Assingment 18

018. `query 18

College Algebra

07-23-2008

......!!!!!!!!...................................

16:56:16

2.4.30 (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.

......!!!!!!!!...................................

RESPONSE -->

(0,6) (3,10) (6, 14)

I graphed the first point(-3, 2) and then followed the slope rose 4 and run 3 to get the remaining three points on the graph.

confidence assessment: 3

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

......!!!!!!!!...................................

16:56:31

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).

......!!!!!!!!...................................

RESPONSE -->

ok

self critique assessment: 3

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

......!!!!!!!!...................................

17:01:27

query 2.4.36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.

......!!!!!!!!...................................

RESPONSE -->

m=(2+1)/(2-1)

m=3/1

y-y1=m(x-x1)

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

I found the slope of the line and used the point-slope form equation.

confidence assessment: 3

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

......!!!!!!!!...................................

17:03:38

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.

......!!!!!!!!...................................

RESPONSE -->

I should have done m=(y2-y1)/(x2-x1) and instead switched the two.

self critique assessment: 3

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

......!!!!!!!!...................................

17:09:23

2.4.46 (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?

......!!!!!!!!...................................

RESPONSE -->

intercepts at (-4,0) and (0, 4)

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

y-o=4/4(x+4)

y=1(x+4)

found the slope and then used point-slope form

confidence assessment: 2

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

......!!!!!!!!...................................

17:09:34

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.

......!!!!!!!!...................................

RESPONSE -->

ok

self critique assessment: 3

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

......!!!!!!!!...................................

17:12:40

2.4.56 (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?

......!!!!!!!!...................................

RESPONSE -->

slope is 2/1 and y-intercept is 1/2

it was already in slope intercept form.

confidence assessment: 3

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

......!!!!!!!!...................................

17:12:49

** 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.**

......!!!!!!!!...................................

RESPONSE -->

ok

self critique assessment: 3

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

&#Your work looks very good. Let me know if you have any questions. &#