Download Homework.8

Homework.8
Definition (READ IT Before you do the questions): any complex number
c is of the form:
c = a + bi
where a is the coefficient of the real part and b is the coefficient of the imaginary part, i means imaginary. The pair of numbers (a, b) is the coordinates
(see Example#2 below)
Example #1:Express the following
numbers in terms of i(Using the
√
definition: i2 = −1, and i = −1):
√ √
√
−4 = 4 · −1 = 4 · i = 2i
√
√
13 = 13 + 0i
√
Question #1: Express the
√ following numbers in terms of i (Using the definition: i2 = −1, and i = −1):
√
−81
√
− −81
√
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Example #2: Spot x = −1 − i in graph below:
(NOTICE: The graph below is NOT a X-Y PLANE, it’s only for imaginary numbers)
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Explanation:
x = −1 − i is just a point on this graph! it’s not a line at all, it could be
written as x = (−1) + (−1)i, the coefficient for the real part is (−1) (for the
HORIZONTAL AXIS)and the coefficient for the imaginary part is also (−1)
(for the VERTICAL AXIS), so the pair of coordinates is (−1, −1), thus, the
complex number x = −1 − i is the blue point shown above.
Question #2: Spot 2 + 4i, −1 + 3i and 2 (which is 2 + 0i) in the following
graph.(Using the the method given in Example#2):
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Example#3: Find the axis of symmetry of the graph of f (x) = x2 +2x−2;
Solution:
(Using the special product x2 + 2x + 1 = (x + 1)2 )
f (x) = x2 + 2x − 2 = x2 + 2x + 1 − 1 − 2
= (x2 + 2x + 1) − 3
= (x + 1)2 − 3
The square term (x + 1)2 is the most important because:
Proposition: if there’s a (x − c)2 in the function, then x = c is the axis of symmetry.
In this case, f (x) = (x + 1)2 − 3, there’s (x + 1)2 = (x − (−1))2 , so x = −1
is the axis of symmetry.
Question#3. a): Find the axis of symmetry of the graph of g(x) =
x2 + 6x + 6;
(Hint: use the special product x2 + 6x + 9 = x3 + 2 · 3 · x + 32 = (x + 3)2 )
Question#3. b): Find the axis of symmetry of the graph of h(x) =
x2 + 6x + 12;
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Example#4: Use the quadratic formula√ on Page247 (The solutions of
ax + bx + c = 0, a 6= 0 are given by x = −b±2a ∆ where ∆ = b2 − 4ac ) to find
the solutions of f (x) = x2 + 2x − 2 = 0, can these solutions be graphed on
the real number line?
Solution:
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f (x) = x2 + 2x − 2
compare this with
f (x) = ax2 + bx + c
so a = 1, b = 2, c = −2. calculate the discriminant first:
√
∆ = b2 − 4ac = 22 − 4 · 1 · (−2) = 12
√
√
√ √
√
∆ = 12 = 4 · 3 = 4 · 3 = 2 3
then the solutions of f (x) = x2 + 2x − 2 = 0 are:
√
√
√
−b + ∆
−2 + 2 3
x1 =
=
= −1 + 3
2a
2
≈ 0.7321
√
√
√
−2 − 2 3
−b − ∆
=
= −1 − 3
x2 =
2a
2
≈ −2.7321
these two solutions are real numbers, so they can be spotted on a real number
line.
Question#4. a): Find the solutions of function g(x) = x2 + 6x + 6 = 0.
Question#4. b): Find the solutions of function h(x) = x2 + 6x + 12 = 0.
Example#5: In Example#3, we talked about the axis of symmetry of
function f (x) = x2 + 2x − 2, now find the bottom point of this function (on
Page 261 the text book call it the vertex of a parabola).
Solution:
In the solution of Example#3, we already know that
f (x) = x2 + 2x − 2 = (x + 1)2 − 3, there’s (x + 1)2
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since (x + 1)2 ≥ 0,
we have (x + 1)2 − 3 ≥ 0 − 3
which is (x + 1)2 − 3 ≥ −3
so f (x) = (x + 1)2 − 3 ≥ −3
which is f (x) ≥ −3
so the least value of f (x) is −3, and f (−1) = −3 so the bottom point(or
vertex ) is (−1, −3).
NOTICE: the reason why I know when x = −1, f (x) will take the least
value is that x = −1 is the axis of symmetry. bottom point (or vertex ) of
parabola is on the axis of symmetry.
Question#5 a):Using the results you get from Question#3. a), find the
vertex of g(x) = x2 + 6x + 6
Question#5 b):Using the results you get from Question#3. b), find the
vertex of h(x) = x2 + 6x + 12
Example#6: Graph the function f (x) = x2 + 2x − 2.
Solution:
From the results of Example #3, #4, #5, we’ve gathered the three most
important features of function f (x) = x2 + 2x − 2:
1)axis of symmetry: x = −1;
2)two solutions that make f (x) = 0:
√
x1 = −1 + 3 ≈ 0.7321
√
x2 = −1 − 3 ≈ −2.7321
NOTICE: that the solutions are the points where the graph of function hits
the x-axis, if the solutions are not real, then the graph of function does not
hit the x-axis.
3)the vertex of f (x): (−1, −3)
now we can draw the graph:
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Question#6 a):Using the results you get from Question#3. a) #4. a)
#5. a), Graph g(x) = x2 + 6x + 6, point out where is the axis, vertex, and
where doest the graph hits the x-axis(if the graph hits the x-axis).
* Question#6 b):Using the results you get from Question#3. b) #4. b)
#5. b), Graph h(x) = x2 + 6x + 12, point out where is the axis, vertex, and
DOES THE GRAPH HITS THE X-AXIS?
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