© the visual classroom 3.7 standard form of a quadratic relation y = a(x – s)(x – t)factored...
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© The Visual Classroom
3.7 Standard Form of a Quadratic Relation
y = a(x – s)(x – t) factored form
y = ax2 + bx + c standard form(expanded form)
Example: Expand these expressions
1. (x + 4)(x – 6) Use the distributive property
= x2
= x2 – 2x – 24
+ 4x Collect like terms– 6x – 24
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= (w – 5)(w – 5)
2. (w – 5)2
Expand and simplify
Use the distributive property
= w2
= w2 – 10w + 25
– 5w Collect like terms– 5w + 25
3. (a – 8)(a + 8)
= a2 + 8a
= a2 – 64
– 8a – 64 Collect like terms
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4. (2x – 5y)(3x + 7y)
= 6x2 + 14xy
= 6x2 – xy – 35y2
– 15xy – 35y2 Collect like terms
Expand and simplify
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5. – 3(m – 2n)(m + 8n)
= – 3m2 – 18mn + 48n2
Multiply the brackets, then multiply by – 3
Expand and simplify
= – 3[(m – 2n)(m + 8n)]
= – 3[ m2 + 8mn – 2mn – 16n2]
= – 3[ m2 + 6mn – 16n2]
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Determine the missing information.
a) x2 – 2x – 15 = (x – 5)( ? ? )
= (x – 5)(x + 3)
= (2x + 5)(3x – 4)
b) 6x2 + 7x – 20 = (2x + 5)( ? ? )
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Determine the expanded form of the equation of the parabola.
y = a(x – s)(x – t)
y = a(x + 1)(x – 3)
4 = a(1 + 1)(1 – 3)
4 = a(2)(– 2)
4 = a(– 4)
– 1 = a
y = – (x + 1)(x – 3)
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y = – (x + 1)(x – 3)
y = – [(x + 1)(x – 3)]
y = – [x2 – 3x + x – 3)]
y = – [x2 – 2x – 3)]
y = – x2 + 2x + 3
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Write an expression for the area.
2a + 5
A = (2a + 5)2
= (2a + 5)(2a + 5)
= 4a2 + 10a
= 4a2 + 20a + 25
+ 10a + 25
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A stone is dropped from a bridge that is 20 m above a river below. The table gives the height of the stone as it falls.
Time 0 0.5 1 1.5 2
Height 20.000 18.775 15.100 8.975 0.400
a) Create a scatter plot and draw a graph of best fit. b) Find the approximate time when the stone hits the water. c) Use (0, 20) as the vertex and indicate the other zero.
d) Determine an algebraic expression, in standard form, that models the data. e) Use a graphing calculator to determine the quadratic regression equation for the data.
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b) Approximate time the stone hits the water is 2.1 sec.
time (sec)he
ight
(m
)c) The other zero would be (-2.1, 0).
d) y = a(x – s)(x – t)
y = a(x – 2.1)(x + 2.1)
20 = a(0 – 2.1)(0 + 2.1) 20 = a (– 4.41)
– 4.535 = a y = – 4.535 (x – 2.1)(x + 2.1)
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y = – 4.535 (x – 2.1)(x + 2.1)y = – 4.535(x2 – 2.1x + 2.1x – 4.41)
y = – 4.535(x2 – 4.41)
y = – 4.535x2 + 20