8.1 completing a square given: x 2 = u x = +√(u) or x = -√(u) e.g. given: x 2 = 3 x = √(3)...
TRANSCRIPT
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8.1 Completing a Square
Given: x2 = ux = +√(u) or x = -√(u)
E.g.Given: x2 = 3x = √(3) or x = -√(3)
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Completing the Square Solve: x2 + 6x + 4 = 0 x2 + 6x = -4 How to make the left side a perfect square? x2 + 6x + 9 = -4 + 9 (x + 3)2 = 5 x + 3 = √5 or x + 3 = -√5 x = -3 + √5 or x = -3 - √5 Check:
(-3 + √5)2 + 6(-3 +√5) + 4 = 0 ?9 - 6√5 + 5 - 18 + 6√5 + 4 = 0 ?(9 + 5 – 18 + 4) + (-6√5 + 6√5 ) = 0 yes
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Completing the Square Solve: 9x2 – 6x – 4 = 0 9x2 – 6x = 4 (9x2 – 6x )/9 = 4/9 x2 – (6/9)x = 4/9 x2 – (2/3)x + (1/9) = 4/9 + (1/9) (x – 1/3)2 = 5/9 x – 1/3 = √(5/9) or x – 1/3 = -√(5/9) x = 1/3 + √(5)/√(9) or x = 1/3 - √(5)/√(9)
x = 1/3 + √(5)/3 or x = 1/3 - √(5)/3 x = (1 + √(5))/3 or x = (1 - √(5))/3
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Your Turn
Solve by completing the square:x2 + 3x – 1 = 0
x2 + 3x = 1 x2 + 3x + 9/4 = 1 + 9/4 (x + 3/2)2 = 13/4 x + 3/2 = ± √(13/4) x = -3/2 ± √(13)/2 x = (-3 ± √(13)) / 2
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Your Turn Solve by completing the square:
3x2 + 6x + 1 = 0 3x2 + 6x = -1 (3x2 + 6x) / 3 = -1/3
x2 + 2x = -1/3 x2 + 2x + 1 = -1/3 + 1
(x + 1)2 = 2/3 x + 1 = ±√(2/3) x = -1 ±√(2/3) x = -1 ±√(2/3) √(3)/√(3) x = -1 ±√((2)(3 ))/ 3 x = -1 ±√(6) / 3
x = (-3 ±√(6) ) / 3
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Review Solve by completing the square.
(4x – 1)2 = 15 16x2 – 8x + 1 = 15 16x2 – 8x = 14 (16x2 – 8x)/16 = 14/16 x2 – (1/2)x = 7/8 x2 – (1/2)x + 1/16 = 7/8 +1/16 (x – ¼)2 = 15/16 x – ¼ = ±√(15/16) x = ¼ + √(15/16) = ¼ + √(15)/4 = (1 + √(15))/4 x = ¼ - √(15/16) = ¼ - √(15)/4 = (1 - √(15))/4
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8.2 Quadratic Formula Given: ax2 + bx + c = 0, where a > 0. ax2 + bx = -c (ax2 + bx)a = -c/a x2 + (b/a)x = -c/a x2 + (b/a)x + (b/2a)2 = -c/a + (b/2a)2
(x + (b/(2a))2 = -c/a + b2/(4a2) (x + (b/(2a))2 = -(c/a)(4a)/(4a) + b2/(4a2) (x + (b/(2a))2 = (-4ac) + b2) / (4a2) x + b/(2a) = ±√ ((b2 – 4ac)/(4a2))
x = (-b /(2a) ±√ ((b2 – 4ac)/(2a)x = (-b ±√ (b2 – 4ac)) / (2a)
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Quadratic Formula Given: ax2 + bx + c = 0, where a > 0.
-b ± √(b2 – 4ac)x = ----------------------- 2a
E.g., if 3x2 – 2x – 4 = 0 a = 3; b = -2; c = -4
2 ±√(4 + 48) 2 ± √(52) 1 ± √(13) x = ---------------- = ------------ = --------------- 6 6 3
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Application The number of fatal vehicle crashes has been
found to be a function of a driver’s age. Younger and older driver’s tend to be involved in more fatal accidents, while those in the 30’s and 40’s tend to have the least such accidents.
The number of fatal crashes per 100 million miles, f(x), as a function of age, x, is given by f(x) = 0.013x2 – 1.19x + 28.24
What age groups are expected to be involved in 3 fatal crashes per 100 million miles driven?
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Fatal Crashes vs Age of Drivers
Age of US Drivers and Fatal Crashes
17.7
9.5
6.24.1
2.8 2.4 3.0 3.8
8.0
16.3
0.02.04.06.08.0
10.012.014.016.018.020.0
16 18 20 25 35 45 55 65 75 79
Age of Drivers
Fa
tal
Cra
sh
es
/ 1
00
Mil
lio
n
Mil
es
of
Dri
vin
g
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Solution f(x) = 0.013x2 – 1.19x + 28.24 3 = 0.013x2 – 1.19x + 28.24 0.013x2 – 1.19x + 25.24 = 0
a = 0.013; b = -1.19; c = 25.24 -(-1.19) ±√((-1.19)2 – 4(0.013)(25.24))
x = -------------------------------------------- 2(0.013)
x ≈ ((1.19) ± √(0.104)) / (0.26)
x ≈ ((1.19) ± 0.322) / (0.26) x ≈ 33; x ≈ 58
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Your Turn Solve the following using the quadratic formula1. 4x2 – 3x = 6
4x2 – 3x - 6 = 0 a = 4; b = -3; c = -6 3 ±√((9 – 4(4)(-6)) 3 ±√(105)
x = ------------------------ = -------------- (2)(4) 8
2. 3x2 = 8x + 7 3x2 - 8x - 7 = 0 a = 3; b = -8; c = -7 8 ±√(64 – 4(3)(-7)) 8 ±√(148) 8 ±√(4 ·37) x = ---------------------------- = -------------- = ----------------- 6 6 6 4 ±√(37) = ------------ 3
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Discriminant Given: ax2 + bx + c = 0, where a > 0. -b ± √(b2 – 4ac)
x = ----------------------- 2a
If (b2 – 4ac) >= 0, x are real numbersIf (b2 – 4ac) < 0, x are imaginary numbers.
(b2 – 4ac) is called a discriminant. Thus,
If (b2 – 4ac) >= 0, solution set is real numbers.If (b2 – 4ac) , 0, solution set is complex numbers.
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Your Turn
Compute the discriminant and determine the number and type of solutions.
1. x2 + 6x = -9
2. x2 + 2x + 9 = 0
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8.3 Quadratic Function & Their Graphs
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Quadratic function f(x) = x2 - x - 2
x
f(x)
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Characteristics of Quadratic Function Graph f(x) = ax2 + bx + c
Shape is a parabolaIf a> 0, parabola opens upwardIf a < 0, parabola opens downwardVertex is the lowest point (when a > 0), and
the highest point (when a < 0)Axis of symmetry is the line through the
vertex which divides the parabola into two mirror images.
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To Sketch a Graph of Quadratic Function Given: f(x) = a(x – h)2 + k Characteristics
If a > 0, parabola opens upwardVertex is at (h, k)If h > 0, graph is shifted to right by h; if h < 0, to
the leftIf k > 0, graph is shifted up by k; if k < 0,
downward by kAxis of symmetry: x = hFor x-intercepts, solve f(x) = 0
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Graph of f(x) = -2(x – 3)2 + 8
f(x) = -2(x - 3)^2 + 8
-250
-200
-150
-100
-50
0
50
-7 -6 -5 -4 -3 -2 -1 -0 1 2 3 4 5 6 7 8 9 10 11
x
f(x)
x f(x)
-7 -192
-6 -154
-5 -120
-4 -90
-3 -64
-2 -42
-1 -24
-0 -10
1 0
2 6
3 8
4 6
5 0
6 -10
7 -24
Excel
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Graph of f(x) = -2(x – 3)2 + 8 Graph the function: f(x) = -2(x – 3)2 + 8 Solution:
Parabola opens downward (a = -2)Vertex: (3, 8)X-intercepts:
-2(x – 3)2 + 8 = 0(x – 3)2 = -8/-2x – 3 = ± √(4)x = 5; x = 1
y-interceptf(0) = -2(0-3)2 + 8 = -2(9) + 8 = -10
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Graph of f(x) = (x + 3)2 + 1
x f(x)
-7 17.0
-6 10.0
-5 5.0
-4 2.0
-3 1.0
-2 2.0
-1 5.0
0 10.0
1 17.0
2 26.0
3 37.0
4 50.0
5 65.0
6 82.0
7 101.0
f(x) = (x + 3)^2 + 1
0.0
20.0
40.0
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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x
f(x)
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Graph of f(x) = (x + 3)2 + 1 Graph the function: f(x) = (x + 3)2 + 1 Solution:
Parabola opens upward (a = 1)Vertex: (-3, 1)X-intercepts:
(x + 3)2 + 1 = 0(x + 3)2 = -1x – 3 = ± ix = 3 + i; x = 3 - i(This means no x-intercepts)
Y-interceptf(0) = (3)2 + 1 = 10
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Graphing f(x) = ax2 + bx + c A Solution:
f(x) = a(x2 + (b/a)x) + c = a(x2 + (b/a)x + (b2/4a2)) + c – a(b2/4a2) = a(x + (b/2a))2 + c – b2/(4a)
Comparing to f(x) = a(x – h)2 + k,h = -b/(2a); k = c – b2/(4a)
f(x) = a(x – (-b/(2a))2 + (c – b2/(4a))
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Graphing f(x) = 2x2 + 4x - 3 a = 2; b = 4; c = -3 f(x) = a(x – (-b/(2a))2 + (c – b2/(4a))
= 2(x – (-4/(4))2 + (-3 – 16/8) = 2(x + 1)2 – 5
Opens upward Vertex: (-1, -5) X-intercept:
-1 - √(10)/2, -1 + √(10)/2 Y-intercept:
f(0) = -3
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Graphing f(x) = 2x2 + 4x - 3
f(x) = 2x^2 + 4x - 3
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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
x
f(x)