Assignment 19

course MTH 158

??????????assignment #019

019. `query 19

College Algebra

07-05-2008

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17:06:03

2.5.22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE -->

To get it into the proper form...

x - 2y = -5

-x -x

-2y = -x -5 , divide all by -2...

y = - 1/2 x + 5/2

And now with the point slope formula we get...

y - 0 = 1/2 (x-0)

y = 1/2 x

confidence assessment: 2

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17:06:10

** The equation x - 2y = -5 can be solved for y to give us

y = 1/2 x + 5/2.

A line parallel to this will therefore have slope 1/2.

Point-slope form gives us

y - 0 = 1/2 * (x - 0) or just

y = 1/2 x. **

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RESPONSE -->

ok

self critique assessment: 3

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17:08:54

2.5.28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE -->

To get it into the proper form...

x - 2y = -5

-x -x

-2y = -x -5 , divide all by -2...

y = 1/2 x + 5/2

Take the negative reciprocal of the slope, and with the point slope formula we get...

y - 0 = -2 (x-0)

y = -2x

confidence assessment: 2

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17:10:35

** The equation x - 2y = -5 can be solved for y to give us

y = 1/2 x + 5/2.

A line perpendicular to this will therefore have slope -2/1 = -2.

Point-slope form gives us

y - 4 = -2 * (x - 0) or

y = -2 x + 4. **

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RESPONSE -->

I thought the two points were (0, 0), not (0, 4)... That's what it says in my book for question 28

self critique assessment: 1

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17:36:57

2.3.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?

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RESPONSE -->

Center point is (1, 3)

Raduis is 1

Equation is (x-1)^2 + (y-3) ^2 = 1^2

confidence assessment: 2

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17:41:10

** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1.

The equation (x-h)^2 + (y-k)^2 = r^2 becomes

(x-1)^2 + (y-2)^2 = r^2.

Substituting the coordinates of the point (0, 1) we get

(0-1)^2 + (1-2)^2 = r^2 so that

r^2 = 2.

Our equation is therefore

(x-1)^2 + (y - 2)^2 = 2.

You should double-check this solution by substituting the coordinates of the point (2, 3). **

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RESPONSE -->

I thought that the points that you filled in the circle equation with ( (x-h)^2 + (y-k)^2 = r^2 ) were the coordinates of the center, not the coordinates of the others??

self critique assessment: 1

h and k are the coordinates of the center, r is the radius.

x and y are the coordinates of points on the circle.

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17:43:37

2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?

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RESPONSE -->

(x-1)^2 = (y-0)^2 = 3^2

confidence assessment: 2

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17:43:56

** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r.

In this example we have (h, k) = (1, 0). We therefore have

(x-1)^2 +(y - 0)^2 = 3^2.

This is the requested standard form.

This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get

x^2 - 2x +1+y^2 = 9

x^2 - 2x + y^2 = 8.

However this is not the standard form.

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RESPONSE -->

ok

self critique assessment: 3

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17:49:13

2.3.24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

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RESPONSE -->

x^2 + (y-1)^2 = 1 is the same as

(x + 0)^2 = (y-1)^2 = 1^2

The center of the circle is apparently (0, 1) and the radius is apparently 1.

The circle is in quadrants I and IIII, and it intercepted at points (0, 0) and (0, 2)

confidence assessment: 2

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17:52:36

** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r.

In this example the equation can be written as

(x - 0)^2 + (y-1)^2 = 1

So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, 1) and r = sqrt(r^2) = sqrt(1) = 1.

The x intercept occurs when y = 0:

x^2 + (y-1)^2 = 1. I fy = 0 we get

x^2 + (0-1)^2 = 1, which simplifies to

x^2 +1=1, or

x^2=0 so that x = 0. The x intercept is therefore (0, 0).

The y intercept occurs when x = 0 so we obtain

0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that

(y-1) = +-1.

If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y = 0. So the y-intercepts are

(0,0) and (0,2)

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RESPONSE -->

I don't understand why the center points are (0, -1) not (0, 1)

one of the y-intercepts could also be (0, 2), correct?

self critique assessment: 1

You are correct.

There were errors in the given solution. I just corrected them.

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18:19:03

2.3.34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

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RESPONSE -->

2 x^2 + 2 y^2 + 8 x + 7 = 0

(2x^2 = 8x) + 2y^2 = -7

(x^2 + 4x) + y^2 = - 7/2

(x^2 + 4x + 4) + y^2 = -7/2 + 4

(x + 2)^2 + (y + 0)^2 = 1/2

Center is apparently (2, 0) and radius is 1/2.

Quadrants I and II

for the intercepts... setting x equal to 0, we get y = sqrt(1/2)

and setting y equal to 0, we get x = -2 + sqrt(1/2)

confidence assessment: 2

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18:21:25

** Starting with

2x^2+ 2y^2 +8x+7=0 we group x and y terms to get

2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get

x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get

x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain

(x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that

the center is (-2,0)

the radius is sqrt (1/2).

To get the intercepts:

We use (x+2)^2 + y^2 = 1/2

If y = 0 then we have

(x+2)^2 + 0^2 = 1/2

(x+2)^2 = 1/2

(x+2) = +- sqrt(1/2)

x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx.

x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx

If x = 0 we have

(0+2)^2 + y^2 = 1/2

4 + y^2 = 1/2

y^2 = 1/2 - 4 = -7/2.

y^2 cannot be negative so there is no y intercept. **

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RESPONSE -->

why is the center (-2, 0) instead of (2, 0)?

x + 2 = 0 when x = -2, not when x = +2.

I see that I forgot the whole (0+2)^2 part in the setting x equal to 0 equation to find hte y intercept.

self critique assessment: 2

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18:54:18

2.3.30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.

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RESPONSE -->

Just so you know, this question in my book (7th edition) is 2.3 40, not 2.3 30.

The points 4,3 and 0,1 give us

(4+0) / 2 , (3+1) / 2 = (2, 2) for the center point

Using the distance formula from one of the points in the circle (4,3) to the center point (2, 2), we get

sqrt { (4-2)^2 + (3 -2)^2 }

so the radius is sqrt (5)

(x - 2)^2 + (y -2)^2 = (sqrt5)^2

x^2 - 4x + 4 + y^2 - 4y + 4 = 5

x^2 + y^2 - 4x - 4y + 3 = 0

confidence assessment: 2

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18:54:41

** The center of the circle is the midpoint between the two points, which is ((4+0)/2, (3+1)/2) = (2, 2).

The radius of the circle is the distance from the center to either of the given points. The distance from (2, 2) to (0, 1) is sqrt( (2-0)^2 + (2-1)^2 ) = sqrt(5).

The equation of the circle is therefore

(x-2)^2 + (y-2)^2 = (sqrt(5))^2 or

(x-2)^2 + (y-2)^2 = 5. **

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RESPONSE -->

ok

self critique assessment: 3

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