1. ## Slope and Y-intercept

State the slope and y-intercept of each line.

e) -(x+4)=2(y-3)

I dont get how that ans is m=-1/2 b=1

2. Originally Posted by Skoz
State the slope and y-intercept of each line.

e) -(x+4)=2(y-3)

I dont get how that ans is m=-1/2 b=1
change to slope-intercept form ... $y = mx + b$

3. It looks okay, I guess. If you would show your work, I could tell you if you did anything besides looking in the back of the book.

4. I did look in the back of the book I dont understand how the answer could be that.

5. Originally Posted by Skoz
State the slope and y-intercept of each line.

e) -(x+4)=2(y-3)

I dont get how that ans is m=-1/2 b=1
this question assumes that you know the definition of "slope" and "y- intercept". Do you?

The slope of a straight line is $\frac{y_1- y_0}{x_1- x_0}$ where $(x_0, y_0)$ and $(x_1, y_1)$ are any two points on the line. If x= 0, then -(0+4)= 2(y- 3) so -4= 2(y-3). Dividing both sides by 2, -2= y- 3. Adding 3 to both sides, 1= y. (0, 1) is a point on the line. If x= 1, -(1+ 4)= 2(y-3) so -5= 2(y-3) and $-\frac{5}{2}= y- 3$. $y= -\frac{5}{2}+3= \frac{1}{2}$. [tex](1, \frac{1}{2})[tex] is also a point on the line. The slope is $\frac{1-\frac{1}{2}}{1- 0}= \frac{-1}{2}$.

The "y- intercept" is where the graph crosses ("intercepts") the y-axis which is at x= 0. Putting x= 0 into -(x+ 4)= 2(y- 3) gives -(0+ 4)= -4= 2(y- 3). Dividing both sides by 2, -2= y- 3. Adding 3 to both sides, 1= y. The "y-intercept" is either the value y= 1 or the point (0, 1) depending upon which convention your class is using.

6. Originally Posted by Skoz
I did look in the back of the book I dont understand how the answer could be that.
You are right. I misread another one. I must need another break.

You still didn't show ANY work.