#$&*
course Mth 158
12/9/12 3:38 pm
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of
what you do or do not understand about it. This response should be given, based on the work you did in completing the
assignment, before you look at the given solution.
030. * 30
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Question: * 4.3.36 / 4.3.42 (4.1.42 7th edition) (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
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Your solution:
The function is in the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3.
The graph of this function will open upwards, because a is a positive number.
The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1.
To find the vertex I plugged in 1 for x.
y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4).
For the x intercepts I plugged in 0 for x.
f(x) = x^2 - 2 x - 3 = 0.
Then I factored,
(x - 3) ( x + 1) = 0 so then I set each one up to 0.
x - 3 = 0 OR x + 1 = 0, which gave me:
x = 3 OR x = -1.
So the x intercepts are (-1, 0) and (3, 0).
The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).
The domain is the set of all real numbers.
The lowest possible y value is -4. So the range is y > -4.
confidence rating #$&*: 3
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Given Solution:
* * The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3.
The graph of this quadratic function will open upwards, since a > 0.
The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 *
1 - 3 = -4. So the the vertex is at the point (1, -4).
The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the
quadratic formula. Here will will factor to get
(x - 3) ( x + 1) = 0 so that
x - 3 = 0 OR x + 1 = 0, giving us
x = 3 OR x = -1.
So the x intercepts are (-1, 0) and (3, 0).
The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).
The function can be evaluated for any real number x, so its domain is the set of all real numbers.
The graph is a parabola opening upward, with vertex (1, -4). So the lowest possible y value is -4, and all values greater
than -4 will occur. So the range is y > -4.
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Self-critique (if necessary): OK
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Self-critique Rating: 3
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Question: * 4.3.51 / 4.3.57. graph of parabola vertex (1, -3), point (3, 5)
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Your solution:
First of all the graph is a parabola. It's vertex is (1, -3).
The formula that we use is:
f(x) = a ( x - h)^2 +k (h,k) h = 1 and k = -3.
Therefore I plug in h and k and get.
f(x) = a (x - 1)^2 + (-3)
I get this far and then I get confused. I have watched the Math XL videos but I get confused on where I need to go next.
I looked forward at your solution, because I thought it might help me understand. But it just confused me more because
you used the formula:
(y - k) = a ( x - h)^2
Which is different from the examples in the book and in my math lab. What am I missing here? I feel like this should be
very easy and I am missing the boat.
confidence rating #$&*: 1
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Given Solution:
* * The graph is a parabola with vertex (1, -3), so it has form
(y - k) = a ( x - h)^2, with h = 1 and k = -3.
Thus we have
y - (-3) = a (x - 1)^2,
and the only unknown is a. To find a we substitute the coordinates of (3, 5) for x and y, obtaining the equation
(5 - (-3)) = a ( 3 - 1)^2, or{}(5 + 3) = a * 2^2, or{}8 = a * 4,
with the obvious solution
a = 2.
Thus the equation of the parabola is
y + 3 = 2 ( x - 1)^2.
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Self-critique (if necessary):
Still confused :-(
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Self-critique Rating: 1
@&
You're doing well so far. Your equation
f(x) = a (x - 1)^2 + (-3)
is the same as the equation obtained in the given solution:
y - (-3) = a (x - 1)^2
All you need to do is let y = f(x) and your equation becomes
y = a (x - 1)^2 + (-3).
If you subtract -3 from both sides you then get
y - (-3) = a (x - 1)^2.
There is actually no reason for you to rearrange your equation to this form. y = a (x - 1)^2 + (-3) will work just fine.
So we have
y = a (x - 1)^2 + (-3).
And we know that the graph goes through (3, 5). This means that when x = 3, we have y = 5. This is sufficient to determine the value of a.
Substituting x = 3 and y = 5 into your equation we obtain
5 = a ( 3 - 1)^2 + (-3).
You can easily solve this equation for a.
*@
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Question: * Extra Problem (4.1.67 7th edition) (was 4.1.50). What are your quadratic functions whose x intercepts are -5
and 3, for the values a=1; a=2; a=-2; a=5?
Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3).
If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15.
If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30.
If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30.
If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.
Does the value of a affect the location of the vertex?
In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x
coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.
The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following:
For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12.
For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24.
For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24.
For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60.
So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).
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Self-critique (if necessary):
OK
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Self-critique Rating: 3
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Question: * Extra Problem (4.1.67 7th edition) (was 4.1.50). What are your quadratic functions whose x intercepts are -5
and 3, for the values a=1; a=2; a=-2; a=5?
Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3).
If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15.
If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30.
If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30.
If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.
Does the value of a affect the location of the vertex?
In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x
coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.
The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following:
For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12.
For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24.
For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24.
For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60.
So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).
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Self-critique (if necessary):
OK
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Self-critique Rating: 3
#*&!
&#Good responses. See my notes and let me know if you have questions. &#