Assignment 27

course Mth 158

4/8/10

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

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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)

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Your solution:

D= sqrt[(0-x)^2 + (-1 – (x^2-8)^2]

Sqrt[X^2 + (-1-x^2 + 8)^2]

Sqrt[x^2 – (x^2-7)^2]

Sqrt[x^2-(x^2-7)(x^2-7)]

Sqrt[x^2-x^4 - 14x^2 +49

Sqrt[ x^4-13x^2+49]

confidence rating #$&*2

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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?

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Your solution:

D=sqrt[(0)^4 – 13(0) + 49]

D= sqrt[ 0-0 + 49]

D=sqrt[49]

D=7

D=sqrt[(-1)^4 – 13(-1) + 49]

D=sqrt[1+13+49]

D=sqrt[63]

confidence rating #$&*

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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( 63).

sqrt(64) = 8, so sqrt(63) is a little less than 8 (turns out that sqrt(63) is about 7.94).

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 verify that these distances make sense. **

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Self-critique (if necessary):

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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?

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Your solution:

confidence rating #$&* 0 Looked at the problem in the book and it seemed to differ from yours. Not sure how to answer it.

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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. **

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Self-critique (if necessary):

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.

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Self-critique rating #$&*2

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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.

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Your solution:

Sqrt(x^2 + y^2) =

Sqrt[(2+ 30t)^2 + (3 + 40t)^2]=

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

sqrt[4 + 60 + 900t^2 + 9 + 80t + 1600t^2]=

sqrt[13 + 140t + 2500t^2]

confidence rating #$&*2

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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

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Self-critique (if necessary):

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Self-critique rating #$&*Ok

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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.

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Your solution:

Graph opens up.

a = 1

b=-2

c=-3

x-coordinate =

x= [-(-2)]/2(1) = 1

y-coordinate=

y=f(x)=x^2-2x-3

y=[(1)^2-2(1)-3

y=-4

vertex: (1,-4)

x-intercept:

x^2-2x-3=0

(x-3)(x+1)=

X=(3,-1)

Y-intercept:

(0)^2-2(0)-3=

(-3)

[-(-2) + - sqrt (-2)^2 – 4(1)(-3)]/2(1) =

[2 + - sqrt[4 + 12]/2 =

[2 + - sqrt[16]/2 =

[2 + - 4]/2 =

3, -1

Domain: { -inf, inf}

Range: {-4<= x <= inf}

Decreasing on the interval (-inf,-4). Increasing on interval (-4,inf)

confidence rating #$&*2

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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).

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Self-critique (if necessary):

the x intercepts are (-1, 0) and (3, 0).

The y intercept is the point (0, -3).

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Self-critique rating #$&*2

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

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Your solution:

F(x)= a[x-(1)]^2 +(-3)

5 = a[(3) – (1)]^2 -3

5 = a[2]^2 -3

5 = 4a-3

8=4a

2=a

(2)[(x-(1)]^2-3=

(2)[(x-1)(x-1)]-3=

2x^2-4x+2-3=

2x^2-4x-1

confidence rating #$&*2

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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):

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Self-critique rating #$&*OK

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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?

1(x+5)(x-3)=0

X^2+2x-15

2(x+5)(x-3)=0

2(x^2+2x-15)=0

2x^2+4x-30=0

-2(x+5)(x-3)=0

-2(x^2+2x-15)=0

-2x^2-2x+30=0

5(x+5)(x-3)=0

5(x^2+2x-15)=0

5x^2+10x-75)=0

Solution:

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):

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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