Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

Choose the graph below that corresponds to the equation y - 6 = -one half(x + 1).

OpenStudy (anonymous):

OpenStudy (anonymous):

@jim_thompson5910 @Hero @Mertsj ?

jimthompson5910 (jim_thompson5910):

what's the slope of the given line?

OpenStudy (anonymous):

I tried working the problem out, but I got a little confused.

jimthompson5910 (jim_thompson5910):

the number that is sitting out front of the parenthesis is your slope

jimthompson5910 (jim_thompson5910):

so your slope is _____

OpenStudy (anonymous):

- 1/2?

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

that means that your line will slope downwards as you go from left to right

jimthompson5910 (jim_thompson5910):

so that rules out graphs 2 and 4

jimthompson5910 (jim_thompson5910):

because they slope upwards as you go from left to right

jimthompson5910 (jim_thompson5910):

making sense so far?

OpenStudy (anonymous):

Mhmm :)

jimthompson5910 (jim_thompson5910):

ok the other info given is that we have this line going through (1,6) how do I know this? because the general point slope form is y - y1 = m(x - x1) where m = slope (x1, y1) = point that this line goes through

jimthompson5910 (jim_thompson5910):

oh sry, not (1, 6) (-1, 6)

jimthompson5910 (jim_thompson5910):

y - y1 = m(x - x1) y - 6 = -1/2(x + 1) y - 6 = -1/2(x - (-1)) notice how x1 matches up with -1, so x1 = -1 notice how y1 matches up with 6, so y1 = 6

jimthompson5910 (jim_thompson5910):

so which line goes through (-1,6) graph 1 or graph 3?

OpenStudy (anonymous):

1, right?

jimthompson5910 (jim_thompson5910):

graph 1 goes through (-1,6) you sure?

OpenStudy (anonymous):

yeah pretty sure

jimthompson5910 (jim_thompson5910):

(-1,6) is in the second quadrant (upper left quadrant)

jimthompson5910 (jim_thompson5910):

and (-1,6) close to the y-axis

OpenStudy (anonymous):

ooooh 3.

jimthompson5910 (jim_thompson5910):

better

OpenStudy (anonymous):

Thank you

jimthompson5910 (jim_thompson5910):

you're welcome

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!