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1. Find the slope between (-1, -3) and (4, 0)
a. 5/3
b. -3
c. - 3/5
d. 3/5
2. Given the equation: 3y = 15x + 9, find slope.
a. 5
b. 2/3
d. -2/3
c. Undefined
3. Given a point and a slope, write the equation in slope-intercept form; (1, 2) and m= -3
a. y = -3x + 1
b. y = -3x – 5
c. y = -3x + 5
d. y = -3x – 1
4. Write the equation of the line in slope-intercept form: y-int = (0,3) , m = 2
a. y = 2x + 3
b. y = -2x + 3
c. y = 2x
d. y = 3x +2
5. Given two points, write the equation in point-slope form: (1, 3) and (-2, 12)
a. y – 1 = -3(x – 3)
b. y – 3 = 5(x – 1)
c. y – 3 = -3(x – 1)
d. y + 12 = -3(x – 1)
6. Write the equation of the function graphed below.
a. y= x - 1
b. y= -x - 4
c. y = x - 4
d. y = x +1
7. The table below represents a linear function. What is the slope of this function?
a. 1/3
b. -1/3
c. 3
d. -3

1 Find the slope between 1 3 and 4 0 a 53 b 3 c 35 d 35 2 Given the equation 3y 15x 9 find slope a 5 b 23 d 23 c Undefined 3 Given a point and a slope write the class=
1 Find the slope between 1 3 and 4 0 a 53 b 3 c 35 d 35 2 Given the equation 3y 15x 9 find slope a 5 b 23 d 23 c Undefined 3 Given a point and a slope write the class=

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Answer:

Step-by-step explanation:

1. Slope = (y2-y1) / (x2-x1) = -3-0/ -1-4= -3/-5= 3/5 (d)

2.

Compare to the equation y= mx+b, where m is the slope

3y = 15x+9 divide by 3 each term

y = 15x/3 +9/3

y= 5x + 3 so slope is 5 ( a)

3.

use first the equation point- slope y-y1 =m (x-x1) and rearange it to be in slope-intercept form y=mx +b

y-y1 =m (x-x1)

y-2 = -3( x-1)

y = 2-3x+3

y= -3x +5 ( c)

4.

if the point y- intercept is at (0, 3) then the y-intercept is b=3

y= mx+b

y = 2x +3 (a)

5.

the equation point- slope y-y1 =m (x-x1), find the slope m= (y2-y1) / (x2-x1)

m= 3-12/ 1+2 = -9/3 = -3

y-1 = -3(x-3) is (a)

6.

y-intercept is at -4  ( yet the graph is off a bit -tell your teacher)

slope m = 1

so the equation is y=mx+b and here is y=x -4 (c)

7.

the slope is m= (y2-y1) / (x2-x1) si pich any two points from the table and calculate the slope

for example (0, 3) and (1, 0)

m = (3-0)/ (0-1) = 3/-1 = -3 (d)

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1. Slope = (y2-y1) / (x2-x1) = -3-0/ -1-4= -3/-5= 3/5 (d)

2.

Compare to the equation y= mx+b, where m is the slope

3y = 15x+9 divide by 3 each term

y = 15x/3 +9/3

y= 5x + 3 so slope is 5 ( a)

3.

use first the equation point- slope y-y1 =m (x-x1) and rearange it to be in slope-intercept form y=mx +b

y-y1 =m (x-x1)

y-2 = -3( x-1)

y = 2-3x+3

y= -3x +5 ( c)

4.

if the point y- intercept is at (0, 3) then the y-intercept is b=3

y= mx+b

y = 2x +3 (a)

5.

the equation point- slope y-y1 =m (x-x1), find the slope m= (y2-y1) / (x2-x1)

m= 3-12/ 1+2 = -9/3 = -3

y-1 = -3(x-3) is (a)

6.

y-intercept is at -4  ( yet the graph is off a bit -tell your teacher)

slope m = 1

so the equation is y=mx+b and here is y=x -4 (c)

7.

the slope is m= (y2-y1) / (x2-x1) si pich any two points from the table and calculate the slope

for example (0, 3) and (1, 0)

m = (3-0)/ (0-1) = 3/-1 = -3 (d)

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