course mth 158
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: * 4.3.42 (4.1.42 7th edition) (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
f(x) = x^2 2x 3
Compare this to a quadratic function f(x) = ax^2 + bx + c you have
a = 1 b = -2 c = -3
Since a = 1 and 1 > 0 the graph opens up and the vertex is the lowest point
The x coordinate of the vertex is 1 and can be found by doing the following
h = - b/2a in this case it would be
h = - (-2)/2(1) = 1
The y coordinate of the vertex is -4 and can be found by doing the following
k = f(1) = 1^2 2(1) 3 = -4
So the vertex is at (1, -4)
The axis of symmetry is the line x = 1
Y intercept is the f(0)
0^2 2(0) 3 = -3
y int (0, -3)
The x intercepts are where f(x) = x^2 - 2 x - 3 = 0.
(x+1) (x-3) = 0
x + 1 = 0
x = -1
x 3 = 0
x = 3
(-1,0) (3, 0)
D: {all real numbers} R: {-1 < x < 3}
confidence rating: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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).
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Self-critique (if necessary):
I dont see any discrepancies. Im not sure if my domain and range are correct.
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 Rating: 3
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Question: * 4.3.57. graph of parabola vertex (1, -3), point (3, 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Since we know the vertex is (1, -3) and there is a point (3, 5) we can find h and k in the equation
f(x) = a(x h)^2 + k
h = 1
k = -3
f(x) = a(x 1)^2 3
to find the value of a, we use the fact that f(0) = f(3) = 5 (the y intercept)
f(x) = a(x 1)^2 3
becomes
5 = a(3 1)^2 3
The equation
5 = a(3 1)^2 3 simplifies to
5 = a ( 2 )^2 - 3, or
8 = 4 * a so that
a = 2.
5 = a + 1
5 1 = a
4 = a
f(x) = a(x-h)^2 + k
giving us the quadratic function
4(x-1)^2 -3
= 4x^2 16x - 3
confidence rating: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 1
* * 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):
I tried to solve this one by following along in the book (page 300 example 6) I was doing it right until I tried to solve for a. The book has an example, but obviously it didnt work for this problem.
You had the right equation for a, but your solution went astray. See the notes I inserted into your solution.
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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?
I cannot answer this based on what I knew about the problem. I had to look at the given solution to try and understand the problem.
I see how you came up with the multiple (x+5)(x-3)
so by taking each value of a you get the function
a=1 gives you the function 1(x+5)(x-3) = x^2 + 2x 15
a = 2 2(x+5)(x-3) = 2x^2 + 4x 30
a = -2 -2(x+5)(x-3) = -2x^2 4x + 30
a = 5 5(x+5)(x-3) = 5x^2 + 10x 75
Even though I have already seen the given solution, I wanted to solve the a values on my own so I can get more practice on how to do this type of problem.
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):
I didnt see the question Does the value of a affect the location of the vertex? I dont know if I could have answered this question either. I dont think the value of a would affect the vertex. The y coordinate of the vertex is hard for me to understand how a affects it. I will have to look more closely at this one.
It's not that easy to see the reasons for the relationship between the y coordinate of the vertex and the value of a. That's not something you can really understand fully without the use of calculus.
However the coordinates of the vertices of the functions we see here do show that a has an effect.
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Self-critique Rating: 3
&#Good responses. See my notes and let me know if you have questions. &#
course mth 158
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
*********************************************
Question: * 4.3.42 (4.1.42 7th edition) (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
f(x) = x^2 2x 3
Compare this to a quadratic function f(x) = ax^2 + bx + c you have
a = 1 b = -2 c = -3
Since a = 1 and 1 > 0 the graph opens up and the vertex is the lowest point
The x coordinate of the vertex is 1 and can be found by doing the following
h = - b/2a in this case it would be
h = - (-2)/2(1) = 1
The y coordinate of the vertex is -4 and can be found by doing the following
k = f(1) = 1^2 2(1) 3 = -4
So the vertex is at (1, -4)
The axis of symmetry is the line x = 1
Y intercept is the f(0)
0^2 2(0) 3 = -3
y int (0, -3)
The x intercepts are where f(x) = x^2 - 2 x - 3 = 0.
(x+1) (x-3) = 0
x + 1 = 0
x = -1
x 3 = 0
x = 3
(-1,0) (3, 0)
D: {all real numbers} R: {-1 < x < 3}
confidence rating: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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).
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Self-critique (if necessary):
I dont see any discrepancies. Im not sure if my domain and range are correct.
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 Rating: 3
*********************************************
Question: * 4.3.57. graph of parabola vertex (1, -3), point (3, 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Since we know the vertex is (1, -3) and there is a point (3, 5) we can find h and k in the equation
f(x) = a(x h)^2 + k
h = 1
k = -3
f(x) = a(x 1)^2 3
to find the value of a, we use the fact that f(0) = f(3) = 5 (the y intercept)
f(x) = a(x 1)^2 3
becomes
5 = a(3 1)^2 3
The equation
5 = a(3 1)^2 3 simplifies to
5 = a ( 2 )^2 - 3, or
8 = 4 * a so that
a = 2.
5 = a + 1
5 1 = a
4 = a
f(x) = a(x-h)^2 + k
giving us the quadratic function
4(x-1)^2 -3
= 4x^2 16x - 3
confidence rating: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 1
* * 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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I tried to solve this one by following along in the book (page 300 example 6) I was doing it right until I tried to solve for a. The book has an example, but obviously it didnt work for this problem.
You had the right equation for a, but your solution went astray. See the notes I inserted into your solution.
------------------------------------------------
Self-critique Rating: 3
*********************************************
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?
I cannot answer this based on what I knew about the problem. I had to look at the given solution to try and understand the problem.
I see how you came up with the multiple (x+5)(x-3)
so by taking each value of a you get the function
a=1 gives you the function 1(x+5)(x-3) = x^2 + 2x 15
a = 2 2(x+5)(x-3) = 2x^2 + 4x 30
a = -2 -2(x+5)(x-3) = -2x^2 4x + 30
a = 5 5(x+5)(x-3) = 5x^2 + 10x 75
Even though I have already seen the given solution, I wanted to solve the a values on my own so I can get more practice on how to do this type of problem.
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):
I didnt see the question Does the value of a affect the location of the vertex? I dont know if I could have answered this question either. I dont think the value of a would affect the vertex. The y coordinate of the vertex is hard for me to understand how a affects it. I will have to look more closely at this one.
It's not that easy to see the reasons for the relationship between the y coordinate of the vertex and the value of a. That's not something you can really understand fully without the use of calculus.
However the coordinates of the vertices of the functions we see here do show that a has an effect.
------------------------------------------------
Self-critique Rating: 3
&#Good responses. See my notes and let me know if you have questions. &#