Assign 27

course mth 158

I don't know if I have submitted this before or not. It seems like I remember working the problems, but don't have any confirmation on it so I will submit it again. I went through it and tried to solve what I couldn't do before, but I don't know how much better it is now.

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.

027. `* 27

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

Question: * 3.6.2 / 10. P = (x, y) on y = x^2 - 8.

Give your expression for the distance d from P to (0, -1)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Let P = (x, y) be a point on the graph of y = x^2 - 8

The distance from P to (0, -1) is

d = sqrt(x - 0)^2 + (y – 0)^2 = sqrt(x^2 + y^2)

Since P is a point on the graph of y = x^2 -8, we substitute for y

d(x) = sqrt (x^2 + (x^2 – 8)^2 ) = sqrt(x^4 -8x^2 + 8)

I’m not sure if that’s right. That’s the way I understood it to be from the book.

Then if x=0 the distance d(0) = sqrt(64)

if x = 1 the distance d(1) = sqrt(1)

x = -1 the distance d(-1) = sqrt(1)

confidence rating:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * ** P = (x, y) is of the form (x, x^2 - 8).

So the distance from P to (0, -1) is

sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) =

sqrt(x^2 + (-7-x^2)^2) =

sqrt( x^2 + 49 - 14 x^2 + x^4) =

sqrt( x^4 - 13 x^2 + 49). **

What are the values of d for x=0 and x = -1?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I really don’t know if it’s right. I think the values of x were right.

confidence rating: 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * If x = 0 we have

sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7.

If x = -1 we have

sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8.

Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I guess I got the whole thing wrong.

------------------------------------------------

Self-critique Rating: 0

*********************************************

Question: * 3.6.9 / 18 (was and remains 3.6.18). Circle inscribed in square.

What is the expression for area A as a function of the radius r of the circle?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I was looking at the book trying to understand the problem and follow along with the given solution to get a better understanding of how to solve these problems and realized the book does not match the question. The book has:

for 3.6.9

A rectangle is inscribed in a circle of radius 2. Let P = (x,y) be the point in quadrant I that is a vertex of the rectangle and is on the circle.

a) express the area A of the rectangle as a function of x

Even with that said, I cannot solve the problem. I just don’t understand them.

confidence rating: 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square.

If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2.

The area of the circle is pi r^2.

So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **

What is the expression for perimeter p as a function of the radius r of the circle?

The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

How is the diameter of the circle equal in length to the side of the square? How do you determine that?

The circle is inscribed in the square. This means that it's the largest possible circle that will fit inside the square. The largest possible circle that will fit inside a square has a diameter that's equal to the side of the square.

------------------------------------------------

Self-critique Rating: 1

*********************************************

Question: * 3.6.19 / 27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph

Give your expression for the distance d between the cars as a function of time.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

For the car that is 2 miles south of the intersection approaching at a constant speed of 30 mph it would be 2 + 30t

Then the other car is 3 miles east of the intersection approaching at a constant speed of 40 mph would be 3 + 40t

The distance would be

D = sqrt(x^2 + y^2)

x = 2 + 30t

y = 3 + 40t

sqrt((2 + 30t)^2 + (3 + 40t)^2)

sqrt(4 + 120t + 900t^2 + 9 + 240t + 1600t^2)

sqrt(2500t^2 + 360t + 13)

confidence rating: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * ** At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t.

The position function of the other is 3 + 40 t.

If these are the x and the y coordinates of the position then the distance between the cars is

distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **

qa college algebra part 2

030. * 30

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

There are no discrepancies

------------------------------------------------

Self-critique Rating:

*********************************************

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:

Since it is a quadratic function and a = 1 > 0 it will open upwards.

To find the x coordinate of the vertex you take :

a=1

b=-2

c=-3

-b/2a = -(-2)/2(1) = 1

and the y coordinate of the vertex is f(1) = 1^2 – 2(1) – 3 = -4

The axis of symmetry is x = 1 or the points (1, -4)

If we factor x^2 -2x -3 we get

(x - 3) ( x + 1) = 0 so that

x - 3 = 0 or x + 1 = 0

x = 3

x = -1.

The x intercepts are (-1, 0) and (3, 0)

The y intercept is f(0) = -3

The y intercept is the point (0, -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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

There are no discrepancies.

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: * 4.3.57. graph of parabola vertex (1, -3), point (3, 5)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph is a parabola that opens upwards with vertex (1, -3). If h = 1 and k = -3, then f(x) = a(x – h)^2 + k which is a(x – 1)^2 + (-3) To find the value of a we use the coordinates (3, 5) for the x value to get a(3 – 1))^2 + (-3) = 1

I don’t think this is right. There is an example of how to solve a problem like this in the book, but it has the y intercept as a given point on the graph, which can be used as y = f(x) to determine the value of a. Here the problem only gives one other point and I am not sure how to use that to get the quadratic function of the graph.

confidence rating: 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I can see how it was solved, but I’m not sure where (y-k) come from.

------------------------------------------------

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?

The quadratic function is a multiple. If you have x intercepts of -5 and 3, then x = – 5 would be (x+5) = 0 and x = 3 would be (x-3) = 0

So the values of

a = 1 would be 1(x+5)(x-3) = x^2 + 2x – 15

a = 2 would be 2(x+5)(x-3) = 2x^2 + 4x - 30

a = -2 would be -2(x+5)(x-3) = -2x^2 -4x + 30

a = 5 would be 5(x+5)(x-3) = 5x^2 + 10x - 75

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).

"

&#This looks good. See my notes. Let me know if you have any questions. &#