Which statement is true of the equation below? y - 6 = -3(x+1) A. The slope is 3 and the y-intercept is 1 B. The slope is 3 and the y-intercept is -6 C. The slope is -3 and the y-intercept is 1 D. The slope is -3 and the y-intercept is 3
well if you find the equation is slope intercept form you get y = -3x + 3 and the general form is y=mx+b where m is the slop and b is the y-intercept. Hope this helps!
I have 1 more if you could help me with it?
sure
What are the x- and y-intercepts of the line with equation. 5x - 2y = 8?
an x-intercept is where y = 0 so set y equal to zero and you get 5x = 8 Solve for x and you get your x-intercept and then plug that back in to get your y-value to create your point since you need a point A y-intercept is the same idea. Set x equal to zero and solve for y -2y = 8 Solve for y and plug if back in to get your x-value for a point
D.
Thank You hilbertboy96
anytime!
Also robtobey thanks for telling me the answer I got though :D
In a problem like this one, solve for y.
hilbertboy96 i have a question
For that second one i'm a bit confused
by 2nd bit you mean the y intercept or what?
Like both of it I don't really understand it
ok I will demonstrate The x intercept is when you set y equal to zero So if we do this we get \[5x - 2(0) = 8\] \[5x = 8\] \[x = \frac{ 5 }{ 8 }\] But they want a point so a x-value will not finish the question. You need to plug it back in to solve to y to create a point \[5(\frac{ 5 }{ 8 }) - 2y = 8\] \[\frac{ 25 }{ 8 } - 2y = 8\] \[-2y = (8-\frac{ 25 }{ 8 })\] \[y = \frac{ (8-25/8) }{ -2 }\] and you get a point with (x, y) that make sense now?
Yes I see know.
so it would be -1?
Wait wait wait...I am stupid xD Since they are x and y intercepts the other variable is just 0 so your points are (5/8,0) (0,-4)
Sorry for that lol
so i would divide 5/8?
no 5/8 is a fraction so leave it. Your points are \[(\frac{ 5 }{ 8 },0)\] \[(0,-4)\]
Oh i see now
Yes. Sorry for the whole replugging back in although if you did you should end up with 0 but this assumption is easier
OK thanks :D
Anytime
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