x (price increase)
60-4x (number of sandwiches sold)
Y=(12+x)(60-4x)
Zeros:
X= -12
X= 15
Axis o.s.= -12+15/2 = 1.5
Y= 12+x
Y= 12 +1.5
Y= 13.5