Assignment 18

course Mth 158

????????????assignment #018

018. `query 18

College Algebra

07-08-2008

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20:26:29

query 2.3.34 / 30 (was 2.3.24). Slope 4/3, point (-3,2)

Give the three points you found on this line and explain how you obtained them.

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RESPONSE -->

If x increases by 3, y increases by 4

-3 + 3 = 0 2 +4=6 (0,6)

0 + 3 = 3 6 + 4=10 (3, 10 )

3 +3 = 6 10 + 4 + 14 (6, 14)

The three points are (0,6), (3, 10 ), (6, 14)

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20:29:23

STUDENT SOLUTION:

(-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get

((-3+3), (2+4)), which simplifies to

(0,6)

(-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get

((-3-3), (2-4)) which simplifies to

(-6,-2)

From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get

((0+3), (6+4)), which simplifies to

(3,10). The three points I obtained are

(-6,-2), (0,6), (3,10).

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RESPONSE -->

I moved all my points to the positive direction and did not get any negative points.

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20:44:17

query 2.3.40 / 36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.

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RESPONSE -->

Slope = 2-1/2-(-1) = 1/3

Using y-y1 = m (x-x1)

y-1 = 1/3 (-x-(-1)

y-1 = -1/3x + 1/3

y=1/3x + 4/3

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20:44:34

STUDENT SOLUTION: The slope is m = (y2 - y1) / (x2 - x1) = (2-1)/(2- -1) = 1/3.

Point-slope form gives us

y - y1 = m (x - x1); using m = 1/3 and (x1, y1) = (-1, 1) we get

y-1=1/3(x+1), which can be solved for y to obtain

y = 1/3 x + 4/3.

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RESPONSE -->

Ok

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20:50:49

**** query 2.3.54 / 46 (was 2.3.40). x-int -4, y-int 4 **** What is the equation of the line through the given points and how did you find the equation?

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RESPONSE -->

x intercept = -4

y intercept = 4

(-4,0) (0,4)

m= 4-0/0- (-4) = 4/4 =1

y= mx + b

y=1(x) +4

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20:51:54

STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1.

The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to

y=x+4.

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RESPONSE -->

ok

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20:56:59

**** query 2.3.76 / 56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them?

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RESPONSE -->

Slope is 2/1 and y intercept is 1/2

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20:57:16

** the y intercept occurs where x = 0, which happens when y = 2 (0) + 1/2 or y = 1/2. So the y-intercept is (0, 1/2).

The slope is m = 2.**

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RESPONSE -->

ok

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21:02:00

**** query 2.3.62 / 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 -->

Slope is 1

y - y1= m(x -x1)

y-0 = 1(x-0)

y = 1x

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21:02:42

** 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 -->

Slope is 1/2 instead of 2/1.

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21:07:50

**** query 2.3.68 / 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 -->

x-2y = -5 (0, 4)

slope 1/2

y-y1 = 1/2 (x-x1)

y-4 = 1/2 ( x-0)

y-4 = 1/2 x -0

y= 1/2x +4

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21:09:17

** 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 -->

Somehow I carried my 1/2 all through the equation.

The slope of the given line is 1/2. The slope of the perpendicular line is the negative reciprocal of 1/2, or -2.

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Good work. See my notes and let me know if you have questions. &#