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Use points and substitution to determine a quadratic function in vertex form,
= ( – )2 + , for each parabola.
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• The vertex of 1 is located at (0, 0), so = 0 and = 0. Since the parabola opens
upward, > 0. Then, 1 = ( – 0)2 + 0 or 1 = 2. Substitute the coordinates of another
point on the graph to find . Choose (1, 1).
1 = (1)2
=1
The quadratic function in vertex form is 1 = 2.
• The vertex of 2 is located at (0, 2), so = 0 and = 2. Since the parabola opens
upward, > 0. Then, 2 = ( – 0)2 + 2 or 2 = 2 + 2. Use (1, 6) to find .
6 = (1)2 + 2
=4
The quadratic function in vertex form is 2 = 4 2 + 2.
• The vertex of 3 is located at (0, –2), so = 0 and = –2. Since the parabola opens
upward, > 0. Then, 3 = ( – 0)2 – 2 or 3 = 2 – 2. Use (2, 0) to find .
0 = (2)2 – 2
1
=
2
1 2
The quadratic function in vertex form is 3 =
– 2.
2
• The vertex of 4 is located at (0, –4), so = 0 and = –4. Since the parabola opens
upward, > 0. Then, 4 = ( – 0)2 – 4 or 4 = 2 – 4. Use (2, –3) to find .
–3 = (2)2 – 4
1
=
4
1 2
The quadratic function in vertex form is 4 =
– 4.
4
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MHR •
Chapter 3
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Page 5 of 80
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For parabolas with the same shape and vertex but open downward, multiply the value
of by –1.
1
1
2
2
+ 2, 3 = – 2 – 2, 4 = – 2 – 4
1 = – , 2 = –4
2
4
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For parabolas with the same shape but translated 4 units to left, add 4 to each value of
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= ( + 4)2,
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= 4( + 4)2 + 2,
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2
1
( + 4)2 – 2,
2
=
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4
=
1
( + 4)2 – 4
4
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For parabolas with the same shape but translated 2 units down, subtract 2 from each
value of .
1 2
1 2
2
– 2, 2 = 4 2, 3 =
– 4, 4 =
–6
1=
2
4
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For ( ) = 5( – 15)2 – 100, = 5, = 15, and = –100.
The vertex is located at ( , ), or (15, –100).
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The equation of the axis of symmetry is = , or = 15.
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Since > 0, the graph opens upward.
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Since > 0, the graph has a minimum value of , or –100.
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The domain is { |  R}. Since the function has a minimum value of –100, the range
is { | • –100,  R}.
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Since the graph has a minimum value of –100, which is below the -axis, and opens
upward, there are two -intercepts.
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For = í4 2 + 14, = –4, = 0, and = 14.
The vertex is located at ( , ), or (0, 14).
The equation of the axis of symmetry is = , or = 0.
Since < 0, the graph opens downward and has a maximum value of , or 14.
The domain is { |  R} and the range is { | ” 14,  R}.
Since the graph has a maximum value of 14, which is above the -axis, and opens
downward, there are two -intercepts.
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MHR •
Chapter 3
Page 6 of 80
For = ( + 18)2 í 8, = 1, = –18, and = –8.
The vertex is located at (–18, –8).
The equation of the axis of symmetry is = –18.
Since > 0, the graph opens upward and has a minimum value of –8.
The domain is { |  R} and the range is { | • –8,  R}.
Since the graph has a minimum value of –8, which is below the -axis, and opens
upward, there are two -intercepts.
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For = 6( í 7)2, = 6, = 7, and = 0.
The vertex is located at (7, 0).
The equation of the axis of symmetry is = 7.
Since > 0, the graph opens upward and has a minimum value of 0.
The domain is { |  R} and the range is { | • 0,  R}.
Since the graph has a minimum value of 0, which is on the -axis, and opens upward,
there is one -intercept.
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1
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For = í ( + 4)2 í 36, = í , = –4, and = –36.
9
9
The vertex is located at (–4, –36).
The equation of the axis of symmetry is = –4.
Since < 0, the graph opens downward and has a maximum value of –36.
The domain is { |  R} and the range is { | ” –36,  R}.
Since the graph has a maximum value of –36, which is below the -axis, and opens
downward, there are no -intercepts.
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Since the vertex is located at (–3, –4), = –3 and = –
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4.
So, the function is of the form = ( + 3)2 – 4. Substitute
(–2, –3) and solve for .
–3 = (–2 + 3)2 – 4
–3 = – 4
=1
The quadratic function in vertex form is = ( + 3)2 – 4.
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MHR •
Chapter 3
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Page 7 of 80
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Since the vertex is located at (1, 12), = 1 and = 12.
So, the function is of the form = ( – 1)2 + 12.
Substitute
(0, 10) and solve for .
10 = (0 – 1)2 + 12
10 = + 12
= –2
The quadratic function in vertex form is
= –2( – 1)2 + 12.
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Since the vertex is located at (3, 1), = 3 and = 1.
So, the function is of the form = ( – 3)2 + 1. Substitute
(1, 3) and solve for .
3 = (1 – 3)2 + 1
3=4 +1
1
=
2
The quadratic function in vertex form is
1
= ( – 3)2 + 1.
2
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Since the vertex is located at (–3, 4), = –3
and = 4.
So, the function is of the form = ( + 3)2 + 4.
Substitute (–1, 3) and solve for .
3 = (–1 + 3)2 + 4
3=4 +4
1
= 4
The quadratic function in vertex form is
1
= ( + 3)2 + 4.
4
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MHR •
Chapter 3
Page 8 of 80