graphing a line

[justan4cat]

I think I have this one, but after the last question I posted, I'm not so sure. Can you please tell me if I'm correct? Here's the problem:

Graph the line for the equation: y+2=2(x+2)

First, I put it in linear equation form, y=mx+b and got y=2x+2
Then got the x and y intercepts:

y=0+2; y=2 so (0,2) is one point
0=2x+2; -2=2x; -1=x so (-1,0) is the other point and I just graph them.

The reason why I'm hesitating is because when I graph these two points and look at the slope, it is rise/run and that comes out to be 1/2, not 2.

 

[Deleted member 4993]

I think I have this one, but after the last question I posted, I'm not so sure. Can you please tell me if I'm correct? Here's the problem:

Graph the line for the equation: y+2=2(x+2)

First, I put it in linear equation form, y=mx+b and got y=2x+2
Then got the x and y intercepts:

y=0+2; y=2 so (0,2) is one point
0=2x+2; -2=2x; -1=x so (-1,0) is the other point and I just graph them.

The reason why I'm hesitating is because when I graph these two points and look at the slope, it is rise/run and that comes out to be 1/2, not 2.

You have calculated the points correctly.

The slope should come out to be 2.

The rise between these two points (-1,0) & (0,2) is the difference between two y's.

rise = 2 - 0

The run between these two points (-1,0) & (0,2) is the difference between two x's.

run = 0 - (-1) = 1

then rise/run = 2/1 = 2 <<< as you had expected

 

[BigGlenntheHeavy]

\(\displaystyle y+2 \ = \ 2(x+2)\)

\(\displaystyle y+2 \ = \ 2x+4\)

\(\displaystyle y \ = \ 2x+2, \ \implies \ m \ = \ 2 \ = \ \frac{2}{1} \ = \ \frac{rise}{run}, \ and \ y \ intercept \ = \ 2, \ (0,2)\)

\(\displaystyle Hence, \ (one \ way) \ start \ at \ (0,2), \ go \ up \ 2 \ and \ to \ the \ right \ 1 \ or \ \bigg(m=\frac{-2}{-1}\bigg)\)

\(\displaystyle down \ 2 \ and \ 1 \ to \ the \ left, \ see \ graph.\)

[attachment=0:395ied01]ggg.jpg[/attachment:395ied01]